🧠:Okay, let's try to tackle this problem step by step. The problem is to evaluate the expression:[frac{1}{2^{n}}left(1 - 3 binom{n}{2} + 3^{2} binom{n}{4} - 3^{3} binom{n}{6} + cdots + 3^{994} binom{n}{1988} - 3^{995} binom{n}{1990} right)]where ( n = 1990 ). First, I need to understand the structure of the expression. Let's break it down. The main components are alternating terms with powers of 3 and binomial coefficients with even indices. Specifically, the general term seems to be ( (-3)^k binom{n}{2k} ) for ( k = 0 ) to ( 995 ), since ( 2 times 995 = 1990 ). Then, the entire sum is divided by ( 2^{n} ).Hmm. So the expression can be rewritten as:[frac{1}{2^{1990}} sum_{k=0}^{995} (-3)^k binom{1990}{2k}]But I need to verify if the signs are alternating correctly. The original expression starts with a positive 1, then subtracts 3 times the binomial coefficient, adds 3 squared times another binomial coefficient, and so on. So yes, the sign alternates as ( (-1)^k ), and the coefficient is ( 3^k ), multiplied by ( binom{n}{2k} ). So the sum is indeed ( sum_{k=0}^{995} (-3)^k binom{1990}{2k} ), and then divided by ( 2^{1990} ).Now, I recall that sums involving binomial coefficients with even indices often relate to evaluating some expression at roots of unity, or using generating functions. Alternatively, perhaps using the binomial theorem with certain substitutions. Let me think.The standard binomial theorem is ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k ). But here, we are only summing over even indices, with alternating signs and powers of 3. So maybe there's a generating function that can be manipulated to get this sum.Another approach: generating functions for even terms. Remember that the sum of the even binomial coefficients can be obtained by ( frac{(1 + x)^n + (1 - x)^n}{2} ). Similarly, the sum of the odd binomial coefficients is ( frac{(1 + x)^n - (1 - x)^n}{2} ). But here, we have more than just summing even terms; each term is multiplied by ( (-3)^k ), which complicates things.Let me formalize this. Let's let ( x = sqrt{3} ), but wait, not sure yet. Alternatively, consider that each term ( (-3)^k binom{n}{2k} ) can be thought of as substituting ( x^2 = -3 ), but perhaps I need to consider generating functions.Wait, here's an idea. Let's consider the generating function ( (1 + x)^{1990} ). If I substitute ( x ) with a certain value, maybe I can extract the even terms multiplied by some factor. Let's recall that the sum over even k of ( binom{n}{k} x^k ) is ( frac{(1 + x)^n + (1 - x)^n}{2} ). So in our case, if we want the sum over even k of ( binom{n}{2k} (-3)^k ), that would be equivalent to substituting ( x^2 = -3 ), so ( x = sqrt{-3} ), but perhaps complex numbers.Wait, let me make this precise. Let me denote ( x = sqrt{-3} ). Then, ( x^2 = -3 ), so ( x^{2k} = (-3)^k ). Therefore, the sum ( sum_{k=0}^{995} binom{1990}{2k} (-3)^k ) is equal to ( sum_{k=0}^{995} binom{1990}{2k} (x^2)^k ). But since ( 2k ) goes up to 1990 (since n=1990 is even), this sum is exactly the sum over all even k of ( binom{1990}{2k} x^{2k} ). Which, as per the generating function identity, is equal to ( frac{(1 + x)^{1990} + (1 - x)^{1990}}{2} ).Therefore, substituting ( x = sqrt{-3} ), the sum becomes:[frac{(1 + sqrt{-3})^{1990} + (1 - sqrt{-3})^{1990}}{2}]But ( sqrt{-3} ) can be written as ( isqrt{3} ), where ( i ) is the imaginary unit. Therefore, ( 1 + sqrt{-3} = 1 + isqrt{3} ), and similarly ( 1 - sqrt{-3} = 1 - isqrt{3} ).Interesting. So the sum ( sum_{k=0}^{995} (-3)^k binom{1990}{2k} ) is equal to ( frac{(1 + isqrt{3})^{1990} + (1 - isqrt{3})^{1990}}{2} ).Therefore, the entire expression given in the problem is:[frac{1}{2^{1990}} times frac{(1 + isqrt{3})^{1990} + (1 - isqrt{3})^{1990}}{2} = frac{(1 + isqrt{3})^{1990} + (1 - isqrt{3})^{1990}}{2^{1991}}]Now, I need to compute ( (1 + isqrt{3})^{1990} + (1 - isqrt{3})^{1990} ). To handle these complex numbers, it might be helpful to express them in polar form. Let's recall that complex numbers can be written in terms of magnitude and angle.First, compute the modulus and argument of ( 1 + isqrt{3} ).The modulus is ( sqrt{1^2 + (sqrt{3})^2} = sqrt{1 + 3} = sqrt{4} = 2 ).The argument ( theta ) is ( arctanleft( frac{sqrt{3}}{1} right) = frac{pi}{3} ). Therefore, ( 1 + isqrt{3} = 2 left( cos frac{pi}{3} + i sin frac{pi}{3} right) ).Similarly, ( 1 - isqrt{3} ) has modulus 2 and argument ( -frac{pi}{3} ).Therefore, using De Moivre's theorem, ( (1 + isqrt{3})^{1990} = 2^{1990} left( cos left( 1990 times frac{pi}{3} right) + i sin left( 1990 times frac{pi}{3} right) right) ).Similarly, ( (1 - isqrt{3})^{1990} = 2^{1990} left( cos left( 1990 times left( -frac{pi}{3} right) right) + i sin left( 1990 times left( -frac{pi}{3} right) right) right) ).But when we add these two expressions, the imaginary parts (which are sine terms) will cancel out because sine is an odd function. Specifically, ( sin(-x) = -sin x ), so the imaginary parts become ( i sin(1990 times frac{pi}{3}) ) and ( -i sin(1990 times frac{pi}{3}) ), which sum to zero. The real parts, which are cosine terms, will add up because cosine is an even function. Therefore, the sum is:[2^{1990} times 2 cos left( 1990 times frac{pi}{3} right ) = 2^{1991} cos left( frac{1990 pi}{3} right )]Therefore, going back to the original expression:[frac{2^{1991} cos left( frac{1990 pi}{3} right ) }{2^{1991}} = cos left( frac{1990 pi}{3} right )]So the entire expression simplifies to ( cos left( frac{1990 pi}{3} right ) ).Now, I need to compute this cosine value. Let's simplify the angle ( frac{1990 pi}{3} ). Since cosine has a period of ( 2pi ), we can subtract multiples of ( 2pi ) to find an equivalent angle between 0 and ( 2pi ).First, compute how many times ( 2pi ) fits into ( frac{1990 pi}{3} ):Let's compute ( frac{1990}{3} div 2 = frac{1990}{6} = 331.666... ). So, ( 331 times 2pi = 662pi ). Let's subtract ( 331 times 2pi = 662pi ) from ( frac{1990 pi}{3} ):First, compute ( frac{1990}{3} - 662 ). Let me compute ( 1990 ÷ 3 ): 3 × 663 = 1989, so 1990 ÷ 3 = 663 + 1/3 ≈ 663.333... So, ( 663.333... pi - 662 pi = 1.333... pi = frac{4}{3} pi ).Therefore, ( frac{1990 pi}{3} ) is equivalent to ( frac{4pi}{3} ) modulo ( 2pi ).But wait, let me verify that calculation again. Wait, 331 × 2π = 662π. Then, subtract 662π from ( frac{1990π}{3} ).Convert 662π to thirds: 662π = ( frac{1986π}{3} ). Therefore, ( frac{1990π}{3} - frac{1986π}{3} = frac{4π}{3} ). Yes, that's correct.But ( frac{4π}{3} ) is still greater than ( 2π )? Wait, no. Wait, ( 2π = frac{6π}{3} ), so ( frac{4π}{3} ) is less than ( 2π ). Therefore, the angle ( frac{1990π}{3} ) is co-terminal with ( frac{4π}{3} ).Therefore, ( cos left( frac{1990 pi}{3} right ) = cos left( frac{4pi}{3} right ) ).But ( frac{4pi}{3} ) is in the third quadrant, and the cosine of ( frac{4pi}{3} ) is equal to ( cos(pi + frac{pi}{3}) = -cos(frac{pi}{3}) = -frac{1}{2} ).Therefore, ( cos left( frac{4pi}{3} right ) = -frac{1}{2} ).Therefore, the entire expression simplifies to ( -frac{1}{2} ).Wait, but let me double-check this calculation because signs can be tricky. Let me confirm:( frac{4pi}{3} ) is π/3 more than π, so in the third quadrant. The reference angle is π/3. Cosine in the third quadrant is negative, so yes, ( cos(4π/3) = -1/2 ). Correct.Therefore, the value of the original expression is ( -1/2 ).But let me just recap to ensure I didn't skip any steps or make a mistake:1. Recognize the sum as the even terms of a binomial expansion with coefficients involving powers of -3.2. Use the identity for even binomial coefficients, leading to expressions involving complex numbers.3. Convert the complex numbers to polar form, apply De Moivre's theorem, simplify the expression by recognizing periodicity.4. Reduce the angle modulo 2π, compute the cosine, leading to -1/2.Everything seems to check out. The key insight was connecting the sum to the binomial expansion evaluated at complex numbers, then using polar form and periodicity to simplify the expression.Alternatively, another approach might involve generating functions or using roots of unity filters, but the method above seems valid.Therefore, the final answer is ( -frac{1}{2} ).Final Answerboxed{-dfrac{1}{2}}

✅:Given ( n = 1990 ), we need to find the value of the expression:[frac{1}{2^{n}}left(1 - 3 binom{n}{2} + 3^{2} binom{n}{4} - 3^{3} binom{n}{6} + cdots + 3^{994} binom{n}{1988} - 3^{995} binom{n}{1990} right)]First, we recognize that the expression can be rewritten as:[frac{1}{2^{1990}} sum_{k=0}^{995} (-3)^k binom{1990}{2k}]We use the identity for the sum of even binomial coefficients multiplied by powers of a variable. By substituting ( x = sqrt{-3} ), we convert the sum into a form involving complex numbers. The sum of even terms in the binomial expansion is given by:[frac{(1 + x)^{1990} + (1 - x)^{1990}}{2}]Substituting ( x = isqrt{3} ), we get:[frac{(1 + isqrt{3})^{1990} + (1 - isqrt{3})^{1990}}{2}]Converting these complex numbers to polar form, we find that ( 1 + isqrt{3} ) and ( 1 - isqrt{3} ) have modulus 2 and arguments ( frac{pi}{3} ) and ( -frac{pi}{3} ) respectively. Using De Moivre's theorem, we raise these to the 1990th power:[(1 + isqrt{3})^{1990} = 2^{1990} left( cos left( 1990 times frac{pi}{3} right) + i sin left( 1990 times frac{pi}{3} right) right)][(1 - isqrt{3})^{1990} = 2^{1990} left( cos left( 1990 times left( -frac{pi}{3} right) right) + i sin left( 1990 times left( -frac{pi}{3} right) right) right)]Adding these, the imaginary parts cancel, and the real parts sum to:[2^{1991} cos left( frac{1990 pi}{3} right)]Simplifying the angle ( frac{1990 pi}{3} ) modulo ( 2pi ), we find it is equivalent to ( frac{4pi}{3} ). The cosine of this angle is:[cos left( frac{4pi}{3} right) = -frac{1}{2}]Thus, the value of the original expression is:[boxed{-dfrac{1}{2}}]

🔑:1. The given problem involves determining the value of the expression: frac{1}{2^{n}} left( 1 - 3 C_{n}^{2} + 3^{2} C_{n}^{4} - 3^{3} C_{n}^{6} + cdots + 3^{994} C_{n}^{1988} - 3^{995} C_{n}^{1990} right) where n = 1990.2. To simplify the analysis, let us introduce a substitution. Let sqrt{3} = t, then the expression can be rewritten as: A = frac{1}{2^{n}} left( 1 - t^{2} C_{n}^{2} + t^{4} C_{n}^{4} - t^{6} C_{n}^{6} + cdots + t^{1988} C_{n}^{1988} - t^{1990} C_{n}^{1990} right) 3. To recognize the pattern, observe that the given sum resembles a portion of the binomial expansion of (1 - t^2)^{n}. We will construct the missing part involving odd powers of t: B = frac{1}{2^{n}} left( -t C_{n}^{1} + t^{3} C_{n}^{3} - t^{5} C_{n}^{5} + cdots + t^{1987} C_{n}^{1987} - t^{1989} C_{n}^{1989} right) 4. Combining A and B gives us the complete binomial expansion: A + B mathrm{i} = frac{1}{2^{n}} left( (1 - mathrm{i} t)^{n} right) 5. Substitute t = sqrt{3} and simplify: A + B mathrm{i} = frac{1}{2^{n}} (-1 + sqrt{3} mathrm{i})^{n} 6. Recognize the term inside the binomial to convert to polar form: -1 + sqrt{3} mathrm{i} = 2 left( -frac{1}{2} + frac{sqrt{3}}{2} mathrm{i} right) = 2 left( cos frac{2pi}{3} + mathrm{i} sin frac{2pi}{3} right) 7. Thus, frac{1}{2^{n}} (-1 + sqrt{3} mathrm{i})^{n} = left( -frac{1}{2} + frac{sqrt{3}}{2} mathrm{i} right)^{n} = left( cos frac{2pi}{3} + mathrm{i} sin frac{2pi}{3} right)^{n} 8. By De Moivre's theorem, we have: left( cos frac{2pi}{3} + mathrm{i} sin frac{2pi}{3} right)^{n} = cos left( frac{2npi}{3} right) + mathrm{i} sin left( frac{2npi}{3} right) 9. Since A is the real part of the above expression, we find: A = cos left( frac{2npi}{3} right) 10. For n = 1990: frac{2 cdot 1990 pi}{3} = frac{3980 pi}{3} which simplifies to: frac{3980 pi}{3} equiv frac{2 pi}{3} pmod{2pi} 11. Consequently: A = cos left(frac{2pi}{3}right) = -frac{1}{2} Conclusion:boxed{-frac{1}{2}}

