🧠:Okay, let me try to figure out this problem step by step. It's about sequences and figuring out the maximum possible value of a sum involving both the sequence terms and some function b_n. Let me start by understanding the problem statement thoroughly.We have a sequence of integers a_1, a_2, a_3, ... where each term is at least as big as the previous one (non-decreasing) and the sequence is unbounded. That means the terms will eventually grow beyond any fixed number. The sequence b_n is defined such that b_n = m if a_m is the first term in the a-sequence that is equal to or exceeds n. So, for each n, b_n tells us the position in the a-sequence where n is first reached or surpassed.Given that a_{19} = 85, we need to find the maximum possible value of the sum S = a_1 + a_2 + ... + a_{19} + b_1 + b_2 + ... + b_{85}.Hmm. So, we have to maximize the sum of the first 19 terms of the a-sequence plus the sum of the first 85 terms of the b-sequence. The constraints are that the a-sequence is non-decreasing, unbounded, and a_{19} = 85. Since the a-sequence is unbounded, but a_{19} is 85, that means the terms after a_19 must eventually go to infinity, but we're only concerned up to a_{19} here. Wait, no—the problem is only about a_1 to a_{19} and b_1 to b_{85}. So maybe the rest of the a-sequence beyond a_{19} isn't directly relevant, except that the entire sequence is unbounded, so after a_{19} = 85, the terms must continue increasing to infinity. But since we need the first occurrence where a term exceeds n, for n up to 85, those first occurrences must all happen by a_{19} at the latest. Because a_{19} is 85, which is the first term to reach 85. But for n less than 85, the first a_m >= n could be at m <= 19. So, the b_n terms for n from 1 to 85 are determined by the positions in the a-sequence where each n is first reached or exceeded.We need to maximize the sum S. To do this, we need to consider how the a-sequence and b-sequence are related. Since b_n is determined by the a-sequence, we can model the relationship between the two sequences.Let me think. For each n from 1 to 85, b_n is the smallest m such that a_m >= n. Therefore, the a-sequence defines a partition of the integers from 1 to 85 into groups where each group corresponds to the n's for which b_n = m. That is, for each m, the numbers n where the first a_k >= n occurs at k = m. So, the values of n assigned to m are those n where a_{m-1} < n <= a_m (assuming a_0 = 0 for simplicity). Therefore, the number of n's assigned to m is a_m - a_{m-1}, since each integer from a_{m-1} + 1 up to a_m will have b_n = m.Wait, but a_m is the first term >= n. So if a_{m} is the first term >= n, then for all k < m, a_k < n. Therefore, the range of n's for which b_n = m is (a_{m-1}, a_m]. But since a_{m} is an integer, it's actually n's from a_{m-1} + 1 up to a_m. Therefore, the number of n's where b_n = m is a_m - a_{m-1}. Therefore, the sum of b_n from n=1 to 85 is equal to the sum over m of m*(a_m - a_{m-1}), but only for the m's such that a_{m} contributes to n <=85. However, since a_19 = 85, and the sequence is non-decreasing, the maximum n we have is 85, which is achieved at m=19. Therefore, all n from 1 to 85 must have b_n <= 19, because a_{19}=85 is the first term to reach 85. But maybe some n's beyond a certain point might have b_n =19 as well.Wait, for example, if a_{18} is, say, 80, then n from 81 to 85 would have b_n=19. Similarly, n=1 would have b_n=1 if a_1 >=1, but since a_1 is at least 1 (because the sequence starts at a_1 and is non-decreasing, and it's a sequence of integers). Wait, actually, the problem states 0 < a_1 <= a_2 <= ..., so a_1 is at least 1. Therefore, n=1 would have b_n=1 because a_1 >=1. Similarly, if a_1 is 1, then n=1 is covered by a_1, but n=2 would be covered by the first a_m >=2. If a_2 is 2, then b_2=2, etc.So, in general, for each m from 1 to 19, the numbers n in the interval (a_{m-1}, a_m] will have b_n = m. Here, we can take a_0 = 0 for convenience. Therefore, the sum over b_n from n=1 to 85 is sum_{m=1}^{19} m*(a_m - a_{m-1}), but we have to be careful because a_19 =85, so the upper limit is 85. Therefore, the total sum for b_n is sum_{m=1}^{19} m*(a_m - a_{m-1}).But let's check this. If a_1 =1, then n=1 is covered by m=1. If a_2=2, then n=2 is covered by m=2, etc. If a_m increases by 1 each time, then each m covers exactly one n. But if a_m increases by more than 1, then m covers multiple n's. For example, if a_1=3, then m=1 covers n=1,2,3. Then a_2 could be 5, covering n=4,5. Then m=2 contributes 2*(5-3)=4, which is n=4 and 5. So the formula seems correct.Therefore, the sum b_1 + ... + b_{85} = sum_{m=1}^{19} m*(a_m - a_{m-1}), where a_0=0 and a_19=85.Then the total sum S is sum_{m=1}^{19} a_m + sum_{m=1}^{19} m*(a_m - a_{m-1}).Let me write that out:S = (a_1 + a_2 + ... + a_{19}) + [1*(a_1 - a_0) + 2*(a_2 - a_1) + ... + 19*(a_{19} - a_{18})]Since a_0=0, the second sum becomes:1*a_1 + 2*(a_2 - a_1) + 3*(a_3 - a_2) + ... + 19*(a_{19} - a_{18})If we expand this, we can see telescoping:=1*a_1 + 2*a_2 - 2*a_1 + 3*a_3 - 3*a_2 + ... + 19*a_{19} - 19*a_{18}Combine like terms:For a_1: 1*a_1 - 2*a_1 = -a_1For a_2: 2*a_2 - 3*a_2 = -a_2...For a_{18}: 18*a_{18} -19*a_{18} = -a_{18}Then the last term is 19*a_{19}So the entire second sum simplifies to:19*a_{19} - (a_1 + a_2 + ... + a_{18})Therefore, the total sum S is:(a_1 + ... + a_{19}) + [19*a_{19} - (a_1 + ... + a_{18})] = a_{19} + 19*a_{19} = 20*a_{19}But wait, that can't be right. Let me check the algebra again.Original S = sum_{m=1}^{19} a_m + sum_{m=1}^{19} m*(a_m - a_{m-1})Let me compute the second sum first:sum_{m=1}^{19} m*(a_m - a_{m-1}) = sum_{m=1}^{19} (m*a_m - m*a_{m-1})This can be rewritten as sum_{m=1}^{19} m*a_m - sum_{m=1}^{19} m*a_{m-1}Let's index shift the second sum. Let m' = m -1. Then when m=1, m'=0. When m=19, m'=18. So the second sum becomes sum_{m'=0}^{18} (m' +1)*a_{m'}But m' starts at 0, which is a_0=0. So:sum_{m=1}^{19} m*a_{m-1} = sum_{m'=0}^{18} (m' +1)*a_{m'} = sum_{k=0}^{18} (k +1)*a_k where k = m'Therefore, the second sum is sum_{k=0}^{18} (k +1)*a_kSo the original expression for the second sum is:sum_{m=1}^{19} m*a_m - sum_{k=0}^{18} (k +1)*a_k= sum_{m=1}^{19} m*a_m - sum_{k=0}^{18} (k +1)*a_k= [sum_{m=1}^{18} m*a_m + 19*a_{19}] - [sum_{k=0}^{18} (k +1)*a_k]= 19*a_{19} + sum_{m=1}^{18} m*a_m - sum_{k=0}^{18} (k +1)*a_kBut note that in the second sum, when k=0, we have 1*a_0 = 0, since a_0=0. Then for k=1 to 18, we have (k+1)*a_k. So:sum_{k=0}^{18} (k +1)*a_k = 0 + sum_{k=1}^{18} (k +1)*a_kTherefore:= 19*a_{19} + sum_{m=1}^{18} m*a_m - sum_{k=1}^{18} (k +1)*a_k= 19*a_{19} + sum_{m=1}^{18} [m*a_m - (m +1)*a_m]= 19*a_{19} + sum_{m=1}^{18} (-a_m)= 19*a_{19} - sum_{m=1}^{18} a_mTherefore, the second sum (sum of b_n) is 19*a_{19} - sum_{m=1}^{18} a_m.Then the total sum S is:sum_{m=1}^{19} a_m + [19*a_{19} - sum_{m=1}^{18} a_m] = (sum_{m=1}^{19} a_m - sum_{m=1}^{18} a_m) + 19*a_{19} = a_{19} + 19*a_{19} = 20*a_{19}But wait, that's a huge simplification! So regardless of the values of a_1 to a_{18}, the total sum S is 20 times a_{19}, which is given as 85. Therefore, S = 20*85 = 1700. But that can't be right because the problem is asking for the maximum possible value. If the sum is always 20*85 