🧠:Okay, let's see. Dana is drawing a rectangular spiral, and each segment he draws has lengths that go 1,1,2,2,3,3,4,4,... and so on. So it's like each number is repeated twice. The problem says that his pen runs out of ink when the total length he's drawn is 3000 cm. We need to find the length of the longest line segment he drew before that happened. The options are given, so maybe after solving, we can check against those. Let me think step by step.First, the spiral is made by drawing line segments in a rectangle, turning each time, right? But maybe the exact shape isn't important here. The key is the total length. Each pair of segments (since each number is used twice) adds up to two times that number. For example, 1+1=2, then 2+2=4, 3+3=6, etc. So the total length after n pairs would be the sum of 2*1 + 2*2 + 2*3 + ... + 2*n. Which can be factored as 2*(1 + 2 + 3 + ... + n). The sum of the first n integers is n(n + 1)/2, so the total length would be 2*(n(n + 1)/2) = n(n + 1). Wait, that's interesting. So if Dana completes n pairs, the total length would be n(n + 1). But the problem says the total is 3000 cm. However, he might not complete all the pairs up to some n. He might stop in the middle of a pair.Wait, but each pair is two segments. So for example, first 1 cm twice, then 2 cm twice, etc. So the total after k complete pairs (each pair being two segments of the same length) is n(n + 1), as above. But if he stops partway through a pair, say after drawing one segment of length m, then the total would be n(n + 1) + m, where n is the number of complete pairs. So we need to find the maximum m such that n(n + 1) + m = 3000. Then the longest segment would be m, unless m is part of the next pair, which would be m+1. Hmm, maybe I need to clarify.Wait, actually, the sequence is 1,1,2,2,3,3,..., so each number is drawn twice. So if he stops after drawing a certain number of segments, the total length is the sum of each segment up to that point. The total ink used is 3000 cm. So maybe he completes some number of pairs, and then starts the next pair but doesn't finish both segments. So, for example, suppose he completes n pairs (each pair being two segments of length n). Then the total length is 2*(1 + 2 + ... + n) = n(n + 1). Then, if he starts the next pair, which is two segments of length n+1. If he draws one segment of length n+1, the total becomes n(n + 1) + (n + 1) = (n + 1)(n + 1) = (n + 1)^2. If he draws both, it's (n + 1)(n + 2). Wait, let's check:Wait, sum of first n pairs: 2*(1 + 2 + ... +n) = 2*(n(n +1)/2) = n(n +1). Then, if he draws one more segment after that, which is the first segment of the next pair, which is n +1, so total length is n(n +1) + (n +1) = (n +1)(n +1) = (n +1)^2. If he draws two more segments, both n +1, then it's n(n +1) + 2(n +1) = (n +1)(n + 2). Hmm, so in that case, the total sum after n pairs plus k segments (k=1 or 2) in the next pair is n(n +1) + k*(n +1). So if he stops at 3000, we need to find n such that n(n +1) ≤ 3000 < (n +1)(n +2). But wait, maybe not exactly, because the next pair is two segments of (n +1). So after n pairs, sum is S = n(n +1). Then, if he draws one segment of (n +1), the sum becomes S + (n +1) = n(n +1) + (n +1) = (n +1)^2. Then, if he draws the second segment of (n +1), the sum becomes (n +1)^2 + (n +1) = (n +1)(n +2). So the total sum can be in three places: n(n +1), (n +1)^2, or (n +1)(n +2). So if 3000 is between n(n +1) and (n +1)^2, then he only drew one segment of (n +1). If it's between (n +1)^2 and (n +1)(n +2), he drew two.But wait, the problem says the total of all lengths is 3000. So perhaps he didn't finish the spiral at the end of a pair, but had to stop in the middle. Therefore, we need to find n where n(n +1) ≤ 3000 < (n +1)(n +2). Then, the total sum would be n(n +1) plus some remaining segments. Wait, no. Wait, actually, each pair is two segments. So after n pairs (each with two segments), the sum is n(n +1). Then, each subsequent segment is part of the next pair. So if he draws m more segments (each of length (n +1)) after n pairs, then the total sum is n(n +1) + m*(n +1). But since each pair is two segments, he can't have more than two segments in the next pair. So m is either 0,1,2. Therefore, to reach 3000, we need to find n such that:n(n +1) ≤ 3000 < (n +1)(n +2)But if that's the case, then after n pairs, he hasn't started the next pair. Wait, but actually, if he starts the next pair, he can have 1 or 2 segments. Wait, maybe the way to think about it is:Total sum S = sum_{k=1}^{n} 2k + m*(n +1) where m is 0,1, or 2. Then S = n(n +1) + m(n +1) = (n +1)(n + m). So 3000 = (n +1)(n + m), where m can be 0,1,2. Wait, but m is the number of segments drawn in the next pair. Each pair is two segments, so m can be 0,1, or 2. But in terms of the formula, since each segment in the pair is (n +1) cm, the total added is m*(n +1). Therefore, 3000 = n(n +1) + m*(n +1) = (n +1)(n + m). So we need to solve (n +1)(n + m) = 3000, where m is 0,1, or 2. So possible values of m are 0,1,2. Therefore, 3000 must be equal to (n +1)(n + m) where m is 0,1,2. So 3000 must be a product of two integers differing by at most 2.Wait, this might be a way to approach. Let's think: 3000 needs to be factored into two numbers, say a and b, where a = n +1 and b = n + m, with m ∈ {0,1,2}. Therefore, the difference between a and b is (n + m) - (n +1) = m -1. Since m ∈ {0,1,2}, the difference between a and b can be -1, 0, or 1. So, 3000 needs to be expressed as a product of two integers that are either consecutive, equal, or one apart. So, we need to find factors of 3000 that are either equal (perfect square) or differ by 1 or 2. Hmm. Let's check.First, check if 3000 is a perfect square. The square of 54 is 2916, 55 squared is 3025. So 3000 is not a perfect square. So m can't be 1 (since that would make a and b equal). Then, check for factors that are consecutive integers (difference 1) or differ by 2.Let me list the factor pairs of 3000:1 x 30002 x 15003 x 10004 x 7505 x 6006 x 5008 x 37510 x 30012 x 25015 x 20020 x 15024 x 12525 x 12030 x 10040 x 7550 x 60... etc. These are the main ones.Looking for pairs that are consecutive or differ by 1 or 2.Looking at 24 x 125: difference 101, nope.25 x 120: difference 95.30 x 100: 70.40 x 75: 35.50 x 60: difference 10. Hmm. Not close.Wait, maybe higher factors:Check 54 x 55. Wait, 54*55=2970, which is less than 3000. 55*56=3080. So between 54 and 55, the product is 2970 and 3080. So 3000 is between those. Therefore, the factors would be 54 and 55.6... which isn't an integer. So no.Wait, so maybe 3000 can't be written as a product of two consecutive integers. Similarly, 50 x 60 is 3000, but those differ by 10. So, no.Wait, but maybe 3000 can be written as a product of two integers differing by 2. Let's see. Let me check for such pairs.Looking at 50 x 60: difference 10. 40 x 75: 35. 30 x 100:70. 25 x 120:95. 24 x 125:101. 20 x150:130. 15x200:185. 12x250:238. 10x300:290. 8x375:367. 6x500:494. 5x600:595. 4x750:746. 3x1000:997. 2x1500:1498. 1x3000:2999.None of these have a difference of 2. So it's not possible to write 3000 as a product of two integers differing by 0,1, or 2. Therefore, this approach might not work. Hmm. Maybe my initial assumption is wrong.Wait, perhaps the error is in how I model the problem. Let me re-examine.The total length is the sum of the segments: 1,1,2,2,3,3,..., so each number k is added twice, except possibly the last one if the total reaches 3000 before completing the pair. So the total sum S is 2*(1 + 2 + ... +n) + m, where m is 0 or k, where k is n+1. Wait, no. If he completes n pairs (1 to n, each twice), the sum is 2*(1 + 2 + ... +n) = n(n +1). Then, if he starts the next pair (n+1), he can draw 0, 1, or 2 segments. So the total sum would be n(n +1) + m*(n +1), where m is 0,1,2. So if he draws m segments of length (n +1), then the total is n(n +1) + m(n +1) = (n +1)(n + m). So solving for (n +1)(n + m) = 3000, where m is 0,1,2. So possible equations:If m=0: (n +1)(n) = 3000 => n² +n -3000 =0. Solve quadratic equation: n = [-1 ± sqrt(1 +12000)]/2 = [-1 ± sqrt(12001)]/2. sqrt(12001) is approx 109.54. So n≈ ( -1 + 109.54 )/2 ≈ 54.27. Not integer.If m=1: (n +1)(n +1) = 3000 => (n +1)^2=3000. Then n +1=sqrt(3000)≈54.77. Not integer.If m=2: (n +1)(n +2)=3000. Solve n² +3n +2=3000 =>n² +3n -2998=0. Discriminant: 9 +4*2998=9 +11992=12001. Same sqrt(12001)=109.54. So n=( -3 +109.54)/2≈53.27. Not integer.Therefore, none of these equations have integer solutions, which means that Dana must have stopped partway through a segment? Wait, but the problem says the total of all the lengths he has drawn is 3000 cm. So he can't stop partway through a segment; he has to finish a segment. So the total length must be exactly 3000, which would be a sum of complete segments. Therefore, the total sum must be exactly 3000, which is not possible as per above, since there's no integer n such that (n +1)(n + m) = 3000 for m=0,1,2. Therefore, maybe my approach is incorrect.Alternative approach: Let's model the total length as the sum of the sequence 1,1,2,2,3,3,... up to some term. Let's denote that the last term drawn is k, and depending on whether he drew one or two of them.Wait, the total sum S can be written as:If he drew up to the first occurrence of k: S = 2*(1 + 2 + ... + (k-1)) + kIf he drew both occurrences of k: S = 2*(1 + 2 + ... +k)So, in the first case, S = 2*( (k-1)k)/2 ) +k = (k-1)k +k = k² -k +k =k²In the second case, S = 2*(k(k +1)/2 )=k(k +1)Therefore, the total sum after drawing both segments of k is k(k +1), and after drawing only one segment of k, it's k².Therefore, the possible total sums are either k² or k(k +1) for some integer k.So, we need to check if 3000 is either a perfect square or one more than a perfect square. Since 54²=2916 and 55²=3025. So 3000 is between 54² and 55². So if he had drawn all segments up to the first 54, the total would be 54²=2916. Then, he starts drawing the pair for 55. Each segment of 55 adds 55 cm. So starting from 2916, he can draw one segment of 55, totaling 2916 +55=2971. Then another segment of 55, totaling 2971 +55=3026. But 3026 exceeds 3000. Therefore, after drawing one segment of 55, the total is 2971, and the next segment would make it 3026. But he runs out at 3000, which is in between. However, the problem states that he stops when the total is 3000. But he can't stop in the middle of a segment; he must finish a segment. Therefore, the total length must be exactly 3000. Since 3000 isn't achievable as k² or k(k +1), but lies between 54²=2916 and 54*55=2970. Wait, no, wait:Wait, actually, let's recast:Wait, if k=54, then:- If he draws both segments for each number up to 54, the total is 54*55=2970.- If he draws one more segment (the first segment of 55), the total becomes 2970 +55=3025, which is 55².But 3025 is greater than 3000, so he can't do that. Wait, but 54*55=2970, then adding 55 once gives 2970 +55=3025, which is 55². But 3025 is 55²=3025, which is more than 3000. So Dana can't get to 3000 by completing a segment. Therefore, maybe my initial model is wrong.Wait, perhaps there is a miscalculation here. Let's step back.The total length after n pairs (i.e., 2*1, 2*2,...,2*n) is Sum = 2*(1 +2 +...+n) = 2*(n(n +1)/2)=n(n +1). Then, if he proceeds to draw the next pair, which is two segments of (n +1). So after drawing the first segment of (n +1), total is n(n +1) + (n +1) = (n +1)^2. After drawing the second segment of (n +1), total is (n +1)^2 + (n +1) = (n +1)(n +2). Therefore, the totals after each step are either n(n +1), (n +1)^2, or (n +1)(n +2). So 3000 must be either equal to n(n +1), (n +1)^2, or (n +1)(n +2) for some integer n. But we saw that none of these equal 3000. Therefore, this suggests that Dana cannot reach exactly 3000 cm by completing any segment. Therefore, the problem must be interpreted that Dana stops when the total reaches or exceeds 3000, but the answer is the length of the longest segment drawn. Wait, but the problem says "the total of all the lengths he has drawn is 3000 cm". So he must have drawn segments whose total is exactly 3000. But according to the above, 3000 is not achievable. Therefore, perhaps there's a mistake in my model.Wait, another thought: Maybe the spiral is constructed by adding two segments per turn, but each turn increases the length by 1 each time. For example, first right 1, up 1, then left 2, down 2, then right 3, up 3, etc. So each "loop" of the spiral consists of four directions, but each pair of segments (horizontal and vertical) have the same length. Wait, but in that case, the sequence would be 1,1,2,2,3,3,4,4,... So each number is used twice, but each loop consists of two numbers. Wait, maybe each full cycle (four sides) would have two pairs? Wait, perhaps not. Let me visualize:Start at a point, draw right 1 cm, then up 1 cm, then left 2 cm, then down 2 cm, then right 3 cm, up 3 cm, left 4 cm, down 4 cm, etc. So in this case, each "loop" around the spiral adds two numbers: the first loop is 1,1; the second loop is 2,2; the third loop is 3,3; etc. But each loop actually consists of four segments: right, up, left, down. But the lengths are 1,1,2,2 for the first two loops? Wait, no:Wait, first, to make a rectangle spiral, you go right 1, up 1, left 1+1=2, down 2, then right 2+1=3, up 3, etc. Wait, maybe the lengths increase every two segments. Wait, maybe the standard rectangular spiral has sides increasing by 1 every two segments. So for example:Right 1, Up 1, Left 2, Down 2, Right 3, Up 3, Left 4, Down 4, etc. So in this case, the lengths are 1,1,2,2,3,3,4,4,... which matches the problem's description. Therefore, each "pair" of segments (right/up, left/down, etc.) have the same length, increasing by 1 each time. Therefore, the total length after n pairs would be 2*(1 +2 +...+n). But as before, that's n(n +1). Then, after that, if he continues, the next pair is n +1, so two segments of n +1 cm. So if he starts drawing the next pair, the total becomes n(n +1) + 2*(n +1) = (n +1)(n + 2). But again, same as before. So perhaps the error is that in the problem statement, the spiral is made by turning each time, so each side is drawn once per direction, but the lengths follow 1,1,2,2,3,3,... So each two sides have the same length. Therefore, the total length after k segments would be sum_{i=1}^{k} a_i, where a_i is 1,1,2,2,3,3,...So the total sum S is the sum of the sequence up to the k-th term. So if he stops at some term, whether it's the first or second in the pair, the total would be accordingly. Therefore, the problem says the total is 3000, so we need to find the largest term in the sequence up to that point.Therefore, to model this, we need to find n such that the sum S = sum_{i=1}^{m} a_i = 3000, where a_i is the sequence 1,1,2,2,3,3,... and we need the maximum a_i in that sum.So, for each pair (1,1), (2,2), etc., the sum contributed by pair k is 2k. Therefore, the total sum after n complete pairs is 2*(1 +2 +...+n) = n(n +1). Then, if he draws an additional segment after that, it's of length n +1, so total sum is n(n +1) + (n +1) = (n +1)(n +1) = (n +1)^2. Then, the next segment would be another n +1, making the total (n +1)(n +2).Therefore, possible total sums are either n(n +1), (n +1)^2, or (n +1)(n +2). Since 3000 is not equal to any of these, but lies between n(n +1) and (n +1)^2 for some n, that suggests that he must have completed n(n +1) and then started the next pair, drawing some segments of length n +1 until the total reaches 3000. But since he can't stop mid-segment, he can only finish a segment if the total doesn't exceed 3000. So the problem must be that the total is exactly 3000, but given that 3000 is not one of those totals, he must have drawn extra segments beyond the n(n +1) sum, but only as much as needed to reach 3000.Wait, but how is that possible? Let me think numerically.Let's compute n(n +1) for various n:n=54: 54*55=2970n=55:55*56=3080So 2970 < 3000 <3080. So he completes 54 pairs, total 2970 cm. Then, he starts the 55th pair, which are two segments of 55 cm each. Each segment is 55 cm. So after the first segment, total is 2970 +55=3025, which is over 3000. But he can't complete that segment because it would exceed 3000. Therefore, he can't draw any part of the next segment, so he must have stopped at 2970 cm. But 2970 is less than 3000. So there's a contradiction here. Therefore, the problem must be interpreted differently.Wait, perhaps the error is in assuming that each "pair" corresponds to two segments. But in the spiral, each turn is 90 degrees, so to complete a full loop, you need four segments. However, the problem states that the lengths are 1,1,2,2,3,3,4,4,... which suggests that each two consecutive segments have the same length, but after two segments, the length increases by 1. So regardless of the direction, the lengths follow 1,1,2,2,3,3,...Therefore, the total length is the sum of this sequence up to some term. So if the total is 3000, then the number of terms m is such that sum_{k=1}^{m} a_k = 3000, where a_k is 1,1,2,2,3,3,...To find m, we can note that every two terms, the length increases by 1. So for each even number of terms, the sum is 2*(1 +2 +...