regardless of the sequence, then the maximum is 1700. But the problem says "maximum possible value", implying that the sum can vary depending on the sequence. Therefore, there must be a mistake in my reasoning.Wait a second, this conclusion suggests that the sum S is fixed once a_{19}=85 is given, which contradicts the problem statement asking for the maximum possible value. So where did I go wrong?Let me re-examine the steps. When I combined the sums, I assumed that the sum over b_n is 19*a_{19} - sum_{m=1}^{18} a_m. Then adding the sum over a_m (from 1 to 19) gives a_{19} + 19*a_{19} = 20*a_{19}. But this seems to cancel out the other terms. However, perhaps the error is in the way the sum over b_n was calculated.Wait, let's take a concrete example. Suppose the sequence a_1, a_2, ..., a_{19} is such that a_1=1, a_2=2, ..., a_{19}=85. Then each a_m = m. But since the sequence is non-decreasing and unbounded (though in this case, it's strictly increasing but finite here). Wait, but a_{19}=85, so after that, it continues to infinity. Anyway, for this sequence, each b_n would be n, because the first term a_m >= n is a_n, since a_n =n. So the sum of b_n from 1 to85 would be sum_{n=1}^{85} n = (85)(86)/2 = 3655. Then the sum of a_1 to a_{19} is sum_{m=1}^{19} m = (19)(20)/2 = 190. So total S = 190 + 3655 = 3845. But according to the previous conclusion, S should be 20*85=1700, which is way lower. Therefore, my previous reasoning is incorrect. There's a mistake here.Therefore, the mistake must be in the algebraic manipulation. Let's go back.Original:sum_{m=1}^{19} m*(a_m - a_{m-1}) = sum_{m=1}^{19} m*a_m - sum_{m=1}^{19} m*a_{m-1}Then, changing the index for the second sum:sum_{m=1}^{19} m*a_{m-1} = sum_{k=0}^{18} (k +1)*a_kThus,sum_{m=1}^{19} m*(a_m -a_{m-1}) = sum_{m=1}^{19} m*a_m - sum_{k=0}^{18}(k +1)*a_k= sum_{m=1}^{19} m*a_m - sum_{k=0}^{18}k*a_k - sum_{k=0}^{18}a_k= [sum_{m=1}^{19} m*a_m - sum_{k=1}^{18}k*a_k] - sum_{k=0}^{18}a_k= [19*a_{19} + sum_{m=1}^{18}m*a_m - sum_{k=1}^{18}k*a_k] - sum_{k=0}^{18}a_k= 19*a_{19} - sum_{k=0}^{18}a_kWait, since sum_{m=1}^{18}m*a_m - sum_{k=1}^{18}k*a_k =0, they cancel. Then we have 19*a_{19} - sum_{k=0}^{18}a_k. But a_0=0, so sum_{k=0}^{18}a_k = sum_{k=1}^{18}a_k.Therefore, the sum over b_n is 19*a_{19} - sum_{k=1}^{18}a_k.Then the total sum S is sum_{k=1}^{19}a_k + 19*a_{19} - sum_{k=1}^{18}a_k = a_{19} +19*a_{19} =20*a_{19}. But this contradicts the concrete example where S was 3845, not 1700.Wait, in my concrete example, a_{19}=85, so according to this formula, S should be 20*85=1700, but when I computed it, S was 3845. Therefore, this shows the formula is incorrect.Therefore, my mistake is in the algebra. Let me check again.Let's re-express the sum over b_n:sum_{n=1}^{85} b_n = sum_{m=1}^{19} m*(a_m -a_{m-1})But in the concrete example where a_m =m for m=1 to19, then a_m -a_{m-1}=1 for each m. Therefore, sum_{m=1}^{19} m*1 = sum_{m=1}^{19} m = (19)(20)/2=190. Then the sum over a_m is sum_{m=1}^{19} m=190. So total S=190+190=380, but wait, earlier I thought the sum over b_n was sum_{n=1}^{85} n=3655. There's a discrepancy here. Wait, no—if a_m=m, then b_n is the smallest m with a_m >=n. So for n=1, b_1=1. For n=2, b_2=2. ... For n=19, b_{19}=19. Then for n=20 to85, since a_{19}=85, the first a_m >=20 is a_{20}=20, but wait, in our problem, a_{19}=85, so actually, if the sequence jumps from a_1=1, a_2=2,...,a_{18}=18, a_{19}=85, then for n=19 to85, b_n=19. Therefore, sum_{n=1}^{85} b_n= sum_{n=1}^{18}n + sum_{n=19}^{85}19 = (18*19)/2 + (85-18)*19= 171 + 67*19=171 +1273=1444. Then sum of a_1 to a_{19}=sum_{1}^{18}m +85= (18*19)/2 +85=171 +85=256. So total S=1444 +256=1700, which matches 20*a_{19}=20*85=1700. But in my previous wrong example, I assumed a_m=m up to 19, but that contradicts a_{19}=85. So my mistake was in the initial example where I incorrectly assumed a_{19}=19, which contradicts the given a_{19}=85. Therefore, actually, in the concrete example where a_1=1, a_2=2,...,a_{18}=18, a_{19}=85, the sum over b_n is 1444 and the sum over a_m is256, totaling1700. Therefore, the formula S=20*a_{19} is correct.But then the problem states "what is the maximum possible value of...", implying that different sequences a_1,...,a_{19}=85 can give different S, and we need the maximum. But according to the formula S=20*85=1700, which is fixed. So this suggests that regardless of the sequence, as long as a_{19}=85, the sum is always1700. Therefore, the answer is1700. But this seems counterintuitive because if we arrange the a-sequence differently, perhaps we can get a higher sum.Wait, let's test another example. Suppose a_1=85, and a_2=85, ..., a_{19}=85. So all terms from a_1 to a_{19} are85. Then b_n=1 for all n from1 to85, because a_1=85 is the first term >=n for all n<=85. Then sum of b_n=85*1=85. Sum of a_1 toa_{19}=19*85=1615. So total S=1615+85=1700. Same result.Another example: Let a_1=1, a_2=85, and a_3 toa_{19}=85. Then for n=1, b_1=1. For n=2 to85, b_n=2. So sum of b_n=1 +84*2=1 +168=169. Sum of a_m=1 +85*18=1 +1530=1531. Total S=1531 +169=1700. Again same result.Therefore, regardless of how we arrange the a-sequence (as long as it's non-decreasing and a_{19}=85), the total sum S is always20*85=1700. Therefore, the maximum possible value is1700, which is fixed. But the problem says "maximum possible value", implying that it could vary. But according to these examples, it's always1700. Therefore, perhaps the answer is1700.But let's check with a different a-sequence. Suppose we make a_1 as large as possible. For example, suppose a_1=85, and the rest also85. As before, sum a=19*85=1615, sum b=85*1=85, total1700. If we make a_1=84, a_2=85, and the rest85. Then sum a=84 +18*85=84 +1530=1614. For b_n: n=1-84 have b_n=1 (since a_1=84), n=85 has b_n=2. So sum b=84*1 +1*2=86. Total S=1614+86=1700. Still the same. If we make a_1=83, a_2=85, rest85. Then sum a=83 +1*85 +17*85=83 +85 +1445=1613. Sum b: n=1-83: b_n=1, n=84-85: b_n=2. So sum b=83*1 +2*2=83 +4=87. Total S=1613+87=1700. Same again.Therefore, no matter how we arrange the a-sequence (as long as non-decreasing and a_{19}=85), the sum S remains1700. Therefore, the maximum possible value is1700. Hence, the answer is boxed{1700}.But wait, the problem says "0 < a_1 ≤ a_2 ≤ ..." and "an unbounded sequence of integers". But we're only given a_19=85. However, the sequence must be unbounded, so beyond a_{19}, the terms must go to infinity. But since we are only summing up to a_{19} and b_1 to b_{85}, which are determined by the first 19 terms, the rest of the sequence doesn't affect the sum. Therefore, regardless of how we set a_1 to a_{19}, as long as they are non-decreasing and a_{19}=85, the sum S=20*85=1700. Therefore, the maximum possible value is1700.But why does the problem mention that the sequence is unbounded? Because if the sequence were bounded, then after some term, all terms would be equal, but since it's given as unbounded, the terms after a_{19} must increase without bound. However, since our sum only involves up to a_{19} and b_1 to b_{85}, which are covered by a_1 to a_{19}, the unboundedness beyond a_{19} doesn't affect the sum. Therefore, the sum S is fixed at1700, and hence the maximum is1700.Therefore, the answer is boxed{1700}.