+n), and for an odd number of terms, it's 2*(1 +2 +...+n) + (n +1). So for example, if m is even, m=2n, then sum is n(n +1). If m is odd, m=2n +1, sum is n(n +1) + (n +1)= (n +1)(n +1)= (n +1)^2.Therefore, the total sum can be either n(n +1) or (n +1)^2. Since 3000 is neither, we need to see between which two sums it falls. As before, n=54: 54*55=2970; n=55:55*56=3080. So 2970 <3000 <3080. Therefore, if he completes 54 pairs (108 segments), the total is 2970. Then, the next segment is 55 cm, which would bring the total to 2970 +55=3025. But 3025 is over 3000. Therefore, he can't draw that full segment. But the problem states that he stops when the total is 3000. Therefore, this suggests that he must have drawn part of the 55 cm segment. However, the problem says he draws line segments, so he can't stop mid-segment. Therefore, there must be an error in the problem statement, or my interpretation.Alternatively, perhaps the total ink used is exactly 3000, which is achievable by summing up segments up to a certain point. But given that 3000 isn't a triangular number times 2 or a square, maybe the problem allows for partial segments, but the answer would still be the last full segment drawn. Wait, but the problem says "the total of all the lengths he has drawn is 3000 cm". So he must have drawn complete segments. Therefore, 3000 must be achievable as sum_{k=1}^m a_k where a_k is the sequence as defined. But given the previous calculations, it's not possible. Therefore, there must be a miscalculation.Wait, let's check the total sum with more precise steps. Let's compute cumulative sums step by step until we reach 3000.Start with 1,1,2,2,3,3,... Each pair adds 2k for k=1,2,3,...Total after first pair (1,1): 2 cmAfter second pair (2,2): 2 +4=6 cmThird pair (3,3):6 +6=12 cmFourth pair (4,4):12 +8=20 cmWait, so after n pairs, the total is 2 +4 +6 +...+2n =2*(1 +2 +3 +...+n)=n(n +1). So as before. Therefore, cumulative sums after n pairs: n(n +1).Let me compute n where n(n +1) is closest to 3000:n=54: 54*55=2970n=55:55*56=3080Therefore, after 54 pairs, total is 2970. Then, the next segments are two of 55 cm each. Drawing one gives 2970 +55=3025, which exceeds 3000. So he can't draw that. Therefore, the total sum can't reach 3000 with complete segments. But the problem states that the total is 3000. Therefore, this suggests that perhaps the problem allows partial segments, but the question is about the length of the longest complete segment drawn. So if he had 2970, and needs 30 cm more to reach 3000, he starts drawing a 55 cm segment but only completes 30 cm of it. However, the problem says "the total of all the lengths he has drawn is 3000 cm". If partial segments are allowed, then the longest complete segment is 54 cm, and the partial segment is 30 cm. But the question asks for the length of the longest line segment that Dana draws. If partial segments are considered, the longest would be 55 cm (even though only part of it is drawn), but the answer choices don't include 55 with a caveat. Wait, the answer choices are 38,39,54,55,30. So 54 and 55 are options. Wait, but according to the previous calculation, the last complete segment he could have drawn is 54 cm, since 55 would exceed the total. Therefore, the answer should be 54. However, the options include 55, which is confusing.Alternatively, perhaps the problem counts the incomplete segment as part of the drawn segments, even if not completed. But the problem states "the total of all the lengths he has drawn is 3000 cm". If he only drew 30 cm of the 55 cm segment, then the total would be 2970 +30=3000, but the length of that segment is 30 cm, which is option E. However, 30 is much lower than the previous segments. But the problem asks for the longest line segment drawn. If he drew segments up to 54 cm, and a partial segment of 55 cm (30 cm), then the longest complete segment is 54 cm. Therefore, answer C)54. But the options also include 55. Hmm.Wait, perhaps my entire approach is wrong. Maybe the problem isn't about pairs but about the spiral having alternating horizontal and vertical segments, each time increasing by 1 after two turns. Wait, let me read the problem again:"Dana creates a 'rectangular spiral' by drawing line segments of lengths, in cm, of 1,1,2,2,3,3,4,4,..."So the lengths follow 1,1,2,2,3,3,4,4,... So each two consecutive segments have the same length, then increase by 1. So regardless of direction, the lengths are 1,1,2,2,3,3,... So the total number of segments is even if he completes pairs, or odd if he stops mid-pair.Therefore, the total length is sum_{k=1}^{m} a_i, where a_i is the sequence [1,1,2,2,3,3,...]. So if m is even, say m=2n, then the sum is 2*(1 +2 +...+n)=n(n +1). If m is odd, m=2n +1, then sum is n(n +1) + (n +1)= (n +1)^2.Therefore, possible sums are n(n +1) and (n +1)^2. Since 3000 is neither, Dana must have stopped partway through a segment. But the problem says the total is exactly 3000. This is a contradiction unless the problem allows partial segments. However, the problem says "drawing line segments", implying that each segment is completed. Therefore, there must be an error in the problem statement or perhaps a miscalculation.Alternatively, maybe the initial modeling is incorrect. Let's consider that in a rectangular spiral, each "loop" consists of four sides: right, up, left, down. Each loop adds two to the length of the sides. For example, first loop: right 1, up 1, left 1, down 1. But that would be lengths 1,1,1,1. But the problem says lengths are 1,1,2,2,3,3,... So maybe each two sides in the spiral have the same length. Let's try:First, move right 1, up 1.Then, left 2, down 2.Then, right 3, up 3.Then, left 4, down 4.Etc. So in this case, each "loop" consists of two pairs: (right, up), (left, down). Each pair has lengths increasing by 1 each time. So the sequence of lengths is 1,1,2,2,3,3,4,4,... Thus, the total number of segments is even for each full loop. But the total length would then be 2*(1 +2 +3 +...+n) for n loops. Each loop corresponding to a pair of numbers. So after n loops, total length is n(n +1). Then, the next loop would start with n+1, but if he can't complete it, he might have partial segments.But if he stops at total 3000, which isn't a multiple of n(n +1), then he must have partial segments. But since he can't have partial segments, the total must be the sum up to some completed segments. Therefore, the total would be the largest n(n +1) less than or equal to 3000, which is 54*55=2970. Then, the next segment he tries to draw is 55 cm, which would exceed 3000. Therefore, he cannot draw it, so the longest segment he drew is 54 cm. Therefore, the answer is C)54. But why is 55 an option? Maybe I made a mistake.Wait, but let's check if there's a different way. Suppose the spiral is constructed such that each full cycle (four sides) increases the length by 1 each time. For example:Right 1, Up 1, Left 1, Down 1,Right 2, Up 2, Left 2, Down 2,Right 3, Up 3, Left 3, Down 3,... etc.In this case, the lengths would be 1,1,1,1,2,2,2,2,3,3,3,3,... But the problem states the lengths are 1,1,2,2,3,3,... So this doesn't match. Therefore, the initial interpretation of two segments per number seems correct.Another approach: Let's calculate the cumulative sum step by step until we pass 3000.We know that after 54 pairs (108 segments), total is 54*55=2970.Next pair is 55, so two segments of 55 cm each.First segment of 55: total becomes 2970 +55=3025, which is over 3000. Therefore, he can't draw this segment. Therefore, the maximum total he can reach without exceeding 3000 is 2970, with the longest segment being 54 cm. Therefore, answer C)54. But option D)55 is also there. Why?Wait, maybe the problem is considering that when he starts drawing the 55 cm segment, even if he can't finish it, the fact that he started it means that the longest segment he attempted was 55 cm. But the problem says "the length of the longest line segment that Dana draws". If he didn't complete the 55 cm segment, then he didn't draw it. So the longest segment is 54 cm.But let's verify with an example. Suppose the total allowed is 2970 + x =3000, where x is the length of the next segment. x=30 cm. But the next segment is supposed to be 55 cm. If he only draws 30 cm of it, does that count as a segment? The problem says "drawing line segments of lengths...". If a line segment is defined as a straight continuous line, then he can't have a segment of 30 cm in the middle of a 55 cm segment. Therefore, he must have completed the previous segment, which was 54 cm, and the next segment he starts (55 cm) he can't complete, so it doesn't count. Therefore, the longest segment is 54 cm. So answer C.However, the presence of option D)55 is confusing. Maybe there's a different way to model the total length.Wait, let's think differently. Suppose that instead of pairs, each number is used for one horizontal and one vertical segment alternately, hence forming the spiral. For example:Right 1, Up 1, Left 2, Down 2, Right 3, Up 3, Left 4, Down 4, etc.In this case, the sequence of lengths is 1,1,2,2,3,3,4,4,... So same as before. The total length after n segments would be sum_{k=1}^n a_k, where a_k alternates between horizontal and vertical. But the total sum calculation remains the same: after 2n segments, sum is 2*(1 +2 +...+n) =n(n +1). After 2n +1 segments, sum is n(n +1) + (n +1).But since 3000 is not equal to either n(n +1) or (n +1)^2, he must have completed 2n segments with total n(n +1) and then started the next segment of (n +1) cm, but couldn't finish it. Therefore, the longest completed segment is n cm.But with n=54, sum=2970, next segment would be 55 cm, which can't be completed. Thus, the longest segment is 54. Answer C.But why is 55 an option? Maybe I need to check the problem statement again.The problem says: "the total of all the lengths he has drawn is 3000 cm". So he must have drawn segments whose total is exactly 3000. If that's not possible with complete segments, but the problem still expects an answer, perhaps there's a miscalculation.Alternatively, maybe the total is allowed to exceed 3000, and the ink runs out at 3000, so the last segment is partially drawn, but the question is about the longest segment he did draw, which would be 55 cm even if only part of it was drawn. However, the problem states "the total of all the lengths he has drawn is 3000 cm". If "drawn" means completed, then the partial segment isn't counted. If "drawn" includes any part, then the length would be 55, but the total would be 2970 + part of 55. But the problem says the total is 3000, so the length of the partial segment would be 3000 -2970=30 cm. Therefore, the lengths drawn are up to 54 cm, plus a 30 cm segment. So the longest complete segment is 54 cm, and there's a partial segment of 30 cm. Therefore, the answer is 54 cm, option C.But why is 55 an option? Maybe the question is considering that the last segment he was drawing counts as a segment of 55 cm, even if it wasn't completed. But in standard terminology, a line segment has a definite length. If you start drawing a 55 cm segment but only complete 30 cm, then you have a 30 cm segment, not a 55 cm one. Therefore, the answer should be 54 cm.Given the answer choices, C)54 is an option, and D)55 is also there. To resolve this, perhaps there's a different way to compute.Wait, let's think in terms of total segments. The total length is the sum S =1 +1 +2 +2 +3 +3 +...+k +k = 2*(1 +2 +...+k) =k(k +1). If he completes k pairs, total is k(k +1). If this exceeds 3000, then k is too big. If he stops before completing a pair, say after one segment of k +1, then the total is k(k +1) + (k +1) = (k +1)^2. So we need to find k where k(k +1) ≤3000 < (k +1)(k +2). For k=54, 54*55=2970, 55*56=3080. So 2970 <3000 <3080. So the total length is between 54*55 and 55*56. Therefore, he must have drawn some segments beyond the 54 pairs. But since each additional segment is 55 cm, he can draw up to m segments of 55 cm such that 2970 +55*m ≤3000. So m=(3000 -2970)/55=30/55=6/11. Which is less than 1. So he can't even complete one segment of 55 cm. Therefore, the total length remains 2970, which is less than 3000. Therefore, this suggests that the problem might have a typo, or the answer is 54. However, the given options include 55. Maybe the problem is designed to consider that even though he couldn't complete the 55 cm segment, the longest length he attempted was 55. But that's not standard. Typically, the length of a segment is as drawn. Therefore, the answer should be 54. But since 55 is an option, maybe there's a different approach.Wait, perhaps the spiral increases the length every two turns, but in a different pattern. For instance, right 1, up 1, left 2, down 2, right 3, up 3, etc. So each time, the horizontal and vertical segments increase by 1 every two turns. So the lengths are 1,1,2,2,3,3,... So after each two segments, the length increases by 1. Therefore, the total length is sum_{k=1}^{n} k*2. So sum=2*(1 +2 +...+n) =n(n +1). But if you have an extra segment after that, it's n +1, so total is n(n +1) + (n +1)= (n +1)^2.But 3000 is not a perfect square. The nearest squares are 54²=2916 and 55²=3025. Therefore, 2916 <3000 <3025. Therefore, he would have completed the square up to 54²=2916, which is sum after 54 pairs plus one segment of 55, totaling 2916 +55=2971. Wait, no, 54²=2916 is the sum after 54 pairs plus one segment of 55? Wait, no:Wait, let's re-express. If after n pairs, the total is n(n +1). Then, adding one segment of (n +1) gives (n +1)^2. So for example, n=54: 54*55=2970. Then, adding one segment of 55: 2970 +55=3025=55². But 3025 is larger than 3000, so he can't do that. Therefore, the total must be 2970, which is less than 3000. But the problem states the total is exactly 3000. This is a contradiction unless partial segments are allowed.If partial segments are allowed, then the total length is 2970 +30=3000, with the last segment being 30 cm of the 55 cm intended. Therefore, the longest complete segment is 54 cm, and there's a partial segment of 30 cm. Therefore, answer is 54, option C.But why is 55 an option? Maybe the question is from an Olympiad where they consider the attempted segment as part of the count, but that seems against standard practice. Alternatively, perhaps there's an error in the problem's answer options.Alternatively, maybe the initial model is incorrect. Let's consider that the total sum is 3000, and we need to find the maximum k such that the sum of the series up to k is less than or equal to 3000. The series is 1,1,2,2,3,3,..., so for each k, the cumulative sum is k(k +1) if we have two of each up to k, or k(k +1) + (k +1) if we have one more.So solving k(k +1) ≤3000. The largest integer k where k(k +1) ≤3000 is k=54, since 54*55=2970 ≤3000, and 55*56=3080 >3000. Therefore, the maximum complete pairs are k=54, total sum 2970. Then, the remaining length is 3000 -2970=30 cm, which is a single segment of 30 cm. Wait, but the next segments after 54 are 55 cm each. Therefore, he can't draw a segment of 30 cm, as the next segments are 55 cm. Therefore, the remaining 30 cm can't be achieved by a complete segment. Therefore, the total sum must remain at 2970, which is less than 3000. This suggests that the problem is ill-posed, or there's a different pattern.Wait, perhaps the pattern isn't two of each number, but rather the spiral has a different structure. For example, in a typical rectangular spiral, each side alternates directions and the length increases every two sides. For example:Right 1, Up 1, Left 2, Down 2, Right 3, Up 3, Left 4, Down 4, ..., which gives lengths 1,1,2,2,3,3,4,4,...In this case, every two sides, the length increases by 1. Therefore, each "loop" consists of four sides with two different lengths. But the total length would still be cumulative.But regardless, the total length contributed by each number k is 2k (except possibly the last one). Therefore, the same as before.Given that, and the previous calculation, the answer should be 54. Therefore, option C. But the presence of 55 as an option is confusing. Let me check the calculation again.Compute 54*55=2970.3000 -2970=30.The next segment is 55 cm. So he needs 30 cm more. Since he can't draw a partial segment, he cannot reach 3000. Therefore, the total remains 2970, and the longest segment is 54.Alternatively, if the problem allows the last segment to be 55 cm, considering that the total would be 2970 +55=3025, which is over 3000, but the problem says the total is exactly 3000, this is not possible. Therefore, the answer must be 54. So option C.But why is 55 an option? Maybe the question is from a source where the answer is 55, considering that he started drawing the 55 cm segment, hence it's counted as the longest, even though it wasn't completed. But in standard interpretation, the longest segment drawn would be the longest completed one, which is 54. Therefore, 54 is the answer, option C.Perhaps the original problem had a different ink total, like 3025, which would make the answer 55, but as it is, 3000 leads to 54.Therefore, after thorough analysis, the answer should be C)54.

✅:Dana draws a rectangular spiral with line segments following the sequence (1, 1, 2, 2, 3, 3, ldots). The total length of all segments drawn is (3000 , text{cm}). To find the longest segment:1. Sequence Analysis: Each integer (k) appears twice, contributing (2k) to the total length. After (n) pairs, the total length is (2(1 + 2 + cdots + n) = n(n + 1)).2. Critical Calculation: - Solve (n(n + 1) leq 3000) for the largest integer (n). - (54 times 55 = 2970) (total after 54 pairs). - The next segment would be (55 , text{cm}), but (2970 + 55 = 3025 > 3000).3. Conclusion: Dana cannot complete the (55 , text{cm}) segment without exceeding (3000 , text{cm}). The longest completed segment is (54 , text{cm}).Answer: (boxed{C})

🔑:1. Identify the sequence formula: Dana draws line segments of lengths 1, 1, 2, 2, 3, 3, ldots. We notice the lengths are repeated twice in sequence. Therefore, the lengths of the line segments can be represented as (1+1)+(2+2)+(3+3)+ ldots + (n+n).2. Sum formula for repeated lengths: The sum of the first n natural numbers is given by: [ 1 + 2 + 3 + ... + n = frac{n(n+1)}{2} ] To adapt this to our double sequence, where each number appears twice, we multiply the sum by 2: [ 2 left(1 + 2 + 3 + ... + nright) = 2 cdot frac{n(n+1)}{2} = n(n+1) ]3. Setting up the inequality: We are given that the total length drawn is 3000 , text{cm}, so we need to find n such that: [ n(n+1) leq 3000 ]4. Estimate initial value for ( n ): Start with sqrt{3000} as an approximate value to find a starting point: [ sqrt{3000} approx 54.77 ]5. Testing n = 54: [ n(n+1) = 54 times 55 = 2970 ] Since 2970 < 3000, n = 54 is feasible.6. Testing n = 55: [ n(n+1) = 55 times 56 = 3080 ] Since 3080 > 3000, n = 55 is not feasible.7. Conclusion of longest segment: Therefore, the longest line segment that Dana draws corresponds to ( n = 54 ): [ text{Longest segment length} = n = 54 , text{cm} ][boxed{text{C}}]

🧠:Okay, let's try to tackle this problem step by step. First, I need to understand the definitions given here. We have a sequence ( v_1, v_2, ldots, v_n ) defined recursively. The first term ( v_1 = 2 ), then each subsequent term is the product of all previous terms plus one. So, ( v_2 = v_1 + 1 = 2 + 1 = 3 ), ( v_3 = v_1 cdot v_2 + 1 = 2 cdot 3 + 1 = 7 ), ( v_4 = v_1 cdot v_2 cdot v_3 + 1 = 2 cdot 3 cdot 7 + 1 = 43 ), and so on. The general formula is ( v_n = v_1 v_2 cdots v_{n-1} + 1 ).Then, ( S_n ) is the sum of the reciprocals of these terms up to ( n ). So, ( S_n = frac{1}{v_1} + frac{1}{v_2} + cdots + frac{1}{v_n} ).The problem asks us to prove or disprove two main statements:1. For any positive integer ( n ), ( S_n < 1 ).2. For any sum ( I_n ) of the reciprocals of ( n ) positive integers that include ( v_1, v_2, ldots, v_{n-1} ), if ( I_n < 1 ), then ( I_n leq S_n < 1 ). Therefore, ( S_n ) is the number in the set ( { I_n mid I_n < 1 } ) that is closest to 1.Let me start with the first part: proving that ( S_n < 1 ) for all positive integers ( n ).Looking at the sequence ( v_n ), it's similar to the Sylvester's sequence, which is defined by ( a_{n} = a_{n-1}(a_{n-1} - 1) + 1 ), starting with ( a_1 = 2 ). Sylvester's sequence has the property that the sum of reciprocals converges to 1. However, in our case, the definition is slightly different. Let me check the terms again:- ( v_1 = 2 )- ( v_2 = 2 + 1 = 3 )- ( v_3 = 2 cdot 3 + 1 = 7 )- ( v_4 = 2 cdot 3 cdot 7 + 1 = 43 )- ( v_5 = 2 cdot 3 cdot 7 cdot 43 + 1 = 1807 )- And so on.Wait, actually, starting from ( v_2 ), the terms are similar to Sylvester's sequence. Let me check Sylvester's sequence:Sylvester's sequence starts with 2, 3, 7, 43, 1807, etc., which matches exactly our ( v_n ). Because Sylvester's sequence is defined by each term being the product of all previous terms plus 1. So, actually, this sequence ( v_n ) is Sylvester's sequence. Therefore, the sum of reciprocals of Sylvester's sequence is known to converge to 1, and each partial sum is less than 1. That would mean ( S_n < 1 ) for all ( n ), which answers the first part.But wait, let me verify this with induction to be thorough, since I need to make sure even if it's Sylvester's sequence, maybe there's a different approach here.Base case: ( n = 1 ). ( S_1 = 1/2 = 0.5 < 1 ). True.Assume ( S_k < 1 ) for some ( k geq 1 ). Now, consider ( S_{k+1} = S_k + 1/v_{k+1} ). But since ( v_{k+1} = v_1 v_2 cdots v_k + 1 ), so ( v_{k+1} - 1 = v_1 v_2 cdots v_k ). Therefore, ( 1/v_{k+1} = 1/(v_1 v_2 cdots v_k + 1) ).But how does this relate to the previous sum? Let's consider the product of ( (1 + 1/v_1)(1 + 1/v_2)cdots(1 + 1/v_k) ). Wait, perhaps an alternative approach.Alternatively, notice that the sum ( S_n = sum_{i=1}^n 1/v_i ).But for Sylvester's sequence, the sum of reciprocals is ( 1 - 1/(v_1 v_2 cdots v_n) ). Let's check:For ( n = 1 ), ( 1 - 1/2 = 1/2 = S_1 ).For ( n = 2 ), ( 1 - 1/(2 cdot 3) = 1 - 1/6 = 5/6 ), and ( S_2 = 1/2 + 1/3 = 5/6 ). Correct.For ( n = 3 ), ( 1 - 1/(2 cdot 3 cdot 7) = 1 - 1/42 = 41/42 ), and ( S_3 = 1/2 + 1/3 + 1/7 = 21/42 + 14/42 + 6/42 = 41/42 ). Correct again.So, it seems that the sum ( S_n = 1 - frac{1}{v_1 v_2 cdots v_n} ). Therefore, since ( v_1 v_2 cdots v_n ) is a positive integer, ( frac{1}{v_1 v_2 cdots v_n} > 0 ), so ( S_n = 1 - ) something positive, hence ( S_n < 1 ). Therefore, by induction, this formula holds for all ( n ), which proves the first part.So, the first statement is proven: ( S_n < 1 ) for all positive integers ( n ).Now, the second part: For any sum ( I_n ) of the reciprocals of ( n ) positive integers that include ( v_1, v_2, ldots, v_{n-1} ), if ( I_n < 1 ), then ( I_n leq S_n < 1 ). Therefore, ( S_n ) is the number in the set ( { I_n mid I_n < 1 } ) that is closest to 1.So, the claim is that if you take any ( n ) positive integers that include ( v_1, v_2, ldots, v_{n-1} ), then the sum of their reciprocals, if less than 1, cannot exceed ( S_n ). Hence, ( S_n ) is the maximum such sum below 1, making it the closest to 1.To approach this, let's first parse the statement. ( I_n ) is a sum of reciprocals of ( n ) positive integers, which include ( v_1, v_2, ldots, v_{n-1} ). So, the ( n ) integers are ( v_1, v_2, ldots, v_{n-1} ), and one more integer, say ( x ). Therefore, ( I_n = S_{n-1} + frac{1}{x} ).But wait, ( S_n = S_{n-1} + frac{1}{v_n} ). So, if we take ( I_n ) as ( S_{n-1} + frac{1}{x} ), where ( x ) is some positive integer different from ( v_n ), then the problem states that if ( I_n < 1 ), then ( I_n leq S_n ). Therefore, ( S_n ) is the maximum possible sum under 1 when choosing ( x = v_n ), so that ( S_n = S_{n-1} + 1/v_n ), and any other choice of ( x ) would result in a smaller sum, hence ( I_n leq S_n ).But the problem says "any sum ( I_n ) of the reciprocals of ( n ) positive integers that include ( v_1, v_2, cdots, v_{n-1} )", so it's not just changing the last term. Wait, perhaps not. The set includes ( v_1, ldots, v_{n-1} ), and one more integer. Wait, but the original ( S_n ) includes ( v_1, ldots, v_n ). So, the difference is that ( S_n ) includes ( v_n ), whereas ( I_n ) includes ( v_1, ldots, v_{n-1} ) plus some other integer.Wait, no. Wait, the problem says: "for any sum ( I_n ) of the reciprocals of ( n ) positive integers that include ( v_1, v_2, cdots, v_{n-1} ), if ( I_n <1 ), then ( I_n leq S_n <1 )".So, ( I_n ) is a sum of reciprocals of ( n ) integers, and these ( n ) integers must include all of ( v_1, ldots, v_{n-1} ), and one more integer. Therefore, ( I_n = frac{1}{v_1} + cdots + frac{1}{v_{n-1}} + frac{1}{x} ), where ( x ) is some positive integer (could be equal to ( v_n ), but not necessarily). However, ( S_n ) is ( frac{1}{v_1} + cdots + frac{1}{v_{n-1}} + frac{1}{v_n} ).So, the problem claims that if such an ( I_n ) is less than 1, then ( I_n leq S_n ). Therefore, ( S_n ) is the maximum possible sum under 1 for such sums. Thus, ( S_n ) is the closest to 1 in the set of such sums.So, to prove this, we need to show that replacing ( v_n ) with any other integer ( x ) in the sum ( S_n ) would result in a smaller sum, provided that the sum remains less than 1.Alternatively, since ( S_n = S_{n-1} + 1/v_n ), and ( I_n = S_{n-1} + 1/x ), we need to show that if ( S_{n-1} + 1/x < 1 ), then ( 1/x leq 1/v_n ), which would mean ( x geq v_n ). Therefore, the reciprocal ( 1/x leq 1/v_n ), hence ( I_n leq S_n ).But is this true? Suppose ( x ) is greater than ( v_n ), then ( 1/x < 1/v_n ), so ( I_n = S_{n-1} + 1/x < S_{n-1} + 1/v_n = S_n ). If ( x ) is less than ( v_n ), then ( 1/x > 1/v_n ), but would the sum ( S_{n-1} + 1/x ) still be less than 1?Wait, but if ( x ) is less than ( v_n ), then maybe adding a larger reciprocal could push the sum over 1. The problem statement says "if ( I_n < 1 )", so even if you choose an ( x < v_n ), as long as the sum remains less than 1, then ( I_n leq S_n ).But how do we know that choosing a smaller ( x ) (i.e., larger reciprocal) doesn't result in a sum still less than 1 but larger than ( S_n )?Wait, that would contradict the statement. So, perhaps we need to show that if you take ( x ) different from ( v_n ), then either ( I_n geq 1 ), or ( I_n leq S_n ).Therefore, if ( x < v_n ), then ( 1/x > 1/v_n ), so ( I_n = S_{n-1} + 1/x > S_{n-1} + 1/v_n = S_n ). But if ( I_n < 1 ), then how can it be greater than ( S_n ), which is already less than 1? Wait, but if ( x ) is smaller than ( v_n ), then adding a larger reciprocal might make the sum larger, but the problem states that if such a sum is still less than 1, then it's actually not larger than ( S_n ). That seems conflicting. So maybe my initial assumption is wrong.Wait, let's test with an example. Take ( n = 2 ).For ( n = 2 ), ( S_2 = 1/2 + 1/3 = 5/6 ≈ 0.833 ). Now, ( I_2 ) must include ( v_1 = 2 ) and one other integer. So, ( I_2 = 1/2 + 1/x ). If we choose ( x = 1 ), then ( I_2 = 1/2 + 1 = 3/2 > 1 ), which is invalid. If ( x = 2 ), then ( I_2 = 1/2 + 1/2 = 1 ), which is not less than 1. If ( x = 3 ), ( I_2 = 5/6 ≈ 0.833 = S_2 ). If ( x = 4 ), ( I_2 = 1/2 + 1/4 = 3/4 = 0.75 < 5/6 ). So, here, when we choose ( x > 3 ), the sum is less than ( S_2 ). If we choose ( x = 3 ), it's equal to ( S_2 ). If we choose ( x < 3 ), then either the sum is not less than 1 or equals 1. So in this case, the maximum possible ( I_n < 1 ) is indeed ( S_n ).Similarly, take ( n = 3 ). ( S_3 = 1/2 + 1/3 + 1/7 ≈ 0.5 + 0.333 + 0.142 ≈ 0.975 ). Now, ( I_3 ) must include ( v_1 = 2 ), ( v_2 = 3 ), and one other integer. So ( I_3 = 1/2 + 1/3 + 1/x ). Let's try different ( x ):- ( x = 1 ): Sum is 1 + 1/2 + 1/3 > 1.- ( x = 2 ): Sum is 1/2 + 1/3 + 1/2 = 1 + 1/3 > 1.- ( x = 3 ): Sum is 1/2 + 1/3 + 1/3 ≈ 0.5 + 0.333 + 0.333 ≈ 1.166 > 1.- ( x = 4 ): Sum is 1/2 + 1/3 + 1/4 ≈ 0.5 + 0.333 + 0.25 = 1.083 > 1.- ( x = 5 ): ≈ 0.5 + 0.333 + 0.2 = 1.033 > 1.- ( x = 6 ): ≈ 0.5 + 0.333 + 0.166 ≈ 1.0.- ( x = 7 ): 0.5 + 0.333 + 0.142 ≈ 0.975 = S_3.- ( x = 8 ): ≈ 0.5 + 0.333 + 0.125 ≈ 0.958 < S_3.So, in this case, choosing ( x = 7 ) gives ( S_3 ), while any ( x > 7 ) gives a smaller sum. If we try ( x = 6 ), the sum is exactly 1, which is not less than 1. So, indeed, the maximum possible sum ( I_3 < 1 ) is ( S_3 ).This suggests that the key is that if you choose ( x ) smaller than ( v_n ), the sum ( I_n ) would be at least 1, hence not in the set ( { I_n mid I_n < 1 } ). Therefore, the only way to have ( I_n < 1 ) is to choose ( x geq v_n ), but then ( 1/x leq 1/v_n ), so ( I_n leq S_n ). Therefore, ( S_n ) is indeed the maximum such sum below 1.To formalize this, suppose we have ( I_n = S_{n-1} + 1/x ), where ( x ) is a positive integer not equal to ( v_n ). We need to show that if ( I_n < 1 ), then ( x geq v_n ), which would imply ( 1/x leq 1/v_n ), hence ( I_n leq S_n ).But how do we show that if ( x < v_n ), then ( I_n geq 1 )?Recall that ( v_n = v_1 v_2 cdots v_{n-1} + 1 ). Therefore, if ( x < v_n ), then ( x leq v_n - 1 = v_1 v_2 cdots v_{n-1} ). Therefore, ( 1/x geq 1/(v_1 v_2 cdots v_{n-1}) ).But ( S_{n-1} + 1/x geq S_{n-1} + 1/(v_1 v_2 cdots v_{n-1}) ).But from the earlier formula, ( S_{n} = S_{n-1} + 1/v_n = 1 - 1/(v_1 v_2 cdots v_n) ). So, ( S_{n} = 1 - 1/(v_1 v_2 cdots v_n) ).But let's compute ( S_{n-1} + 1/(v_1 v_2 cdots v_{n-1}) ).Since ( S_{n-1} = 1 - 1/(v_1 v_2 cdots v_{n-1}) ), as per the formula from earlier. Therefore, adding ( 1/(v_1 v_2 cdots v_{n-1}) ) gives ( S_{n-1} + 1/(v_1 v_2 cdots v_{n-1}) = 1 - 1/(v_1 v_2 cdots v_{n-1}) + 1/(v_1 v_2 cdots v_{n-1}) ) = 1 ).Therefore, if ( x leq v_1 v_2 cdots v_{n-1} ), then ( 1/x geq 1/(v_1 v_2 cdots v_{n-1}) ), so ( I_n = S_{n-1} + 1/x geq S_{n-1} + 1/(v_1 v_2 cdots v_{n-1}) ) = 1 ).Therefore, if ( x < v_n ), then ( x leq v_n - 1 = v_1 v_2 cdots v_{n-1} ), which leads to ( I_n geq 1 ). Therefore, if ( I_n < 1 ), then ( x geq v_n ), which implies ( 1/x leq 1/v_n ), so ( I_n leq S_n ).Thus, we have shown that any sum ( I_n ) that includes ( v_1, ldots, v_{n-1} ) and another integer ( x ), if ( I_n < 1 ), then ( x geq v_n ), hence ( I_n leq S_n ). Therefore, ( S_n ) is the maximum such sum below 1, making it the closest to 1 in the set ( { I_n mid I_n < 1 } ).Therefore, both statements are proven: ( S_n < 1 ) for all ( n ), and ( S_n ) is the closest to 1 among all such sums ( I_n ) that are less than 1.Final Answerboxed{S_n < 1} and the second statement is also true, hence both are proven.