✅:Given an unbounded, non-decreasing sequence of integers (0 < a_1 le a_2 le ldots) where (a_{19} = 85), we need to find the maximum possible value of the sum (a_1 + a_2 + ldots + a_{19} + b_1 + b_2 + ldots + b_{85}). Here, (b_n = m) if (a_m) is the first term in the sequence to equal or exceed (n).1. Understanding (b_n): For each (n), (b_n) is the smallest (m) such that (a_m ge n). This means the values of (n) for which (b_n = m) are those in the interval ((a_{m-1}, a_m]).2. Summing (b_n): The sum (b_1 + b_2 + ldots + b_{85}) can be expressed as (sum_{m=1}^{19} m cdot (a_m - a_{m-1})), where (a_0 = 0). This accounts for the number of (n) values each (m) covers.3. Simplifying the Sum: - The sum of (b_n) terms is (sum_{m=1}^{19} m cdot (a_m - a_{m-1})). - Expanding and telescoping this sum results in (19 cdot a_{19} - sum_{m=1}^{18} a_m).4. Combining Sums: - The total sum (S) is the sum of (a_1) to (a_{19}) plus the sum of (b_1) to (b_{85}). - This simplifies to (sum_{m=1}^{19} a_m + left(19 cdot a_{19} - sum_{m=1}^{18} a_mright)). - Further simplification gives (a_{19} + 19 cdot a_{19} = 20 cdot a_{19}).5. Conclusion: Since (a_{19} = 85), the total sum (S) is (20 cdot 85 = 1700). This result is independent of the specific values of (a_1) to (a_{18}) as long as the sequence is non-decreasing and (a_{19} = 85).Thus, the maximum possible value of the given sum is (boxed{1700}).

🔑:Let's solve the problem step-by-step to find the maximum possible value of the given sum:1. Define the sequence and sum: - We have (0 < a_1 leq a_2 leq a_3 leq ldots) as an unbounded sequence of integers. - Define (b_n = m) if (a_m) is the first member of the sequence to equal or exceed (n). - We need to find the maximum possible value of (a_1 + a_2 + ldots + a_{19} + b_1 + b_2 + ldots + b_{85}).2. Analyze the given constraints: - Given that (a_{19} = 85), we know (a_{19} geq 85). - Since (a_i) is non-decreasing, all previous terms must also be at most 85 with (a_1 leq a_2 leq ldots leq a_{19} = 85).3. Initial assumption: - We start with the assumption (a_i = 85) for all (i) from 1 to 19. This assumption helps us check the maximum possible sum. In this case, the sum of (a_1 + a_2 + ldots + a_{19} = 19 times 85).4. Calculation of (b_n)s: - Since all (a_i = 85), for (n leq 85), (b_n = 1) because the first sequence element (all equal to 85) is reached at (a_1).5. Sum of (b_n)s: - So, the sum (b_1 + b_2 + ldots + b_{85} = 1 + 1 + ldots + 1) (85 ones) equals (85).6. Total initial sum: [ a_1 + a_2 + ldots + a_{19} + b_1 + b_2 + ldots + b_{85} = 19 times 85 + 85 times 1 = 19 times 85 + 85 = 20 times 85 = 1700 ]7. General case and adjustment: - Suppose we increase one (a_i) by 1, then this changes (b_j) for some (j). - Specifically, if we increase (a_i) from (k) to (k+1): - (b_k) becomes (leq i) (earlier it was (i+1) for reaching new (k+1)). - (b_{k+1}) changes from (i+1) to (i), leaving all other (b_j) unchanged. - This does not change the sum of the sequence's terms significantly. - We maintain general increases in (a_i) until all become 85, which confirms the maximum sum remains unchanged because incremental increases balance out terms with strictly adhering constraints.8. Conclusion: - Thus, after converting any initial unbounded sequence under given constraints to equal terms, the maximum possible value of the sum (a_1 + a_2 + ldots + a_{19} + b_1 + b_2 + ldots + b_{85}) remains 1700.[boxed{1700}]

🧠:Okay, so I need to find the roots of the quartic equation (x^4 - 2a x^2 + b^2 = 0) and express them as the sum of two simple square roots. Hmm, quartic equations can be tricky, but maybe this one can be factored or simplified somehow. Let me start by recalling that sometimes quartic equations can be reduced to quadratic form. Since the equation is (x^4 - 2a x^2 + b^2 = 0), it looks like a quadratic in terms of (x^2). Let me try substituting (y = x^2), which would make the equation (y^2 - 2a y + b^2 = 0). Solving this quadratic equation for (y), I can use the quadratic formula: (y = frac{2a pm sqrt{(2a)^2 - 4 cdot 1 cdot b^2}}{2}). Simplifying under the square root: (4a^2 - 4b^2 = 4(a^2 - b^2)), so the square root becomes (2sqrt{a^2 - b^2}). Therefore, (y = frac{2a pm 2sqrt{a^2 - b^2}}{2} = a pm sqrt{a^2 - b^2}). So, (x^2 = a pm sqrt{a^2 - b^2}). Now, to find (x), I need to take the square roots of these expressions. Therefore, the roots are (x = pm sqrt{a + sqrt{a^2 - b^2}}) and (x = pm sqrt{a - sqrt{a^2 - b^2}}). But the problem asks to express the roots as the sum of two simple square roots. Hmm, the current form is a square root of a sum involving another square root. So maybe there's a way to write (sqrt{a pm sqrt{a^2 - b^2}}) as the sum of two square roots. Let me think about that.Suppose I want to write (sqrt{a + sqrt{a^2 - b^2}}) as (sqrt{c} + sqrt{d}). Let me square both sides to see what conditions c and d must satisfy. [(sqrt{c} + sqrt{d})^2 = c + d + 2sqrt{cd} = a + sqrt{a^2 - b^2}]Comparing the terms, the non-radical part must be (c + d = a), and the radical part must be (2sqrt{cd} = sqrt{a^2 - b^2}). Let me square the second equation to eliminate the square root: (4cd = a^2 - b^2). So now we have a system of equations:1. (c + d = a)2. (4cd = a^2 - b^2)I can solve this system for c and d. Let me denote (c + d = a) and (cd = frac{a^2 - b^2}{4}). These are the sum and product of two numbers, which means c and d are roots of the quadratic equation (t^2 - (a) t + frac{a^2 - b^2}{4} = 0). Solving this quadratic equation: discriminant is (a^2 - 4 cdot 1 cdot frac{a^2 - b^2}{4} = a^2 - (a^2 - b^2) = b^2). Therefore, the roots are (t = frac{a pm sqrt{b^2}}{2} = frac{a pm b}{2}). Therefore, (c = frac{a + b}{2}) and (d = frac{a - b}{2}), or vice versa. So, substituting back, we get:[sqrt{a + sqrt{a^2 - b^2}} = sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}}]Similarly, let's check the other term (sqrt{a - sqrt{a^2 - b^2}}). Suppose we want to write this as (sqrt{e} - sqrt{f}) (since if we use a minus sign, squaring would give (e + f - 2sqrt{ef})), let's see.Let’s assume (sqrt{a - sqrt{a^2 - b^2}} = sqrt{e} - sqrt{f}). Squaring both sides:[(sqrt{e} - sqrt{f})^2 = e + f - 2sqrt{ef} = a - sqrt{a^2 - b^2}]So similarly, we get:1. (e + f = a)2. (2sqrt{ef} = sqrt{a^2 - b^2})Which is the same system as before. Therefore, (e = frac{a + b}{2}) and (f = frac{a - b}{2}), so:[sqrt{a - sqrt{a^2 - b^2}} = sqrt{frac{a + b}{2}} - sqrt{frac{a - b}{2}}]Therefore, combining both results, the roots of the original equation can be written as:[x = pm left( sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}} right) quad text{and} quad x = pm left( sqrt{frac{a + b}{2}} - sqrt{frac{a - b}{2}} right)]Simplifying further, notice that (sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}}) and (sqrt{frac{a + b}{2}} - sqrt{frac{a - b}{2}}) can be written as the sum and difference of two square roots, which is exactly what the problem is asking for. Therefore, the roots are expressed as the sum (and difference) of two simple square roots.To verify, let me substitute these expressions back into the original equation. Let's take (x = sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}}). Compute (x^4 - 2a x^2 + b^2) and check if it equals zero.First, compute (x^2):[x^2 = left( sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}} right)^2 = frac{a + b}{2} + frac{a - b}{2} + 2 sqrt{frac{a + b}{2} cdot frac{a - b}{2}} = frac{2a}{2} + 2 sqrt{frac{a^2 - b^2}{4}} = a + sqrt{a^2 - b^2}]Then (x^4 = (x^2)^2 = (a + sqrt{a^2 - b^2})^2 = a^2 + 2asqrt{a^2 - b^2} + (a^2 - b^2) = 2a^2 - b^2 + 2asqrt{a^2 - b^2})Now compute (x^4 - 2a x^2 + b^2):[(2a^2 - b^2 + 2asqrt{a^2 - b^2}) - 2a(a + sqrt{a^2 - b^2}) + b^2 = 2a^2 - b^2 + 2asqrt{a^2 - b^2} - 2a^2 - 2asqrt{a^2 - b^2} + b^2 = 0]So it checks out. Similarly, for the other roots, since the equation has even powers, the negative counterparts will also satisfy the equation. Therefore, the solution is correct.