✅:Given the sequence ( v_1, v_2, ldots, v_n ) defined by ( v_1 = 2 ) and ( v_n = v_1 v_2 cdots v_{n-1} + 1 ) for ( n geq 2 ), we need to prove two statements about the sum ( S_n = frac{1}{v_1} + frac{1}{v_2} + cdots + frac{1}{v_n} ). Proof of ( S_n < 1 ) for all positive integers ( n )1. Base Case: For ( n = 1 ), ( S_1 = frac{1}{2} < 1 ).2. Inductive Step: Assume ( S_k < 1 ) for some ( k geq 1 ). For ( S_{k+1} ), we have: [ S_{k+1} = S_k + frac{1}{v_{k+1}} ] Since ( v_{k+1} = v_1 v_2 cdots v_k + 1 ), we know ( v_{k+1} - 1 = v_1 v_2 cdots v_k ). The sum ( S_n ) can be expressed as: [ S_n = 1 - frac{1}{v_1 v_2 cdots v_n} ] This formula shows ( S_n < 1 ) because ( frac{1}{v_1 v_2 cdots v_n} > 0 ). Proof that ( S_n ) is the closest sum to 1 among all ( I_n < 1 )1. Definition of ( I_n ): ( I_n ) is a sum of reciprocals of ( n ) positive integers including ( v_1, v_2, ldots, v_{n-1} ). Thus, ( I_n = S_{n-1} + frac{1}{x} ) where ( x ) is another positive integer.2. Comparison with ( S_n ): - If ( x < v_n ), then ( x leq v_1 v_2 cdots v_{n-1} ). This implies: [ I_n = S_{n-1} + frac{1}{x} geq S_{n-1} + frac{1}{v_1 v_2 cdots v_{n-1}} = 1 ] Hence, ( I_n geq 1 ), which contradicts ( I_n < 1 ). - If ( x geq v_n ), then ( frac{1}{x} leq frac{1}{v_n} ). Thus: [ I_n = S_{n-1} + frac{1}{x} leq S_{n-1} + frac{1}{v_n} = S_n ] Therefore, any valid ( I_n < 1 ) must satisfy ( I_n leq S_n ). ConclusionBoth statements are proven: ( S_n < 1 ) for all positive integers ( n ), and ( S_n ) is the closest to 1 among all sums ( I_n ) that are less than 1.[boxed{S_n < 1}]

🔑:: 1. Define the Sequence: We start with the sequence defined as: [ v_1 = 2, quad v_2 = 2 + 1 = 3, quad v_3 = 2 cdot 3 + 1 = 7, quad ldots, quad v_n = v_1 v_2 cdots v_{n-1} + 1 ]2. Sum (S_n): Define the sum as: [ S_n = frac{1}{v_1} + frac{1}{v_2} + cdots + frac{1}{v_n} ]3. Calculate Initial Terms: We calculate the initial terms of (S_n) and identify a pattern: [ begin{aligned} S_1 &= frac{1}{2} = 1 - frac{1}{2} = 1 - frac{1}{v_2 - 1} S_2 &= frac{1}{2} + frac{1}{3} = frac{3 + 2}{6} = frac{5}{6} = 1 - frac{1}{6} = 1 - frac{1}{v_3 - 1} S_3 &= frac{1}{2} + frac{1}{3} + frac{1}{7} = frac{21 + 14 + 6}{42} = frac{41}{42} = 1 - frac{1}{42} = 1 - frac{1}{v_4 - 1} end{aligned} ] Here, (v_4 = 2 cdot 3 cdot 7 + 1 = 43). [ S_4 = S_3 + frac{1}{43} = 1 - frac{1}{42} + frac{1}{43} = frac{43 - 1}{43 cdot 42} = 1 - frac{1}{1806} = 1 - frac{1}{v_5 - 1} ] Here, (v_5 = 2 cdot 3 cdot 7 cdot 43 + 1 = 1807).4. Find General Pattern: From these calculations, we conjecture the formula: [ S_n = 1 - frac{1}{v_{n+1} - 1} ]5. Mathematical Induction: We use induction to prove the formula for all (n). - Base Case (n=1): [ S_1 = frac{1}{2} = 1 - frac{1}{2} = 1 - frac{1}{v_2 - 1} ] This holds true. - Induction Hypothesis: Assume it holds for (n = k), i.e., [ S_k = frac{1}{v_1} + frac{1}{v_2} + cdots + frac{1}{v_k} = 1 - frac{1}{v_{k+1} - 1} ] - Induction Step: Show it holds for (n = k + 1). [ S_{k+1} = S_k + frac{1}{v_{k+1}} ] Substitute (S_k): [ S_{k+1} = 1 - frac{1}{v_{k+1} - 1} + frac{1}{v_{k+1}} ] Simplify the right-hand side: [ S_{k+1} = 1 - frac{v_{k+1} - (v_{k+1} - 1)}{(v_{k+1} - 1) v_{k+1}} = 1 - frac{1}{v_1 cdot v_2 cdots v_k cdot v_{k+1}} = 1 - frac{1}{v_{k+2} - 1} ] Hence, by induction, the formula holds for all (n): [ S_n = 1 - frac{1}{v_{n+1} - 1} ] Therefore, (S_n < 1) for all (n).6. Further Proof: To show (S_n) is the closest sum less than 1 involving the reciprocals: - For (n=1), (S_1 = 1 - frac{1}{v_2 - 1}), the conjecture holds. - Assume true for (n=k). Given (I_{k+1} < 1) and considering possible terms (frac{1}{m}): [ I_{k+1} = I_k + frac{1}{m} ] if (m geq v_{k+1}): [ frac{1}{m} leq frac{1}{v_{k+1}} text{ hence } I_{k+1} leq S_{k+1} < 1 ] if (m < v_{k+1}): [ m leq v_{k+1} - 1 implies frac{1}{m} geq frac{1}{v_{k+1}-1}, text {and} I_{k+1} geq S_{k+1} = 1 text{ (contradiction with (I_{k+1}<1))} ] Thus, we conclude: [ S_n = sum_{i=1}^n frac{1}{v_i} < 1 implies S_n ] is the closest sum to 1 under the given conditions. (boxed{S_n < 1, forall n})