✅:The roots of the equation (x^4 - 2a x^2 + b^2 = 0) can be expressed as the sum and difference of two square roots through the following steps:1. Substitute (y = x^2) to convert the quartic equation into a quadratic: [ y^2 - 2a y + b^2 = 0 ]2. Solve the quadratic using the quadratic formula: [ y = a pm sqrt{a^2 - b^2} ]3. Express (x) by taking square roots: [ x = pm sqrt{a pm sqrt{a^2 - b^2}} ]4. Decompose the nested square roots into sums/differences of simpler square roots. For (sqrt{a + sqrt{a^2 - b^2}}) and (sqrt{a - sqrt{a^2 - b^2}}): [ sqrt{a + sqrt{a^2 - b^2}} = sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}} ] [ sqrt{a - sqrt{a^2 - b^2}} = sqrt{frac{a + b}{2}} - sqrt{frac{a - b}{2}} ]Final Roots:[x = pm left( sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}} right) quad text{and} quad x = pm left( sqrt{frac{a + b}{2}} - sqrt{frac{a - b}{2}} right)]Verification: Substituting these expressions back into the original equation confirms they satisfy (x^4 - 2a x^2 + b^2 = 0), ensuring correctness.

🔑:1. Given the equation ( x^{4} - 2ax^{2} + b^{2} = 0 ), we can perform a substitution to simplify this quartic equation. Let ( y = x^2 ). Thus, the equation becomes: [ y^2 - 2ay + b^2 = 0 ] This is a quadratic equation in terms of ( y ).2. To solve the quadratic equation for ( y ), we use the quadratic formula: [ y = frac{2a pm sqrt{(2a)^2 - 4 cdot 1 cdot b^2}}{2 cdot 1} ] Simplifying the expression inside the square root: [ y = frac{2a pm sqrt{4a^2 - 4b^2}}{2} = frac{2a pm 2sqrt{a^2 - b^2}}{2} = a pm sqrt{a^2 - b^2} ]3. Thus, the values for ( y ) are: [ y_{1} = a + sqrt{a^2 - b^2} quad text{and} quad y_{2} = a - sqrt{a^2 - b^2} ] Since ( y = x^2 ), we have: [ x^2 = a + sqrt{a^2 - b^2} quad text{or} quad x^2 = a - sqrt{a^2 - b^2} ]4. Solving for ( x ) in each of these equations, we get four possible values for ( x ): [ x = pm sqrt{a + sqrt{a^2 - b^2}} quad text{or} quad x = pm sqrt{a - sqrt{a^2 - b^2}} ]5. To express one of the solutions as a sum of two simpler square roots, let's consider: [ sqrt{a + sqrt{a^2 - b^2}} = sqrt{u} + sqrt{v} ]6. Squaring both sides, we obtain: [ a + sqrt{a^2 - b^2} = u + v + 2sqrt{uv} ]7. Equating the linear and square root terms, we get: [ u + v = a quad text{and} quad 2 sqrt{uv} = sqrt{a^2 - b^2} Rightarrow uv = frac{a^2 - b^2}{4} ]8. ( u ) and ( v ) are roots of the quadratic equation: [ z^2 - az + frac{a^2 - b^2}{4} = 0 ] Solving this quadratic equation for ( z ), we have: [ z = frac{a pm sqrt{a^2 - (a^2 - b^2)}}{2} = frac{a pm b}{2} ] Hence: [ u = frac{a + b}{2} quad text{and} quad v = frac{a - b}{2} ]9. Substituting back, we get: [ sqrt{a + sqrt{a^2 - b^2}} = sqrt{frac{a + b}{2}} + sqrt{frac{a - b}{2}} ]10. Similarly: [ sqrt{a - sqrt{a^2 - b^2}} = sqrt{frac{a + b}{2}} - sqrt{frac{a - b}{2}} ]11. Combining these results, the roots of the original equation can be expressed as: [ x = pm left( sqrt{frac{a + b}{2}} pm sqrt{frac{a - b}{2}} right) ]Conclusion:[boxed{x = pm left( sqrt{frac{a + b}{2}} pm sqrt{frac{a - b}{2}} right)}]

🧠:Okay, so I need to prove that for an n-th degree polynomial with all real roots, if the derivative at some point x₀ is not zero, then there's a root x₁ such that the distance between x₀ and x₁ is bounded by n times the absolute value of f(x₀) over f’(x₀). Hmm, let me think step by step.First, since all roots of f(x) are real, the polynomial can be written as f(x) = a(x - r₁)(x - r₂)...(x - rₙ), where a is the leading coefficient and each r_i is a real root. Now, maybe I should recall some properties of polynomials with all real roots. They are also called hyperbolic polynomials. One key property is that the derivative of such a polynomial also has all real roots (by Rolle's theorem), right? Because between any two roots of f(x), there's at least one root of f’(x). So, the derivatives are also hyperbolic.But how does this help with the given inequality? Let me see. The problem is about relating the distance from x₀ to a root in terms of f(x₀) and f’(x₀). This reminds me of Newton's method, where you approximate a root using the tangent line. The formula x₁ = x₀ - f(x₀)/f’(x₀) is the standard iteration. The bound given here is similar but scaled by n. Maybe there's a connection?Wait, the inequality is |x₀ - x₁| ≤ n |f(x₀)/f’(x₀)|. So it's saying that there's a root within n times the Newton step distance from x₀. Since n is the degree, perhaps the multiple accounts for the number of roots? If all roots are real, maybe the polynomial can't be too "flat" away from the roots, so the derivative can't be too small compared to the function value?Let me try to write f(x) in terms of its roots. Let’s say f(x) = a∏_{k=1}^n (x - r_k). Then the derivative f’(x) = a∑_{i=1}^n ∏_{k≠i} (x - r_k). So, f’(x) is the sum over all the products omitting one root each. Alternatively, we can write the logarithmic derivative: f’(x)/f(x) = ∑_{i=1}^n 1/(x - r_i). That seems useful. Because if we take f’(x)/f(x) = ∑_{i=1}^n 1/(x - r_i), then at a point x₀, this is ∑ 1/(x₀ - r_i). So, f’(x₀)/f(x₀) = ∑_{i=1}^n 1/(x₀ - r_i). Let me denote S = ∑_{i=1}^n 1/(x₀ - r_i). Then f’(x₀)/f(x₀) = S, which implies that S = f’(x₀)/f(x₀). Therefore, 1/S = f(x₀)/f’(x₀). But S is the sum of reciprocals of (x₀ - r_i). The problem wants to show that there exists some root r_i (let's call it x₁) such that |x₀ - x₁| ≤ n |f(x₀)/f’(x₀)|. Which is equivalent to |x₀ - x₁| ≤ n / |S|, since |f(x₀)/f’(x₀)| = 1/|S|. So, we need to show that the minimum distance from x₀ to any root is at most n divided by |S|. Alternatively, the minimum |x₀ - r_i| over all i is ≤ n / |S|. Let me think about how to relate the sum S to the minimum term. If S is the sum of 1/(x₀ - r_i), then each term is 1/(x₀ - r_i). Let's suppose for simplicity that x₀ is not equal to any root, which is given because f’(x₀) ≠ 0 (if x₀ were a root, then f(x₀) would be 0, but then f’(x₀) ≠ 0 would mean it's a simple root. Wait, but the problem doesn't state that x₀ is not a root. Wait, if x₀ is a root, then f(x₀) = 0, and the right-hand side is 0. But the inequality would say |x₀ - x₁| ≤ 0, so x₁ must be x₀, which is a root. But the problem states f’(x₀) ≠ 0. If x₀ is a root, then f’(x₀) ≠ 0 implies that x₀ is a simple root, but f(x₀) = 0. Then |f(x₀)/f’(x₀)| is 0, so the inequality requires |x₀ - x₁| ≤ 0, which would force x₁ = x₀. So in that case, x₀ itself is a root, so the statement holds. But the problem probably assumes x₀ is not a root because if x₀ is a root, then the inequality is trivial. So maybe we can assume x₀ is not a root, but the problem doesn't specify. Anyway, let's proceed.Assuming x₀ is not a root, so f(x₀) ≠ 0, and f’(x₀) ≠ 0. Then S = ∑ 1/(x₀ - r_i). Let me denote d_i = |x₀ - r_i|. Let’s consider the absolute value of S: |S| = |∑ 1/(x₀ - r_i)|. Let's suppose that all the terms in the sum have the same sign. For instance, if x₀ is to the right of all roots, then each (x₀ - r_i) is positive, so all terms are positive, and S is the sum of positive terms. Similarly, if x₀ is to the left of all roots, then each (x₀ - r_i) is negative, so each term is negative, and S is the sum of negative terms, so |S| is the sum of |1/(x₀ - r_i)|. However, if x₀ is between some roots, then some terms are positive and some are negative. Hmm, complicating things.But perhaps we can use the triangle inequality. |S| = |∑ 1/(x₀ - r_i)| ≤ ∑ |1/(x₀ - r_i)|. But that might not help directly. Wait, the problem is to show that the minimum d_i is ≤ n / |S|. So, if we can show that the minimum |x₀ - r_i| ≤ n / |S|, then we are done.Let’s suppose that the minimum distance is m = min_i |x₀ - r_i|. Then for each term in the sum |1/(x₀ - r_i)| ≥ 1/m, because |x₀ - r_i| ≤ m would imply 1/|x₀ - r_i| ≥ 1/m. Wait, no. If m is the minimum distance, then for all i, |x₀ - r_i| ≥ m, so 1/|x₀ - r_i| ≤ 1/m. Therefore, each term in the sum ∑ |1/(x₀ - r_i)| ≤ n/m. But then |S| ≤ ∑ |1/(x₀ - r_i)| ≤ n/m. Therefore, |S| ≤ n/m ⇒ m ≤ n/|S|. Which is exactly what we need! Because m is the minimum distance from x₀ to a root, so |x₀ - x₁| = m ≤ n/|S| = n |f(x₀)/f’(x₀)|. That seems to work!Wait, let me verify that again. If m is the minimum of |x₀ - r_i|, then each |x₀ - r_i| ≥ m. Therefore, 1/|x₀ - r_i| ≤ 1/m. So the sum over i of |1/(x₀ - r_i)| ≤ n/m. But |S| = |∑ 1/(x₀ - r_i)| ≤ ∑ |1/(x₀ - r_i)| ≤ n/m. So |S| ≤ n/m ⇒ m ≤ n / |S|. But m is the minimum distance, so that's exactly |x₀ - x₁| ≤ n |f(x₀)/f’(x₀)|, since |f(x₀)/f’(x₀)| = 1/|S|. Therefore, this inequality holds. Therefore, the key idea is using the reciprocal sum of distances from x₀ to each root and relating the minimum distance to the sum via inequalities. Since each term in the sum is bounded by 1/m, the total sum is at most n/m, leading to m ≤ n / |S|. But wait, does this hold even if the terms in S have different signs? For example, if some terms are positive and some are negative, could |S| be smaller than the sum of the absolute values? Yes, but in that case, the inequality |S| ≤ ∑ |1/(x₀ - r_i)| still holds by the triangle inequality. So even if there is cancellation in the sum S, the absolute value |S| is at most the sum of the absolute values, which is ≤ n/m. Therefore, m ≤ n / |S| is still valid. Therefore, regardless of the signs of the terms in S, the inequality holds. Therefore, the minimum distance is indeed bounded by n / |S|, which is n |f(x₀)/f’(x₀)|. Therefore, the proof is complete. Let me recap:Given f(x) with all real roots, write f’(x)/f(x) = ∑ 1/(x - r_i). Let S = ∑ 1/(x₀ - r_i), so |S| = |f’(x₀)/f(x₀)|. Let m = min |x₀ - r_i|. Then each |1/(x₀ - r_i)| ≤ 1/m, so |S| ≤ ∑ |1/(x₀ - r_i)| ≤ n/m. Hence, m ≤ n / |S| = n |f(x₀)/f’(x₀)|. Therefore, there exists a root x₁ with |x₀ - x₁| ≤ n |f(x₀)/f’(x₀)|. QED.I should check if there are any hidden assumptions or steps that might need more justification. For instance, the use of the logarithmic derivative requires that f(x₀) ≠ 0, which is true because if f(x₀) = 0, then x₀ is a root, and as discussed earlier, the inequality would require x₁ = x₀, which is valid. But in the case f(x₀) ≠ 0, the logarithmic derivative is valid. So, the steps hold whether or not x₀ is a root, but if x₀ is a root, the inequality is trivial. Hence, the argument is solid.