🧠:Okay, so I have this problem here: Given 3n points in the plane, no three of which are collinear, is it always possible to form n triangles (with vertices at the points) such that no point lies in more than one triangle? Hmm, interesting. Let me try to work through this step by step.First off, let me parse the problem again. We have 3n points, none three collinear. Need to form n triangles, each using three distinct points, and no two triangles share a point. So essentially, we need to partition the 3n points into n groups of three, each forming a triangle, with the additional condition that the triangles don't overlap in any point. But wait, the problem says "no point in the plane lies in more than one triangle." Wait, does that mean the triangles can't overlap geometrically, or just that the points are partitioned? The wording is a bit ambiguous. Let me check again.Original statement: "form n triangles (with vertices at the points), such that no point in the plane lies in more than one triangle." Hmm. So "no point in the plane" is in more than one triangle. But the triangles are formed by the given points. So if the triangles are non-overlapping, in the sense that their interiors don't intersect, then no point in the plane is in more than one triangle. But maybe the problem is just saying that each point is used in only one triangle, which would be the same as partitioning the 3n points into n triangles. But then why the emphasis on "no point in the plane"? Because if two triangles share a vertex, then that vertex is a point in the plane lying in two triangles. Wait, but the triangles have vertices only at the given points. So if two triangles share a vertex, then that point would be in two triangles. But the problem says "no point in the plane lies in more than one triangle." So does that mean that the triangles must be vertex-disjoint? That is, no two triangles share a vertex? Because if a point (vertex) is part of two triangles, then that point is in two triangles, which is not allowed. So the problem is actually asking for a partition of the 3n points into n triangles such that no two triangles share a vertex. But in that case, it's just a partition into triangles, which is straightforward, since 3n points can be partitioned into n groups of three. But maybe the problem is more about the geometric non-overlapping?Wait, but the problem says "no point in the plane lies in more than one triangle." So even a point that's not one of the original 3n points cannot lie in two triangles. So, for example, if two triangles intersect each other, then the intersection points (which are not necessarily original points) would be points in the plane lying in two triangles. Therefore, the problem is actually requiring the triangles to be non-overlapping, i.e., their interiors do not intersect. So the question is: given 3n points with no three collinear, can we always form n triangles, each with three of the points, such that the triangles are non-overlapping?But wait, the original statement says "with vertices at the points," so the triangles must have their vertices among the given 3n points. Then, if two triangles overlap geometrically (i.e., their interiors intersect), then there are points in the plane (not necessarily the original points) that lie in both triangles. Therefore, the problem is asking whether it's always possible to form n triangles with the given 3n points such that the triangles are non-overlapping. That is, their interiors do not intersect.But the problem statement says "no point in the plane lies in more than one triangle." So even a single point that's in two triangles would violate the condition. But if two triangles share a common edge or a vertex, then points on that edge or vertex would be in both triangles. However, if the triangles are non-overlapping and also do not share edges or vertices, then they are completely disjoint. But wait, the original problem doesn't specify whether the triangles can share edges or vertices. Let me check the exact wording again."Form n triangles (with vertices at the points), such that no point in the plane lies in more than one triangle."So if two triangles share a vertex, then that vertex is a point (one of the given points) lying in two triangles, which violates the condition. Similarly, if two triangles share an edge, the points on that edge are part of both triangles, but since the vertices are only the given points, sharing an edge would require sharing two vertices. So in any case, the triangles must be vertex-disjoint. Because if two triangles share a vertex, that vertex is in two triangles, which is prohibited. Therefore, the problem is equivalent to partitioning the 3n points into n vertex-disjoint triangles. But since the points are in general position (no three collinear), each triangle is non-degenerate.Wait, but if that's the case, then the answer is yes, because you can just partition the 3n points into n groups of three, each forming a triangle. However, the problem might be more complicated because the user mentions "no point in the plane lies in more than one triangle," which, as I thought earlier, could refer to geometric overlapping. But according to the problem's wording, since the triangles are formed by the points, the only points that can be part of multiple triangles are the original points. But the problem says "no point in the plane," which includes all points, not just the original ones. Therefore, the triangles must not only be vertex-disjoint but also their interiors and edges must not intersect. That is, the triangles must be non-overlapping in the geometric sense.So now the problem becomes: Given 3n points in the plane, no three collinear, can we always form n non-overlapping triangles, each with three of the points as vertices?This is a classic problem in combinatorial geometry. I recall that there's a result called the "Erdos-Szekeres theorem," but that's more about finding subsets in general position. Alternatively, maybe it's related to triangulation. A triangulation of a point set divides the plane into triangles such that all given points are vertices, and the triangles only intersect at their edges. But in this case, we need to partition the points into n triangles, not necessarily forming a full triangulation of the convex hull. However, in a triangulation, you can have more triangles than n, depending on the number of points.Wait, let's think differently. For any set of points in general position (no three collinear), can we always find a non-crossing matching? Wait, but here we need non-overlapping triangles. Maybe this is related to matching, but in a geometric sense.Alternatively, perhaps there's a way to order the points and form triangles sequentially. For example, take the convex hull of all points. The convex hull of 3n points will have some number of points on the boundary. Let's say the convex hull has m points. Then, we can form a triangle using three consecutive points on the convex hull. But since the points are in general position, the triangle formed by three consecutive convex hull points will not contain any other points inside it (by the convex hull property). Then, removing those three points and proceeding recursively. However, the problem is that the convex hull might not have exactly three points. If the convex hull has more than three points, we can take any three consecutive points, form a triangle, remove them, and repeat. But when we remove three points from the convex hull, the next convex hull of the remaining points may change.Wait, but if we take three consecutive points on the convex hull, say A, B, C, then the triangle ABC is empty (contains no other points inside it). So if we remove A, B, C, the remaining points are still in general position. Then, we can take the convex hull of the remaining points and repeat the process. But does this always work? Let's test with a small example.Suppose we have 6 points (n=2). Let's arrange them in a convex position, forming a convex hexagon. Then, the convex hull is all six points. If we take three consecutive points, say 1, 2, 3, forming a triangle, and remove them, the remaining points are 4, 5, 6, which form another convex triangle. So in this case, we can form two triangles, which are non-overlapping. But what if the points are not all on the convex hull?Suppose we have 6 points, 3 on the convex hull and 3 inside. Let's say the convex hull is a triangle, and there are three points inside. Then, if we take the three convex hull points to form the first triangle, then we have three interior points left. But three points can form a triangle, but is it non-overlapping with the first one? If the first triangle is the convex hull, then the interior triangle is entirely inside the first triangle. Therefore, the two triangles would overlap. So in this case, this method fails.Therefore, the strategy of taking convex hull points might not work if there are interior points. So perhaps a different approach is needed.Another thought: In graph theory, a 3-regular graph is a graph where each vertex has degree 3. But here, we are dealing with geometric graphs. Maybe we need to connect the points into triangles such that the triangles don't cross. But I'm not sure.Alternatively, perhaps we can use induction. Suppose the statement is true for n-1, and then show it's true for n. For the base case, n=1, we have three points, form one triangle, which is trivial. For n=2, 6 points. How would we do that? Let's try to think. If the 6 points are in convex position, as before, we can split them into two triangles by selecting three non-consecutive points? Wait, but if they form a convex hexagon, any three points selected will form a triangle. But if we take two triangles such that they don't overlap. For example, split the hexagon into two triangles by connecting every other vertex. But a convex hexagon can be divided into four triangles via a triangulation. Wait, perhaps not helpful.Alternatively, in a convex hexagon, if you take three alternate vertices, say 1, 3, 5, forming a triangle, and 2, 4, 6 forming another triangle, these two triangles intersect each other. So that's bad. Alternatively, partition the hexagon into two triangles by drawing a line from vertex 1 to 4, splitting the hexagon into two quadrilaterals, then each quadrilateral can be split into two triangles. But that results in four triangles. Not helpful.Wait, but we need two triangles. How? If the hexagon is convex, you can't split it into two triangles without overlapping, because a convex hexagon cannot be covered by two triangles without overlapping. Therefore, perhaps the answer is no? But the problem states "no three collinear," so maybe the answer is yes, but requires a different approach.Wait, but the problem isn't about covering the convex hull, but rather partitioning the 3n points into n triangles. So even if the convex hull is a hexagon, maybe we can choose triangles that are not necessarily part of the convex hull.Wait, for example, in a convex hexagon, take three non-consecutive points to form a triangle, then the remaining three points form another triangle. But those two triangles intersect each other. So that's overlapping. Alternatively, maybe arrange the triangles such that one is inside the other. But if the inner triangle is entirely inside the outer one, then the outer triangle's interior contains the inner triangle, so points in the plane (the inner triangle's area) are in both triangles. Therefore, that's overlapping.Hmm, so maybe in some configurations, it's impossible? But the problem states "no three collinear," so general position. Wait, but the problem asks "is it always possible," so if there exists even one configuration where it's impossible, the answer would be no. But I need to check whether the answer is yes or no.Alternatively, perhaps there's a theorem that says that any set of 3n points in general position can be partitioned into n pairwise non-overlapping triangles. I vaguely recall something about matching points into non-crossing segments, but triangles are more complex.Wait, here's a possible approach: Use a Ham-Sandwich cut or some recursive partitioning. Divide the plane into two regions, each containing 3n/2 points, recursively form triangles in each half. But since 3n might not be divisible by 2, maybe not. Alternatively, use a line that separates the points into two sets, each with a multiple of 3 points. For example, for 3n points, there exists a line that splits them into 3k and 3(n-k) points on each side, by the Ham-Sandwich theorem or some other result. Then, recursively partition each side into triangles. But how do you ensure the non-overlapping?Wait, if you split the plane with a line into two halves, each containing 3k and 3(n−k) points, then recursively form triangles in each half. The triangles on one side of the line won't interfere with the triangles on the other side, so they are non-overlapping. Then, by induction, you can form all n triangles. But is there always such a line? The Ham-Sandwich theorem in two dimensions says that for any two measures, there's a line that simultaneously bisects both, but here we need a line that splits the set of 3n points into two subsets, each with a multiple of 3 points. This might not always be possible. For example, if 3n is not divisible by 2, but 3k and 3(n−k) would need to sum to 3n. However, for example, if you have 3n=6 points, you can split them into 3 and 3. For 3n=9, split into 6 and 3, etc. But how do you guarantee such a line exists?Alternatively, use a result from discrete geometry: For any set of points in general position, there exists a line that passes through two points and divides the remaining points into two equal halves. Wait, but again, the count here needs to be a multiple of 3. Maybe not straightforward.Alternatively, use convex hull peeling. Take the convex hull of all points, form a triangle with three consecutive points on the convex hull, remove them, and repeat. As I thought earlier, but in the case where removing three convex hull points leaves interior points, the next convex hull may still have enough points to form another triangle. But earlier example with a triangle convex hull and three interior points causes a problem. If we take the convex hull (three points), form a triangle, remove them, leaving three interior points. Then form a triangle with the interior points. But the two triangles would overlap, as the interior triangle is inside the first one. So that's not allowed.Therefore, convex hull peeling might not work because the remaining triangles could be inside the earlier ones, leading to overlapping.Another approach: Connect the points in such a way that the triangles are non-overlapping. Maybe use a greedy algorithm. Start with any three points, form a triangle, then choose the next three points that are "outside" the existing triangles. But how to define "outside"? Maybe use the convex hull again. After forming a triangle, remove it and take the convex hull of the remaining points.Wait, let's try an example. Suppose we have 6 points: three on the convex hull forming a triangle, and three inside. If we first form a triangle with the three convex hull points, then the three interior points form another triangle. But the interior triangle is entirely inside the first triangle, so they overlap. Therefore, this approach fails. Alternatively, if we first form a triangle with two convex hull points and one interior point, maybe the next triangle can be formed with the remaining points without overlapping. Let me visualize.Suppose the convex hull is a triangle ABC, with points D, E, F inside. If I form a triangle ABD. Then, the remaining points are C, E, F. The convex hull of C, E, F would be a triangle CEF (assuming E and F are inside ABC but not colinear with C). Then, triangle ABD and triangle CEF: does ABD overlap with CEF? If ABD is a triangle with vertices A, B (on the original convex hull) and D inside, then ABD is a smaller triangle inside ABC. Then CEF is another triangle inside ABC, but depending on where C is. Wait, C is a vertex of the original convex hull. So if we form triangle CEF, which is inside ABC, but C is a vertex of ABC, so triangle CEF would have vertex C and two interior points. This triangle might not overlap with ABD. For example, ABD is near edge AB, and CEF is near vertex C. Depending on the positions, they might not overlap. But it's not guaranteed. Maybe they do overlap.Alternatively, form triangle ADE, then BCF. But again, if ADE is somewhere inside ABC, and BCF is using B and C on the hull and F inside, maybe these triangles don't overlap. It's getting complicated. Maybe with specific arrangements, you can make it work, but is this always possible?Alternatively, think of it as a graph where each triangle is a node, and edges represent overlapping. Then, finding a matching without overlapping. But I'm not sure.Wait, here's a different idea. Since the points are in general position, we can always find a triangle that doesn't contain any other points inside it. This is known as an empty triangle. Then, if we can repeatedly find empty triangles, remove them, and continue until all points are exhausted. Since there are 3n points, we need n empty triangles. But is it always possible to find n disjoint empty triangles?In computational geometry, there's a concept called "happy ending problem," which states that any five points in general position have an empty convex quadrilateral, but I'm not sure about triangles. Wait, actually, any set of points in general position contains an empty triangle. Because you can take the convex hull, and if the convex hull has at least three points, then any three consecutive points on the convex hull form an empty triangle. But if the convex hull is a triangle, then the only empty triangle is the convex hull itself, and the inner points may not form empty triangles.Wait, for example, if you have four points: three on the convex hull forming a triangle, and one inside. Then, the only empty triangle is the convex hull. The inner point can't form a triangle without including two convex hull points, but such a triangle would contain the inner point. Wait, no. If you take two convex hull points and the inner point, the triangle formed would contain the other convex hull point? No, because the three points are the two on the hull and the inner one. So the triangle is formed by A, B, and D (the inner point). This triangle doesn't contain the third convex hull point C, right? Because A, B, D form a triangle, and C is outside of it since the original convex hull was ABC. So triangle ABD is empty? Wait, no. If D is inside ABC, then triangle ABD is inside ABC, and does it contain any other points? In this case, there's only one inner point D, so triangle ABD is empty. Wait, but if there are multiple inner points, like in the previous example with three inner points, then forming a triangle with two convex hull points and one inner point might leave other inner points inside.Wait, let's formalize this. If we have a set of points in general position, can we always find an empty triangle? Yes, according to the Erdős–Szekeres theorem on empty convex polygons, but for triangles, it's simpler. Specifically, any set of five points in general position contains an empty triangle. But in our case, maybe we can always find an empty triangle, remove it, and repeat.However, in our problem, we have 3n points. If we can find an empty triangle, remove those three points, then we have 3(n-1) points left. By repeating this process, we could partition all points into n empty triangles. Since empty triangles are non-overlapping (because once you remove them, the next empty triangle is from the remaining points), but actually, even if the empty triangles are nested, they might still overlap with previous ones. Wait, no. If an empty triangle is removed, the remaining points are either inside or outside of it. But if you remove three points forming an empty triangle, the remaining points can't be inside that triangle because it was empty. Therefore, all remaining points are outside. Wait, no. If you have a triangle formed by three convex hull points, which is empty, then removing those three points leaves the remaining points inside the convex hull of the original set, but not necessarily inside the removed triangle. Wait, actually, if you remove three convex hull points forming a triangle, the remaining points could still be both inside and outside, but since the triangle was part of the convex hull, the remaining points are either on the convex hull or inside the original convex hull but outside the removed triangle. Wait, maybe not. Let's think.Suppose we have a convex hull with points A, B, C, D, E, F in order. If we remove triangle A, B, C, then the remaining points D, E, F are still on the convex hull, forming a new convex hull. However, if the original convex hull was a hexagon, removing three consecutive vertices leaves the other three as the new convex hull. The triangle D, E, F would not overlap with triangle A, B, C. Therefore, those two triangles are non-overlapping. Similarly, if we have a convex hull with more points, removing three consecutive ones would leave the rest on the convex hull, so their triangles wouldn't overlap.But if the convex hull is a triangle with points inside, like in the earlier example, removing the convex hull triangle leaves the inner points. Then, forming a triangle with the inner points would be inside the original triangle, hence overlapping. Therefore, the problem arises when there are inner points.But in that case, maybe there's a different empty triangle we can remove first. For example, if there's a triangle inside the convex hull that's empty. Suppose we have a convex hull triangle ABC, and points D, E, F inside. If among D, E, F, there's an empty triangle, say DEF, then we can remove DEF first, leaving ABC, which can then be removed as a triangle. But DEF is inside ABC, so if we remove DEF first, then ABC is still there. Then removing ABC would result in two triangles: DEF and ABC. But ABC contains DEF, so their interiors overlap. Therefore, that's not allowed.Alternatively, if we can find an empty triangle that doesn't lie entirely inside another triangle. Wait, but how?Alternatively, use the concept of onion layers. The onion peeling of a point set is the process of repeatedly removing the convex hull. So first convex hull, then the convex hull of the remaining points, etc. Each layer is a convex polygon. For a set of 3n points, the number of convex layers depends on the distribution. If we can peel off triangles from each layer. But convex layers are convex polygons, which can have any number of vertices. If a convex layer has m points, we can partition it into triangles. But if m is divisible by 3, then we can split it into m/3 triangles. However, m might not be divisible by 3. For example, a convex pentagon can't be partitioned into triangles each using three vertices without overlapping.Wait, no. A convex pentagon can be triangulated into three triangles, but those triangles share edges and vertices. But in our problem, the triangles must be vertex-disjoint and non-overlapping. So, for instance, in a convex pentagon, you cannot partition the five vertices into one triangle and two remaining points, which is insufficient. Therefore, onion peeling might not work unless each convex layer has a number of points divisible by 3.Alternatively, perhaps combine points from different layers. For example, take two points from the outer convex hull and one from an inner layer to form a triangle, such that the triangle doesn't contain any other points. Then, remove those three points and continue. But ensuring that the triangle is empty might be difficult.Wait, here's another idea. Use a perfect matching for the points. But a perfect matching in a graph is a set of edges without common vertices, covering all vertices. But here we need a 3-matching, or a triangle factor. In graph theory, a triangle factor is a set of vertex-disjoint triangles covering all vertices. For a complete graph on 3n vertices, a triangle factor exists, but we are dealing with geometric graphs where edges can cross. However, the problem doesn't mention edges, just triangles as polygons. So maybe the question is purely combinatorial: can the points be partitioned into n vertex-disjoint triangles, regardless of geometry? But no, because the problem requires that no point in the plane lies in more than one triangle, which is a geometric condition.Therefore, the answer must take into account the geometric non-overlapping. So it's a geometric problem, not purely combinatorial.Wait, but I recall a result that any set of points in general position can be partitioned into disjoint triangles. Is that true? Let me search my memory. There's a theorem called "Dilworth's theorem," but that's for posets. Alternatively, in graph drawing, there's a concept of non-crossing matchings. For points in general position, it's always possible to form a non-crossing perfect matching. Similarly, maybe for triangles.Ah, here's a paper I recall: "Matching points with triangles" by Hosono et al., but I don't remember the exact result. Alternatively, a more recent result. Wait, but since I can't access external resources, I need to reason it out.Suppose we proceed by induction. For n=1, trivial. Assume it's true for n-1. For a set of 3n points, can we find a triangle that, when removed, leaves 3(n-1) points in general position, allowing the induction hypothesis to apply? The key is to find such a triangle that doesn't enclose any other points, i.e., an empty triangle. If we can always find an empty triangle, then we can remove it and proceed. But can we always find an empty triangle in any set of 3n points?In 1994, Erdős proved that every set of n points contains at least √n/2 empty triangles. But that's a lower bound. However, for our purpose, we need at least one empty triangle in any set of 3n points. Is that guaranteed?Actually, yes. Consider the convex hull of the point set. If the convex hull has three or more points, then any three consecutive convex hull points form an empty triangle. Because by the convex hull property, there are no points inside the convex hull, so the triangle formed by three consecutive points on the convex hull is empty.Therefore, if the convex hull has three or more points, we can take an empty triangle from the convex hull, remove those three points, and the remaining points are still in general position (no three collinear). Then, by induction, we can partition the remaining 3(n-1) points into n-1 triangles. The only problem is if the convex hull has exactly three points, and all other points are inside. Then, removing the convex hull triangle would leave the inner points, which might not have an empty triangle. Wait, but even in that case, the inner points are in general position. So their convex hull (if there are three or more points) would have an empty triangle. Wait, if there are 3(n-1) points inside, which is 3n-3 points, then if n>1, 3n-3 >=3. So the inner points have a convex hull. Take three consecutive points on that inner convex hull, forming an empty triangle. But this inner convex hull is inside the original convex hull. So the triangle formed by the inner convex hull points is empty (with respect to the remaining inner points), but is it empty with respect to all points? Wait, no. Because the original outer convex hull triangle has been removed, so the inner points are all that's left. Therefore, the inner convex hull triangle is empty. Therefore, by induction, we can keep removing convex hull triangles layer by layer.Wait, let's test this with an example. Suppose we have 6 points: 3 on the outer convex hull (triangle ABC), and 3 inside forming another triangle DEF. If we first remove ABC, then DEF is the convex hull of the remaining points. Remove DEF, done. However, ABC and DEF are both triangles, but DEF is inside ABC, so their interiors overlap. But the problem states that no point in the plane can lie in more than one triangle. If we remove ABC first, then DEF is inside the original ABC triangle. When we remove DEF, the triangle DEF's interior is entirely inside ABC's interior. Therefore, points in the plane lying inside DEF are also inside ABC, which violates the condition. Therefore, this approach doesn't work.Ah, right! So the problem is that even though we are partitioning the points into triangles, the geometric locations cause the triangles to overlap. So even though the points are used only once, the triangles themselves overlap in the plane. So the induction approach using convex hulls fails because of overlapping interiors.Therefore, we need another method. Let's think differently.Suppose we connect all the points into a triangulation. A triangulation of a point set is a maximal set of non-crossing edges. In a triangulation, the number of triangles formed is 2n - 2 - h, where h is the number of points on the convex hull. But this might not directly help, as we need exactly n triangles from 3n points.Wait, but in a triangulation of 3n points, the number of triangles is 2(3n) - 2 - h. For example, if all points are on the convex hull (h=3n), then the number of triangles in a triangulation would be 3n - 2. Which is more than n. So that's not helpful.Alternatively, maybe select a subset of triangles from the triangulation that are non-overlapping and cover all points. But since a triangulation includes many triangles that share edges and vertices, this is not directly applicable.Wait, another idea: Use a 3-coloring of the points such that each color class forms a triangle. But I don't see how coloring would help with non-overlapping.Alternatively, use a sweep line algorithm. Sweep the plane from left to right, and whenever three points are encountered, form a triangle. But this is vague.Wait, here's a different approach inspired by non-crossing matchings. For points in general position, it's known that you can always form non-crossing perfect matchings. Maybe extend this to triangles. For example, divide the points into three groups, each group forming a non-crossing matching. But triangles are more complex.Alternatively, use the fact that any set of points can be partitioned into three pairwise disjoint sets whose convex hulls are pairwise disjoint. But I'm not sure about that.Wait, here's a theorem by Bárány and Valtr: Any set of points in the plane can be partitioned into four subsets with disjoint convex hulls. But we need n subsets (triangles) with disjoint convex hulls. Not sure.Alternatively, think of it incrementally. Start with a point, and find two other points such that the triangle formed doesn't enclose any other points. Then, remove them and repeat. But ensuring that each triangle doesn't enclose points is tricky.Wait, perhaps use a 3-Steiner tree or something, but not relevant.Another angle: The problem is equivalent to finding a set of n vertex-disjoint triangles such that their convex hulls are pairwise disjoint. If we can ensure that the convex hulls of the triangles do not intersect, then the triangles themselves do not overlap. Therefore, the problem reduces to partitioning the 3n points into n groups of three such that the convex hulls of the groups are pairwise disjoint.Is this always possible? Let's consider that in any set of 3n points, we can find a triangle with a small enough area that its convex hull doesn't interfere with other points. But this is heuristic.Alternatively, use a recursive partitioning approach with lines. As I thought earlier, split the plane with a line into two regions, each containing a multiple of 3 points, and recurse. This would ensure that triangles in different regions don't overlap. But the problem is guaranteeing the existence of such a line.There's a result called the "Ham-Sandwich Cut," which in two dimensions, for any two measures, there exists a line that simultaneously bisects both. But here, we need to split the point set into two subsets, each with a multiple of 3 points. For example, for 3n points, split into 3k and 3(n−k) points. The question is whether such a split is always possible with a line.This is similar to the "3-divisibility" problem. There's a theorem called the "Stone–Tukey theorem," which is a generalization of the Ham-Sandwich theorem. It states that for any d measurable sets in R^d, there exists a hyperplane that simultaneously bisects all of them. However, applying this to our problem, we need a line that splits the point set into two subsets, each with a multiple of 3 points. But this might not always be possible.For example, consider 3n=3 points. There's no line that splits them into 0 and 3. Unless you allow the line to pass through a point, but since no three are collinear, a line can pass through at most two points. But in that case, splitting into 1 and 2 points, which isn't a multiple of 3. So the theorem doesn't apply here.Alternatively, instead of splitting with a line, use a different shape. But I don't know.Alternatively, use a result from discrete geometry called the "Erdős–Szekeres theorem," but that's about finding subsets in convex position.Alternatively, since the points are in general position, we can always find a line that passes through two points and leaves at least ⌊(3n-2)/2⌋ points on each side. But this doesn't guarantee divisibility by 3.Given that the problem is asking whether it's always possible, and considering that I can't find a definitive theorem in my mind, but recalling that in combinatorial geometry, partitioning points into non-overlapping triangles is a known problem, but I might need to think of a specific method.Wait, here's a possible construction. Order the points by their x-coordinates. Then, group them into triples: first three, next three, etc. Then, for each triple, connect them in a triangle. However, adjacent triples might have overlapping triangles. But if we order them along the x-axis, and connect each three consecutive points, the triangles might be arranged in a way that they don't overlap. Wait, but in general position, points ordered by x-coordinate can be in any y-coordinate. So the triangles could be all over the place and overlapping.Alternatively, use a sweeping line to create triangles incrementally. As the line sweeps, when three points are passed, form a triangle below the line. But this is vague.Wait, here's a different approach inspired by graph planarity. If we can embed a planar graph where each face is a triangle, then the dual graph is a triangulation. But our problem is not directly related.Alternatively, think of the problem as a geometric graph where we need to cover all vertices with disjoint triangles, and the graph is the complete geometric graph, but with edges not crossing.Wait, but covering with vertex-disjoint triangles in a geometric graph without crossings is similar to finding a non-crossing triangle matching. It's known that for any set of points in general position, there exists a non-crossing perfect matching. But for triangles, is this possible?A paper by Aichholzer et al. titled "Non-crossing Hamiltonian paths and cycles in output-polynomial time" might be related, but I don't recall specifics. However, if we can connect the points into a non-crossing cycle and then partition the cycle into triangles, that might work. For example, a polygon triangulation. If the points are in convex position, the polygon can be triangulated into n-2 triangles, but we need n triangles for 3n points. Wait, no. For a convex polygon with 3n points, a triangulation would have 3n - 2 triangles. Which is way more than n.But we need exactly n triangles, each with three points, and no overlapping. So maybe a different approach.Wait, if we can group the points into n groups of three, each group forming a triangle, such that all triangles are pairwise non-overlapping. How to ensure that?Maybe use a greedy algorithm: at each step, pick the leftmost remaining point, then the next two points to form a triangle that doesn't enclose any other remaining points. But I'm not sure.Alternatively, use the fact that in general position, there exists a point with a unique x-coordinate, say the leftmost point. Connect it with the next two leftmost points to form a triangle. Since this triangle is "on the edge," it might not enclose other points. Then, remove these three points and repeat. But this is similar to the convex hull approach. If the leftmost three points are not on the convex hull, their triangle might enclose other points. However, in general position, three points forming a triangle won't have other points on their edges, but they might enclose points inside.Wait, but if we take the leftmost three points, form a triangle, and remove them, even if that triangle contains other points inside, those inner points would still be part of the remaining set. Then, subsequent triangles could be formed without overlapping. But if the first triangle encloses points, then any triangle formed later inside would overlap with the first triangle's interior. So that's not allowed.Therefore, the greedy approach of taking leftmost points might not work.Hmm. This is proving to be quite challenging. Maybe the answer is yes, and it can be done through a recursive partitioning using lines, as follows:1. For a given set of 3n points, find a line that divides them into two subsets, each with 3k and 3(n−k) points.2. Recursively partition each subset into k and (n−k) triangles.3. Since the subsets are separated by a line, the triangles from each subset cannot overlap with those from the other subset.If such a line exists for any 3n points, then the result follows by induction. The base case n=1 is trivial. The problem is whether such a line always exists.But how can we guarantee the existence of a line that splits 3n points into two subsets of sizes divisible by 3? This is similar to the "Ham-Sandwich cut for divisibility" problem.In discrete geometry, there's a result called the "3-cut" theorem, but I'm not sure. Alternatively, use the intermediate value theorem. If we rotate a line from vertical to horizontal, the number of points on each side changes continuously. By the continuity, there must be an angle where the number of points on each side is a multiple of 3. But is this necessarily true?Suppose we have 3n points. As we rotate a line, the number of points on each side varies. The number of points on one side can be any integer from 0 to 3n. We need to find an angle where the number is a multiple of 3. Since there are 3n + 1 possible counts (from 0 to 3n), and 3n is a multiple of 3, the counts that are multiples of 3 are 0, 3, 6, ..., 3n. There are n + 1 such counts. As we rotate the line, the count changes by ±1 when the line crosses a point. Therefore, by the intermediate value theorem, between two angles where the count changes from k to k+1, there must be a position where the count takes all intermediate values. Therefore, for each multiple of 3 between 0 and 3n, there must be a line achieving that count. Therefore, there exists a line that splits the set into two subsets, each with a multiple of 3 points.Wait, this seems plausible. Here's the argument in more detail:Consider any set of 3n points. Choose a direction not aligned with any of the point pairs (possible since there are finitely many points). Sweep a line perpendicular to this direction. Initially, the line will have all points on one side. As we sweep, the number of points on one side decreases from 3n to 0. Each time the line passes a point, the count decreases by 1. By the time the line has swept past all points, the count is 0. Now, consider the count modulo 3. Initially, the count is 3n ≡ 0 mod 3. After passing each point, the count decreases by 1, so modulo 3, it cycles through 2, 1, 0, 2, 1, 0, etc. Therefore, when the count reaches 3(n - 1), which is ≡ 0 mod 3, there must be a position where the line has exactly 3(n - 1) points on one side and 3 on the other. Therefore, such a line exists.Yes, this seems correct. By continuity, as we sweep the line, the count decreases from 3n to 0, passing through every integer. The count modulo 3 cycles every three steps. Since we start at 0 mod 3, after moving past three points, the count is 3n - 3 ≡ 0 mod 3 again. Therefore, there must be a position where the line has exactly 3k points on one side and 3(n - k) on the other, for some integer k. Hence, we can always find such a line.Therefore, using this line, we can split the 3n points into two subsets of 3k and 3(n - k) points. Recursively apply the procedure to each subset. By induction, each subset can be partitioned into k and (n - k) non-overlapping triangles. Since the two subsets are separated by a line, the triangles from each subset cannot overlap with those from the other. Therefore, combining them gives a partition of the original set into n non-overlapping triangles.Thus, the answer is yes, it's always possible.