✅:To prove that for an ( n )-th degree polynomial ( f(x) ) with all real roots, if ( f'(x_0) neq 0 ) at some ( x_0 ), there exists a root ( x_1 ) such that ( |x_0 - x_1| leq n left| frac{f(x_0)}{f'(x_0)} right| ), follow these steps:1. Express ( f(x) ) in Factored Form: Let ( f(x) = a prod_{k=1}^n (x - r_k) ), where ( r_1, r_2, dots, r_n ) are the real roots.2. Compute the Logarithmic Derivative: The derivative of ( f(x) ) is ( f'(x) = a sum_{i=1}^n prod_{k neq i} (x - r_k) ). The logarithmic derivative gives: [ frac{f'(x)}{f(x)} = sum_{i=1}^n frac{1}{x - r_i}. ] At ( x = x_0 ), let ( S = sum_{i=1}^n frac{1}{x_0 - r_i} ), so ( frac{f'(x_0)}{f(x_0)} = S ).3. Relate ( S ) to the Minimum Distance: Let ( m = min_{1 leq i leq n} |x_0 - r_i| ). For each term in ( S ), ( left| frac{1}{x_0 - r_i} right| leq frac{1}{m} ). By the triangle inequality: [ |S| leq sum_{i=1}^n left| frac{1}{x_0 - r_i} right| leq sum_{i=1}^n frac{1}{m} = frac{n}{m}. ]4. Derive the Inequality: Rearranging ( |S| leq frac{n}{m} ) gives: [ m leq frac{n}{|S|} = n left| frac{f(x_0)}{f'(x_0)} right|. ] Since ( m ) is the minimum distance from ( x_0 ) to any root ( r_i ), there exists a root ( x_1 ) such that: [ |x_0 - x_1| leq n left| frac{f(x_0)}{f'(x_0)} right|. ]Conclusion: The inequality holds for any ( x_0 ) where ( f'(x_0) neq 0 ), ensuring a root lies within the specified bound. This leverages the structure of polynomials with all real roots and properties of their derivatives.Final Answerboxed{ left|x_{0} - x_{1}right| leq n left| frac{f(x_{0})}{f^{prime}(x_{0})} right| }

🔑:We need to prove that if ( f ) is a polynomial of degree ( n geq 2 ) with all ( n ) roots real, and ( f^prime(x_0) neq 0 ) at some point ( x_0 ), then there exists a root ( x_1 ) of ( f ) such that[|x_0 - x_1| leq n left| frac{f(x_0)}{f^prime(x_0)} right|.]# Step-by-Step Solution:1. Consider the polynomial ( f(x) = x^n + a_1 x^{n-1} + cdots + a_n ) with roots ( x_1, x_2, ldots, x_n ). 2. If ( x_0 ) is exactly one of the roots, say ( x_0 = x_i ), then ( f(x_0) = 0 ) and ( f^prime(x_i) neq 0 ). Thus, the inequality holds trivially as both sides are zero.3. Assume ( x_0 ) is not equal to any of the roots ( x_i ). Since ( f(x_0) neq 0 ), the problem reduces to proving the equivalent statement:[left| frac{1}{x_0 - x_i} right| = left| frac{1}{n} frac{f^prime(x_0)}{f(x_0)} right|,]for some index ( i ).4. Express ( f(x) ) in terms of its roots:[f(x) = (x - x_1)(x - x_2) cdots (x - x_n).]5. Differentiate ( f(x) ):[f^prime(x) = sum_{k=1}^n left[ (x - x_1) cdots cancel{(x - x_k)} cdots (x - x_n) right] = sum_{k=1}^n frac{f(x)}{x - x_k},]where each term is obtained by differentiating one of the factors and multiplying by the others.6. Evaluate ( f^prime(x) ) at ( x_0 ):[f^prime(x_0) = sum_{k=1}^n frac{f(x_0)}{x_0 - x_k}.]7. Consequently,[left| frac{f^prime(x_0)}{f(x_0)} right| = left| frac{1}{x_0 - x_1} + frac{1}{x_0 - x_2} + cdots + frac{1}{x_0 - x_n} right|.]8. Using the triangle inequality,[left| frac{f^prime(x_0)}{f(x_0)} right| leq sum_{k=1}^n left| frac{1}{x_0 - x_k} right|.]9. To bound this sum, note that each term ( left| frac{1}{x_0 - x_k} right| ) can be no larger than their maximum:[left| frac{f^prime(x_0)}{f(x_0)} right| leq n left| frac{1}{x_0 - x_k} right|,]for some ( k ), the term within the sum that has the largest magnitude.10. Rearranging this, we get:[left| x_0 - x_k right| leq n left| frac{f(x_0)}{f^prime(x_0)} right|.]Therefore, there exists a root ( x_k ) such that the given condition holds true. Conclusion:[boxed{left| x_0 - x_1 right| leq n left| frac{f(x_0)}{f^prime(x_0)} right|}]

🧠:Okay, so I need to prove that the sum from k = n+1 to 2n of 1/k is greater than 1/2 for any natural number n. Hmm, let's start by understanding what this sum actually represents. It's the harmonic series starting from 1/(n+1) up to 1/(2n). I remember that harmonic series terms decrease as k increases, so each term in this sum is less than or equal to 1/(n+1). Wait, but adding up these terms... How many terms are there exactly?Let me count. If we start at n+1 and go up to 2n, that's 2n - (n+1) + 1 = n terms. Right, there are n terms in this sum. So we're adding n terms, each of which is between 1/(2n) and 1/(n+1). If all the terms were equal to 1/(2n), the sum would be n*(1/(2n)) = 1/2. But since each term is actually larger than 1/(2n), the sum should be greater than 1/2. Wait, is that the case?Wait, hold on. Let's check with n=1. If n=1, the sum is from 2 to 2, which is just 1/2. So 1/2 > 1/2? No, that's equal. But the problem states n is a natural number. If n=1, 1/2 is not greater than 1/2. Hmm, maybe the problem has a typo or maybe n starts from 2? Wait, let me check the original problem again. It says "if n is a natural number," so n=1 is allowed. Then the inequality would not hold for n=1. That's a problem. Wait, maybe the original problem has a non-strict inequality? Let me confirm.Wait, the original problem says "then [sum] > 1/2". So for n=1, the sum is exactly 1/2, which is not greater. So maybe there's a mistake here. Wait, perhaps the problem is stated correctly, but n is supposed to be a natural number greater than 1? Or maybe I made a mistake in counting the terms. Let me check again for n=1.For n=1, the sum is from 1+1=2 to 2*1=2, which is just the term 1/2. So indeed, 1/2 is not greater than 1/2. Therefore, the inequality as stated does not hold for n=1. But the problem says "if n is a natural number," so maybe natural numbers here start from 2? Wait, in some definitions, natural numbers start from 1, in others from 0, but 2 is not standard. Hmm.Alternatively, maybe the problem is correct, and there's a different approach. Maybe my initial thought was wrong. Let me test for n=2. For n=2, the sum is from 3 to 4, so 1/3 + 1/4 = 7/12 ≈ 0.583..., which is greater than 0.5. So for n=2, it works. For n=3: 1/4 + 1/5 + 1/6 = (15 + 12 + 10)/60 = 37/60 ≈ 0.616..., which is also greater than 0.5. So starting from n=2, the inequality holds. But for n=1, it doesn't. Maybe the problem has a typo, or maybe the user intended n to be at least 2. Alternatively, maybe the original problem is correct, and I need to see if there's a different interpretation.Wait, another thought: maybe the sum is from n+1 to 2n inclusive. For n=1, that's 2 terms: n+1=2 and 2n=2. Wait, no, 2n=2, so only one term: 1/2. So n=1 gives 1/2, which is equal. So perhaps the problem should state "for all natural numbers n ≥ 1, the sum is ≥ 1/2, with strict inequality for n ≥ 2." But the problem as stated says "> 1/2". Hmm.Alternatively, maybe I misapplied the initial logic. Let me see. The user wrote: "Prove that if n is a natural number, then [sum] > 1/2". So either the problem is incorrect for n=1, or there's an error in my reasoning. Let me check n=1 again. Wait, if n=1, then 1/(1+1) + ... +1/(2*1) is just 1/2. So exactly 1/2. So the inequality is not strict. Therefore, the problem statement might have an error. Alternatively, maybe in the original problem, n is a natural number greater than 1. Or perhaps the user intended n to be a positive integer, and the sum is from n+1 to 2n, which for n=1 gives 1/2, which is equal. Hmm. Maybe the problem is correct, but the inequality is not strict for n=1. So perhaps the correct statement is that for all natural numbers n ≥ 1, the sum is ≥ 1/2, with strict inequality for n ≥ 2. But the problem states strictly greater. So maybe the problem has a mistake, or perhaps in some sources, natural numbers start at 0? Let's check n=0. If n=0, sum from 1 to 0, which doesn't make sense. So n=0 is invalid.Alternatively, maybe the problem is correct as stated, and there's a different approach. Let me think again. Maybe induction?Let me try mathematical induction. Let's test for n=1: the base case. As we saw, the sum is 1/2, which is not greater than 1/2. So induction might not work here. Alternatively, maybe start induction from n=2.Assume that for some k ≥2, the sum from k+1 to 2k is greater than 1/2. Then for k+1, we need to show that the sum from (k+1)+1 to 2(k+1) is greater than 1/2. Hmm. Let's see. The sum for n=k+1 would be from k+2 to 2k+2. Comparing to the sum for n=k, which is from k+1 to 2k. So the new sum is the old sum minus 1/(k+1) plus 1/(2k+1) + 1/(2k+2). So the difference between S(k+1) and S(k) is [ -1/(k+1) + 1/(2k+1) + 1/(2k+2) ].Let me compute this difference: -1/(k+1) + 1/(2k+1) + 1/(2k+2). Let's combine the terms:First, 1/(2k+1) + 1/(2k+2) = [ (2k+2) + (2k+1) ] / [ (2k+1)(2k+2) ) ] = (4k + 3) / [ (2k+1)(2k+2) )Then subtract 1/(k+1):So total difference is (4k + 3)/[ (2k+1)(2k+2) ] - 1/(k+1). Let's compute a common denominator:The first term has denominator (2k+1)(2k+2), the second term is 1/(k+1) = (2k+1)/[ (k+1)(2k+1) ) ] = (2k+1)/[ (2k+1)(k+1) ) ]So difference:[ (4k +3) - (2k+1) ] / [ (2k+1)(2k+2) ) ] = [4k +3 -2k -1]/[ (2k+1)(2k+2) ) ] = (2k +2)/[ (2k+1)(2k+2) ) ] = 2(k +1)/[ (2k+1)(2k+2) ) ]Simplify numerator and denominator:2(k+1) / [ (2k+1)(2(k+1)) ] = 2(k+1) / [2(k+1)(2k+1) ) ] = 1/(2k+1)Therefore, the difference S(k+1) - S(k) = 1/(2k+1). Since 1/(2k+1) is positive for all natural numbers k, this means S(k+1) = S(k) + 1/(2k+1). Therefore, if S(k) > 1/2, then S(k+1) = S(k) + something positive, so also greater than 1/2.But wait, for induction, we need the base case. If the base case is n=2, where S(2)=1/3 +1/4=7/12≈0.583>0.5, then assuming S(k)>1/2, S(k+1) is S(k) +1/(2k+1), so even larger. Therefore, by induction, the statement holds for all n≥2. But the problem states "if n is a natural number", which typically includes n=1. However, since for n=1 the sum is exactly 1/2, the inequality is not strict. Therefore, perhaps the problem has a typo, and n should be a natural number greater than 1. Alternatively, maybe the original problem is correct, and the user expects the answer to recognize that n=1 is an edge case but the inequality holds for n≥2. However, since the problem says "natural number", which usually includes n=1, this is conflicting.Alternatively, maybe my initial approach was wrong, and there's another way to show the sum is greater than 1/2 for all n≥1, including n=1. Wait, but for n=1, it's exactly 1/2, so that's not possible. Therefore, the problem must have an error. Alternatively, maybe the sum is from n+1 to 2n inclusive, but for n=1, that's two terms? Wait, no. For n=1, n+1=2 and 2n=2, so only one term: 1/2. So it's still 1/2. Hmm.Alternatively, maybe the user made a mistake in the problem statement. But assuming that the problem is as stated, and perhaps the user expects us to consider n≥2, or maybe there's another approach where the inequality is actually strict for all n≥1, but that contradicts the n=1 case. Wait, maybe my calculation for the difference in induction step was incorrect?Wait, let me re-examine the induction step. When moving from n=k to n=k+1, the sum changes from 1/(k+1) + ... +1/(2k) to 1/(k+2) + ... +1/(2k+2). So actually, the new sum is the old sum minus the term 1/(k+1) and adding two terms: 1/(2k+1) +1/(2k+2). Therefore, S(k+1) = S(k) - 1/(k+1) + 1/(2k+1) +1/(2k+2). So the difference is indeed [1/(2k+1) +1/(2k+2) -1/(k+1)].Let me compute this difference again. 1/(2k+1) +1/(2k+2) = [ (2k+2) + (2k+1) ] / [ (2k+1)(2k+2) ) ] = (4k +3)/[ (2k+1)(2k+2) )Then subtract 1/(k+1): which is (2k+2)/[ (k+1)(2k+2) ) ] = 2/[2k+2]? Wait, no. Wait, 1/(k+1) = [2k+2]/[ (k+1)(2k+2) ) ] = 2/[2(k+1)] = 1/(k+1). Wait, maybe another approach.Let me compute 1/(2k+1) +1/(2k+2) -1/(k+1):= 1/(2k+1) + [1/(2k+2) -1/(k+1)]= 1/(2k+1) + [ (1 - 2) / (2k+2) ]= 1/(2k+1) -1/(2k+2)= [ (2k+2) - (2k+1) ] / [ (2k+1)(2k+2) )= 1 / [ (2k+1)(2k+2) )So the difference is 1/[ (2k+1)(2k+2) ). Wait, that's different from my previous conclusion. Wait, now I'm confused. Which one is correct?Let me compute 1/(2k+1) +1/(2k+2) -1/(k+1):First, 1/(2k+2) -1/(k+1) = 1/(2(k+1)) -1/(k+1) = -1/(2(k+1)).Therefore, total difference is 1/(2k+1) -1/(2(k+1)).So difference = 1/(2k+1) -1/(2k+2) = [ (2k+2) - (2k+1) ] / [ (2k+1)(2k+2) ) ] = 1/[ (2k+1)(2k+2) ). Therefore, the difference S(k+1) - S(k) = 1/[ (2k+1)(2k+2) ). Wait, that contradicts my previous conclusion. Wait, which is correct?Wait, let's take a specific example. Let's take k=1. Then S(1) = 1/2. S(2) =1/3 +1/4=7/12≈0.583. The difference is 7/12 -1/2=7/12 -6/12=1/12. According to the formula 1/[ (2k+1)(2k+2) ), when k=1: 1/(3*4)=1/12. Correct. So that's the difference. Therefore, S(k+1) = S(k) + 1/[ (2k+1)(2k+2) ). Therefore, the difference is positive, so S(k+1) > S(k). Therefore, if S(k) >1/2, then S(k+1) is even larger, hence also >1/2.But for the base case, n=1: S(1)=1/2. Then S(2)=1/2 +1/12=7/12>1/2. So induction would work if we start from n=1, but since S(1)=1/2, the inductive step would only show that S(n) >=1/2 for all n>=1, but with S(n) strictly increasing. Therefore, for n>=2, S(n) > S(1)=1/2, hence the inequality holds for n>=2. But for n=1, it's equal. So perhaps the problem should state "for all natural numbers n>=2" or the inequality is non-strict. However, the problem as stated says ">1/2" for any natural number n. Which is not true for n=1.Therefore, maybe there's a different approach. Perhaps using integrals to approximate the harmonic series. The harmonic series can be approximated by the natural logarithm. Recall that the sum from k=1 to m of 1/k is approximately ln(m) + gamma, where gamma is Euler-Mascheroni constant. So the sum from k=n+1 to 2n of 1/k is approximately ln(2n) - ln(n) = ln(2). And ln(2)≈0.693>0.5. So this suggests that the sum is around 0.693, which is greater than 0.5. However, this is an approximation. Maybe we can use integrals to bound the sum.We know that for a decreasing function f(x), the sum from k=a to b of f(k) is less than the integral from a-1 to b of f(x) dx. Wait, no, the integral test says that the sum from k=1 to n of f(k) is less than f(1) + integral from 1 to n of f(x) dx. Alternatively, for an decreasing function, the integral from a to b+1 of f(x) dx <= sum from k=a to b of f(k) <= integral from a-1 to b of f(x) dx.Wait, let's recall that for a decreasing function f, we have:integral_{a}^{b+1} f(x) dx <= sum_{k=a}^{b} f(k) <= integral_{a-1}^{b} f(x) dx.So applying this to f(x)=1/x, which is decreasing, the sum from k=n+1 to 2n of 1/k can be bounded below by integral_{n+1}^{2n+1} 1/x dx = ln(2n+1) - ln(n+1) = ln( (2n+1)/(n+1) ) = ln(2 - 1/(n+1)).Similarly, upper bounded by integral_{n}^{2n} 1/x dx = ln(2n) - ln(n) = ln(2).But we need a lower bound. Let's compute the lower bound: ln( (2n+1)/(n+1) ). For n>=1, (2n+1)/(n+1)=2 -1/(n+1). Let's compute ln(2 -1/(n+1)).We need to compare this to 1/2. For n=1: ln( (2*1 +1)/(1+1) )=ln(3/2)≈0.405<0.5. So for n=1, the lower bound is 0.405, but the actual sum is 0.5. So the lower bound is not tight enough.For n=2: ln(5/3)≈0.510>0.5. So for n=2, the lower bound is already greater than 0.5, which would imply the sum is greater than 0.5. But for n=1, the lower bound is less than 0.5, but the actual sum is 0.5. So maybe for n>=1, the sum is >=0.5, with equality at n=1. But the problem states strict inequality. Therefore, maybe combining the integral approach and induction.Alternatively, another idea: compare each term to the average. Since there are n terms, if each term is greater than 1/(2n), then the sum is greater than n*(1/(2n))=1/2. But the terms are 1/(n+1), 1/(n+2), ...,1/(2n). The smallest term is 1/(2n). Wait, no, the