✅:The problem asks whether, given (3n) points in the plane with no three collinear, it is always possible to form (n) triangles such that no point in the plane lies in more than one triangle. This requires the triangles to be vertex-disjoint and geometrically non-overlapping. Key Steps in the Solution:1. Understanding Non-Overlapping: The condition "no point in the plane lies in more than one triangle" implies triangles must be vertex-disjoint and their interiors must not intersect. This ensures no geometric overlap, not just partitioning points into triangles.2. Inductive Approach: Use mathematical induction. For the base case (n=1), three points trivially form one triangle. Assume the statement holds for (k < n).3. Splitting Points with a Line: For (3n) points, use a rotating line argument. By the intermediate value theorem, there exists a line dividing the points into two subsets, each with a multiple of 3 points ((3k) and (3(n-k))). This ensures divisibility for recursive partitioning.4. Recursive Partitioning: Apply the inductive hypothesis to each subset. Since the line separates the subsets, triangles formed in each subset do not overlap with those in the other, maintaining non-overlapping across the entire set.5. Guaranteeing Existence of Splitting Line: Sweep a line across the plane. The count of points on each side cycles modulo 3, ensuring a position where counts are multiples of 3. This guarantees the required division. Conclusion:Yes, it is always possible to form (n) non-overlapping triangles from (3n) points in general position. The proof relies on inductive partitioning using a splitting line, ensuring geometric non-overlapping by construction.Final Answer: boxed{Yes}

🔑:To demonstrate that it is always possible to form ( n ) triangles from ( 3n ) points in the plane, such that no three points are collinear and no point lies in more than one triangle, follow these steps:1. Choose a Slope: Select a line with a specific slope that is different from the slope of any line joining any two of the given points. This ensures that when you project points onto this line, no two points have the same projection, avoiding any collinearity issues.2. Divide the Points: Begin with a line that is parallel to the chosen slope but initially positioned far away from all the points. Imagine moving this line gradually towards and through the points.3. Group the Points: As this parallel line moves, it will intersect the points one at a time. Using this method, partition the set of ( 3n ) points into groups of three.4. Form Triangles: For each group of three points formed in the previous step, construct a triangle using these three points as vertices. Here's how it works: - Start with the initial parallel line far away from the points. - Move the line towards the points gradually. The first three points the line crosses will form the first group. - Continue moving the line and grouping the next set of three points it intersects. - Repeat this process until all points are grouped.5. Ensure No Overlapping Triangles: By creating groups of three points through an orderly process using a line of unique slope, the points within each group are distinct from those in the other groups. Thus, the triangles formed will not overlap. - Since each point is encountered exactly once in this sweeping process, each point will form the vertex of exactly one triangle. - Additionally, no point is shared between two triangles, ensuring that no point lies inside more than one triangle. Conclusion:It is indeed always possible to form ( n ) triangles from ( 3n ) points in the plane, with no three points being collinear, and no point lying in more than one triangle. This concludes the proof.(boxed{text{Yes}})

🧠:Okay, so I need to figure out the number of edges and vertices in a polyhedron that has 20 triangular faces. Hmm, let me start by recalling some basic concepts about polyhedrons. I remember that there's something called Euler's formula that relates the number of vertices (V), edges (E), and faces (F) of a convex polyhedron. The formula is V - E + F = 2. Maybe I can use that here.But first, let me make sure I remember the formula correctly. Yeah, Euler's characteristic formula states that for any convex polyhedron, the number of vertices minus the number of edges plus the number of faces equals 2. So, V - E + F = 2. Got that.Now, the problem says the polyhedron has 20 triangular faces. Since each face is a triangle, I can probably use that information to find the number of edges. But wait, each triangular face has 3 edges, right? So, if there are 20 faces, each with 3 edges, that would give 20 * 3 = 60 edges. But hold on, this counts each edge twice because each edge is shared by two adjacent faces. So, the actual number of edges should be half of that. So, E = 60 / 2 = 30. Is that right?Let me check. If every edge is shared by two faces, then each edge is counted twice when we sum up all the edges of all the faces. Therefore, dividing the total by 2 gives the correct number of edges. That makes sense. So, with 20 triangular faces, we have (20 * 3)/2 = 30 edges. So, E = 30.Now, we need to find the number of vertices. Using Euler's formula: V - E + F = 2. We know E is 30 and F is 20. Plugging those values into the formula: V - 30 + 20 = 2. Simplifying that, V - 10 = 2, so V = 12. Therefore, the polyhedron has 12 vertices.Wait a second, but does this polyhedron actually exist? Because sometimes Euler's formula can give a result, but you have to check if such a polyhedron is possible. For example, in this case, a polyhedron with 20 triangular faces, 30 edges, and 12 vertices. Let me think if such a shape is known.Oh! This seems like an icosahedron, but wait, a regular icosahedron has 20 triangular faces, 12 vertices, and 30 edges. So, yes, that's exactly what we're talking about here. So, it's a regular icosahedron. But the problem didn't specify it's regular, just a polyhedron with 20 triangular faces. However, Euler's formula applies to any convex polyhedron, not just regular ones. So even if it's not regular, as long as it's a convex polyhedron with triangular faces, the formula holds.But just to make sure I didn't make a mistake in the calculations. Let's recap. Each triangular face has 3 edges, 20 faces gives 60 edges counted with multiplicity. Since each edge is shared by two faces, divide by 2 to get 30 edges. Then Euler's formula: V - E + F = 2 => V = 2 + E - F = 2 + 30 - 20 = 12. Yep, that seems right.Alternatively, maybe using another formula. For polyhedrons, there are relationships between the faces and vertices. For example, if all the faces are triangles, it's a deltahedron. But in this case, we can stick with Euler's formula.Another way to check: in a polyhedron where all faces are triangles, each face has 3 edges, and each edge is shared by 2 faces, so total edges E = (3F)/2. So, if F =20, E= (3*20)/2=30, which matches. Then Euler's formula gives V= 2 + E - F = 2 +30 -20=12. So that seems consistent.So, the answer should be 30 edges and 12 vertices. Therefore, the polyhedron has 30 edges and 12 vertices. That makes sense, and since the regular icosahedron is a prime example of such a polyhedron, even if it's not specified to be regular, the counts hold as per Euler's formula.I think that's it. Let me just check once more. Maybe compute the average number of edges per vertex or something. The number of edges can also be related to the vertices by the handshaking lemma, which in graph theory terms, the sum of all vertex degrees equals twice the number of edges. So, if we denote the sum of degrees as Σdeg(V) = 2E. So, Σdeg(V) = 2*30=60. If there are 12 vertices, then the average degree would be 60/12=5. So, each vertex has degree 5 on average. In a regular icosahedron, each vertex is indeed of degree 5, since each vertex is part of five triangular faces. So that checks out. So even if it's not regular, the average degree is 5, which is feasible.Therefore, all the numbers align correctly with Euler's formula and the related graph theory principles. Hence, I can confidently say the polyhedron has 30 edges and 12 vertices.

✅:The polyhedron with 20 triangular faces has 30 edges and 12 vertices.Step-by-Step Explanation:1. Apply Euler's Formula: For any convex polyhedron, Euler's formula states: [ V - E + F = 2 ] where (V) = vertices, (E) = edges, and (F) = faces. Given (F = 20), we need to find (V) and (E).2. Calculate Edges ((E)): Each triangular face has 3 edges, but each edge is shared by 2 faces. Thus: [ E = frac{3 times F}{2} = frac{3 times 20}{2} = 30 ]3. Solve for Vertices ((V)): Substitute (E = 30) and (F = 20) into Euler's formula: [ V - 30 + 20 = 2 implies V - 10 = 2 implies V = 12 ]4. Verification: - The result matches the properties of a regular icosahedron (a known polyhedron with 20 triangular faces, 30 edges, and 12 vertices). - Consistency check: The sum of vertex degrees equals (2E). For (V = 12), the average degree is (60/12 = 5), which aligns with the structure of triangular faces.Final Answer: The polyhedron has (boxed{30}) edges and (boxed{12}) vertices.

🔑:1. Definitions and initial values: - Let ( S ) be the number of vertices (sommets). - Let ( A ) be the number of edges (arêtes). - Let ( F ) be the number of faces (faces) with ( F = 20 ).2. Using double counting on edges and faces: - Each face is a triangle. - Each triangle has 3 edges. - Therefore, the total edge count from the face perspective is ( 3F ).3. Counting pairs ( (a, f) ): - Each edge belongs to exactly 2 faces, meaning every edge is counted twice in ( 3F ). - Therefore, the number of edges ( A ) can be determined by: [ 2A = 3F ] - Substituting ( F = 20 ): [ 2A = 3 times 20 implies 2A = 60 ] - Solving for ( A ): [ A = frac{60}{2} implies A = 30 ]4. Applying Euler's characteristic formula: - Euler's formula for polyhedra states: [ S - A + F = 2 ] - Substituting ( A = 30 ) and ( F = 20 ): [ S - 30 + 20 = 2 ] - Simplifying the equation: [ S - 10 = 2 implies S = 12 ] Conclusion:The polyhedron has ( 12 ) vertices and ( 30 ) edges.[boxed{30 text{ arêtes, } 12 text{ sommets}}]

🧠:Alright, so I have this problem here that says any positive integer n can be expressed as a sum of one or more positive integers in 2^{n-1} ways. For example, they showed that 3 can be represented in four ways, which is 2^{3-1} = 4. That checks out. But how do I prove this for any positive integer n?Let me start by understanding the problem better. When they say "expressed as a sum of one or more positive integers," they mean ordered sums, right? Because for 3, the example includes 1+2 and 2+1 as different, so order matters. So this is about compositions, not partitions. In partitions, the order doesn't matter, so 1+2 and 2+1 would be considered the same. But here, they are different, so we are dealing with compositions.Okay, so compositions of n. And the number of compositions of n is 2^{n-1}. Got it. So I need to show that. How do compositions work? Let me recall. For a composition, you can think of it as breaking the integer n into a sequence of positive integers where the order matters.One way to approach counting compositions is to consider the "gaps" between the units that make up n. For example, if we have n = 3, we can represent it as three ones: 1 1 1. Now, between these ones, there are n - 1 gaps. In the case of n = 3, there are two gaps. For each gap, we can decide whether to put a plus sign or not. If we put a plus, it separates the numbers, and if we don't, it combines them. So for each gap, there are two choices: split or not. Therefore, the total number of compositions would be 2^{n-1}.Wait, that seems straightforward. For each of the n-1 gaps between the units, you choose to split or not, leading to 2^{n-1} compositions. But maybe I need to elaborate more to make sure this is rigorous.Alternatively, another way to think about it is recursively. Let's say the number of compositions of n is C(n). Then, consider how you can form a composition of n. If the first term is 1, then the remaining part is a composition of n-1. If the first term is 2, then the remaining part is a composition of n-2, and so on, up to the first term being n itself. But this might not be the easiest way since it would lead to a recursive formula. Let's check:C(n) = C(n-1) + C(n-2) + ... + C(1) + 1. Because you can have first term 1 followed by any composition of n-1, first term 2 followed by any composition of n-2, etc., and finally the composition that is just n itself. But if we calculate this sum, let's see for small n:For n=1: C(1)=1.n=2: C(2)=C(1)+1=1+1=2.n=3: C(3)=C(2)+C(1)+1=2+1+1=4. Which matches the example.n=4: C(4)=C(3)+C(2)+C(1)+1=4+2+1+1=8.So it seems that C(n) = 2^{n-1}. So the recursive formula gives us C(n) = 2^{n-1}. But how can we show that the recursive formula indeed results in 2^{n-1}?Alternatively, maybe we can find a closed-form solution for the recursion. Let's see:Let's suppose C(n) = 2^{n-1}. Let's verify if this satisfies the recursive formula.Assuming for all k < n, C(k) = 2^{k-1}. Then C(n) would be sum_{k=1}^{n-1} C(k) + 1. Let's compute this sum:sum_{k=1}^{n-1} 2^{k-1} + 1. The sum from k=1 to n-1 of 2^{k-1} is a geometric series. The sum of 2^{0} + 2^{1} + ... + 2^{n-2} = 2^{n-1} - 1. Then adding the +1 gives 2^{n-1} -1 +1 = 2^{n-1}. Therefore, if C(k) = 2^{k-1} for all k < n, then C(n) = 2^{n-1}. Thus, by induction, C(n) = 2^{n-1} for all n ≥ 1.Base case: n=1, C(1)=1=2^{0}=1. Correct.Therefore, this recursive approach works. So the number of compositions is 2^{n-1}.But another way to think about it is through binary choices. Each composition corresponds to a way of placing plus signs between the units of n. Since there are n-1 gaps between the units, and each gap can either have a plus or not, the total number is 2^{n-1}.Wait, this seems similar to the first approach. So both the recursive method and the binary choice method lead to the same conclusion.Alternatively, maybe we can model this using generating functions. The generating function for compositions where order matters is the sum over all compositions of x^{sum}. Since each composition is a sequence of positive integers, the generating function would be:G(x) = sum_{k=1}^infty (x + x^2 + x^3 + ...)^k.But since each term in the composition is at least 1, the generating function for each term is x/(1 - x). Then, since compositions are sequences of one or more terms, the generating function is:G(x) = (x/(1 - x)) + (x/(1 - x))^2 + (x/(1 - x))^3 + ... = (x/(1 - x)) / (1 - (x/(1 - x))) ) = (x/(1 - x)) / ( (1 - x - x)/(1 - x) ) ) = (x/(1 - x)) / ( (1 - 2x)/(1 - x) ) ) = x/(1 - 2x).Therefore, the generating function is x/(1 - 2x), which expands to sum_{n=1}^infty 2^{n-1} x^n. Therefore, the coefficient of x^n is 2^{n-1}, which is the number of compositions of n. Hence, proved.But maybe I need to explain this generating function approach more carefully. Let's see:Each composition can be thought of as a sequence of one or more positive integers. So for each position in the sequence, we have a positive integer. The generating function for a single positive integer is x + x^2 + x^3 + ... = x/(1 - x). Then, since a composition is a sequence of one or more such terms, the generating function is:G(x) = (x/(1 - x)) + (x/(1 - x))^2 + (x/(1 - x))^3 + ... This is a geometric series with first term a = x/(1 - x) and common ratio r = x/(1 - x). So the sum is a / (1 - r) = (x/(1 - x)) / (1 - x/(1 - x)) ) = (x/(1 - x)) / ( (1 - x - x)/(1 - x) ) ) = x/(1 - 2x). Expanding this as a power series: 1/(1 - 2x) is the generating function for 2^n, starting at n=0. But since we have x/(1 - 2x), it's shifted by one. So the coefficient of x^n in x/(1 - 2x) is 2^{n-1}, which is exactly the number of compositions of n. Therefore, the generating function approach confirms that the number is 2^{n-1}.Alternatively, maybe we can use induction. Let's try mathematical induction.Base case: n=1. There is only one composition: 1. 2^{1-1}=1. So it holds.Assume that for some integer k ≥ 1, the number of compositions of k is 2^{k-1}. Now consider k+1.A composition of k+1 can either start with 1 or a number greater than 1. If it starts with 1, the remaining part is a composition of k. If it starts with a number greater than 1, say m ≥ 2, then we can write m as 1 + (m - 1), so the composition becomes 1 followed by a composition of k. Wait, maybe this isn't the right approach.Wait, another way: For the induction step, suppose we have a composition of k+1. If we subtract 1 from the first term, if the first term is at least 2, we get a composition of k where the first term is at least 1. If the first term is exactly 1, then the rest of the composition is a composition of k. Hmm, this might get complicated.Alternatively, perhaps a better way is to note that each composition of k+1 can be obtained by either adding 1 to the first term of a composition of k or by appending a new term 1 to a composition of k. Therefore, the number of compositions of k+1 is twice the number of compositions of k. Hence, if C(k) = 2^{k-1}, then C(k+1) = 2 * C(k) = 2 * 2^{k-1} = 2^{k}, which is 2^{(k+1)-1}. Therefore, by induction, the formula holds.Wait, that seems too simplistic. Let me check with n=3.If n=2 has 2 compositions: 2, 1+1. Then for n=3, adding 1 to the first term of each composition of 2: 3, 2+1. Appending 1 to each composition of 2: 2+1, 1+1+1. Wait, but 2+1 appears in both, so actually, the total is 3, 2+1, 1+2, 1+1+1. So actually, the number doubles. So for each composition of 2, you can either add 1 to the first term or append 1 at the end. However, in the case of appending 1, it's not just appending, because the original composition could have multiple terms. For example, if you have a composition like 1+1 of 2, appending 1 gives 1+1+1, while adding 1 to the first term gives 2+1. Similarly, for the composition 2 of 2, adding 1 to the first term gives 3, and appending 1 gives 2+1. So in total, from 2 compositions of 2, we get 4 compositions of 3, which is double. Hence, this method works. Therefore, if we assume that C(k) = 2^{k-1}, then C(k+1) = 2*C(k) = 2^{k}. Hence, the formula holds by induction.Therefore, this seems like a valid inductive step. So the number of compositions doubles each time n increases by 1. Starting from n=1 with 1 composition, n=2 has 2, n=3 has 4, n=4 has 8, etc., which is 2^{n-1}.Another way to think about it is that each composition corresponds to a binary string of length n-1, where each bit represents whether there is a plus or not between two units. For example, for n=3, the binary strings of length 2 are: 00, 01, 10, 11. Each corresponds to a composition:- 00: no splits, so 3.- 01: split after the first unit, so 1+2.- 10: split after the second unit, so 2+1.- 11: split after both, so 1+1+1.Each binary choice corresponds to a composition, hence 2^{n-1} compositions.This is similar to the first approach where each gap can have a plus or not. So in total, each composition is uniquely determined by the positions of the plus signs, which are n-1 in number, each with two choices.Therefore, all these different methods—recursive, inductive, generating functions, binary representation—lead to the conclusion that the number of compositions of n is 2^{n-1}.But let me check if there's any possible mistake here. For example, does this count the same composition multiple times? Wait, in ordered sums, each different arrangement is considered unique. So 1+2 and 2+1 are different, which is correctly accounted for in the binary choice model because the splits are at different positions.Yes, so in the binary string, each split corresponds to where you insert the plus, so different orders naturally result from different split positions. Therefore, the count is accurate.Another possible concern: Does this account for all possible compositions? For instance, a composition with multiple terms. Let's take n=4. The splits would be at positions 1, 2, 3. The binary string of length 3. For example, 101 would correspond to splits after position 1 and 3, leading to 1+2+1. Yes, that's a valid composition. Similarly, 000 would be 4, 001 is 1+3, 010 is 2+2, 011 is 1+1+2, 100 is 3+1, 101 is 1+2+1, 110 is 2+1+1, 111 is 1+1+1+1. That's 8 compositions, which is 2^3=8. All of them are unique and valid. So this method works.Therefore, it seems that all approaches confirm the number of compositions is 2^{n-1}. Hence, the proof is solid.Final AnswerThe number of ways to express a positive integer ( n ) as a sum of one or more positive integers is ( boxed{2^{n-1}} ).

✅:To show that any positive integer ( n ) can be expressed as a sum of one or more positive integers in ( 2^{n-1} ) ways, we consider the concept of compositions, where the order of terms matters. Here are the key steps:1. Binary Choice Model: - Represent ( n ) as a sequence of ( n ) ones: ( 1 + 1 + cdots + 1 ). - There are ( n - 1 ) gaps between these ones. - For each gap, we can choose to either place a plus sign (split) or not (combine). - Each gap has 2 choices (split or not), leading to ( 2^{n-1} ) possible compositions.2. Recursive Approach: - Let ( C(n) ) be the number of compositions of ( n ). - A composition of ( n ) can start with 1 followed by a composition of ( n - 1 ), or start with a number greater than 1. - This leads to the recursive formula: ( C(n) = 2 cdot C(n-1) ). - Base case: ( C(1) = 1 ). - By induction, ( C(n) = 2^{n-1} ).3. Generating Function: - The generating function for compositions is ( G(x) = frac{x}{1 - 2x} ). - Expanding this, the coefficient of ( x^n ) is ( 2^{n-1} ), confirming the number of compositions.4. Verification with Examples: - For ( n = 3 ), the compositions are ( 3, 1+2, 2+1, 1+1+1 ), which is ( 2^{3-1} = 4 ). - For ( n = 4 ), the compositions are 8, which is ( 2^{4-1} = 8 ).All methods confirm that the number of compositions of ( n ) is ( 2^{n-1} ).[boxed{2^{n-1}}]

🔑:To prove that any positive integer n can be expressed as the sum of positive integers in 2^{n-1} ways, we will use a combinatorial approach involving binary representations.1. Setup the problem: We need to represent an integer n as the sum of one or more positive integers. Let's consider how to construct these sums.2. Consider the integer n as a sequence of ones: Write n as a sequence of n ones, i.e., 1 + 1 + 1 + dots + 1 (total n times). [ underbrace{1, , 1, ldots, 1}_{n text{ ones}} ]3. Identify the gaps between the ones: Observe that there are (n-1) gaps between these ones where a "+" can be inserted to form different sums of integers. For example, for n=4: [ 1 underbrace{,+,}_{1} 1 underbrace{,+,}_{2} 1 underbrace{,+,}_{3} 1 ]4. Binary choices for gaps: Each gap represents a binary choice: either insert a "+" or do not insert anything. Here "+" denotes starting a new addition term, and the absence of "+" means the continuation of the current term. [ underbrace{(n-1)}_{text{gaps}} text{ binary choices} ]5. Number of ways to fill gaps: Each of the (n-1) gaps can have two possibilities: insert a "+" or do not insert anything. Hence, there are 2^{(n-1)} ways to fill these gaps. [ text{Total number of ways} = 2^{(n-1)} ]6. Example: For n=3: - Sequence of ones: 1 , 1 , 1 - Gaps: 1 gap between every pair of ones: - 1 , (gap) , 1 , (gap) , 1 - Possible sums: - Fill the first gap but not the second: 1+ 1+ 1 - Fill the second gap but not the first: 1 + 2 - Fill neither gap: 3 - Fill both gaps: 1 + 1 + 1 Checking this enumeration, we get 2^{(n-1)} = 2^{2} = 4 ways: [ begin{align*} 3 1+2 2+1 1+1+1 end{align*} ]7. Conclusion: Therefore, the number of different ways to express any positive integer n as the sum of positive integers is indeed 2^{n-1}.[boxed{2^{n-1}}]