terms decrease as k increases, so the smallest term is 1/(2n). Therefore, each term is greater than or equal to 1/(2n). Therefore, the sum is greater than or equal to n*(1/(2n))=1/2. But since not all terms are equal to 1/(2n), the sum is actually greater than 1/2. Wait, but for n=1, the sum is exactly 1/2. So this approach would work if all terms are strictly greater than 1/(2n). But when n=1, the only term is 1/2, which is equal. Therefore, this argument shows that the sum is >=1/2, with strict inequality when there's at least one term greater than 1/(2n). Which happens when n>=2, because for n>=2, the first term is 1/(n+1) which is greater than 1/(2n) (since n+1<2n when n>=2). Wait, n+1 <2n when n>=2? Let's check n=2: 3<4 yes. n=3:4<6 yes. So for n>=2, 1/(n+1) >1/(2n), because n+1 <2n. Therefore, the first term is greater than 1/(2n), and the rest are >=1/(2n). Therefore, the sum is greater than 1/2. But for n=1, n+1=2=2n=2, so the only term is 1/2, equal. Therefore, the sum is >=1/2 for all n>=1, with strict inequality for n>=2.But the problem states "if n is a natural number, then [sum] >1/2". So unless n=1 is excluded, the statement is false. However, maybe in the problem's context, natural numbers start at 2, but that's unusual. Alternatively, maybe the user made a typo and the sum starts at n instead of n+1. Let's check: if the sum is from n to 2n, then for n=1, it's from 1 to 2:1 +1/2=3/2>1/2. But that's a different problem. Alternatively, maybe the problem is to show that the sum is greater than 1/2 for all n>=1, but considering that for n=1 it's exactly 1/2, that seems inconsistent.Alternatively, maybe there's another approach. Let's consider pairing terms. For example, pair the first term and the last term: 1/(n+1) +1/(2n). Then the next pair:1/(n+2) +1/(2n-1), etc. Let's see if each pair is greater than some value.Let's compute 1/(n+k) +1/(2n -k +1) for k=1 to n. Each pair would be 1/(n+k) +1/(2n -k +1). Let's compute for k=1:1/(n+1) +1/(2n). For k=2:1/(n+2) +1/(2n-1), etc.Perhaps we can show that each pair is greater than 2/(2n) =1/n. Then the total sum would be greater than n*(1/n)=1. But the actual sum is greater than 1/2, not 1. So maybe this approach isn't directly helpful.Alternatively, consider that each term 1/(n+i) for i=1 to n can be compared to the integral from n+i-1 to n+i of 1/x dx. Since 1/x is decreasing, 1/(n+i) < integral_{n+i-1}^{n+i} 1/x dx. Therefore, sum_{i=1}^n 1/(n+i) < integral_{n}^{2n} 1/x dx = ln(2n) -ln(n) =ln(2). But this gives an upper bound, not helpful for a lower bound.Alternatively, use the integral from n+1 to 2n+1 of 1/x dx < sum_{k=n+1}^{2n} 1/k. This integral is ln(2n+1) -ln(n+1). We need to show that this is greater than 1/2.So need to show that ln( (2n+1)/(n+1) ) >1/2. Solve for n:(2n+1)/(n+1) > e^{1/2} ≈1.6487.Multiply both sides by (n+1):2n+1 >1.6487(n+1)2n +1>1.6487n +1.64872n -1.6487n >1.6487 -10.3513n >0.6487n>0.6487/0.3513≈1.847.Therefore, for n>=2, ln( (2n+1)/(n+1) ) >1/2. For n=2, (2*2+1)/(2+1)=5/3≈1.6667>1.6487, so ln(5/3)≈0.5108>0.5. Therefore, the integral lower bound exceeds 1/2 for n>=2. Therefore, the actual sum is greater than this integral, hence for n>=2, sum>0.5. For n=1, the integral from 2 to 3 of 1/x dx=ln(3)-ln(2)=ln(3/2)≈0.405<0.5, but the actual sum is 0.5. Therefore, combining these results, for n>=1, sum>=0.5, with strict inequality for n>=2. Therefore, if the problem considers n>=2, the inequality holds. If n=1 is included, it's equal.Given this, perhaps the problem intended n to be a natural number greater than or equal to 2, but it's stated as a natural number. However, since the user presented the problem as is, maybe we should proceed under the assumption that n>=2, or note the discrepancy and provide the proof for n>=2, mentioning the n=1 case.Alternatively, perhaps there's another approach without using integrals. Let's consider the arithmetic mean of the terms. The sum is the average of the terms multiplied by the number of terms. There are n terms. The average is [sum_{k=n+1}^{2n}1/k]/n. If we can show that this average is greater than 1/(2n), then the sum is greater than n*(1/(2n))=1/2. But the average is greater than the smallest term? Wait, no. The average is between the smallest and largest term. Since the terms are decreasing, the average is less than the first term 1/(n+1). However, 1/(n+1) >1/(2n) when n+1<2n, which is true for n>=2. But again, this brings us back to the previous conclusion.Alternatively, use the Cauchy-Schwarz inequality. For positive real numbers a_i, (sum a_i)^2 <= (sum 1)^2 * sum a_i^2. Not sure if helpful.Alternatively, use the AM-HM inequality. The harmonic mean. For positive real numbers, the harmonic mean is less than or equal to the arithmetic mean. But again, not sure.Wait, another idea: the sum from k=n+1 to 2n of 1/k is equal to the sum from k=1 to 2n of 1/k minus the sum from k=1 to n of 1/k. So it's H_{2n} - H_n, where H_m is the m-th harmonic number. It is known that H_{2n} - H_n = sum_{k=1}^{2n} 1/k - sum_{k=1}^{n}1/k = sum_{k=n+1}^{2n}1/k.There's a known identity that H_{2n} - H_n = sum_{k=1}^{n} frac{1}{n +k}.Also, it's known that H_{2n} - H_n approaches ln(2) as n approaches infinity. So for large n, the sum approaches ln(2)≈0.693>0.5. Therefore, for sufficiently large n, the sum exceeds 1/2. But we need to show it for all natural numbers n (assuming n>=2).Alternatively, consider using the inequality H_n <= ln(n) + gamma +1/(2n) -1/(12n^2), where gamma is Euler-Mascheroni constant. But this might be too complicated.Alternatively, use mathematical induction starting from n=2. We saw that for n=2, the sum is 7/12>0.5. Assume it holds for n=k, then for n=k+1, as shown earlier, the sum increases by 1/( (2k+1)(2k+2) ), which is positive, so the sum remains above 1/2.Therefore, by induction, the statement holds for all n>=2. Therefore, the answer is that the inequality holds for all natural numbers n>=2, with n=1 being an edge case where equality holds. However, since the problem states "if n is a natural number", which typically includes n=1, there's a contradiction. Therefore, the problem might have an error. However, assuming the problem intended n>=2, the proof via induction works.Alternatively, perhaps the problem is correct as stated, and I made a mistake. Let me check n=1 again. The sum is from 2 to 2: only 1/2. So 1/2 is not greater than 1/2. Hence, the inequality is false for n=1. Therefore, the correct statement is for n>=2. Given that, I will proceed to prove the inequality for n>=2, and mention the n=1 case.But since the user provided the problem as is, perhaps they expect the standard proof which might ignore the n=1 case or consider natural numbers starting at 1. Let me check online for similar inequalities.Upon a quick recall, the inequality sum from n+1 to 2n of 1/k >1/2 is a standard result, often proved using induction or integrals. For example, in some textbooks, this is given as an exercise, and the base case is n=1, but the sum is 1/2, so maybe they use a different approach. Wait, maybe there's a different grouping of terms.Wait, another approach: group the terms in pairs and use inequalities. For example, consider the sum:For n >=1,Sum_{k=n+1}^{2n} 1/k =1/(n+1) +1/(n+2) + ... +1/(2n).Let me pair the terms as follows: (1/(n+1) +1/(2n)), (1/(n+2)+1/(2n-1)), ..., up to the middle term if n is even or odd.For example, if n is even, pair the first and last, second and second-last, etc. Each pair is of the form 1/(n+i) +1/(2n -i +1) for i=1 to n/2.Let me compute one such pair:1/(n+i) +1/(2n -i +1) = [ (2n -i +1) + (n+i) ] / [ (n+i)(2n -i +1) ) ] = (3n +1)/[ (n+i)(2n -i +1) ) ]Hmm, not sure if that helps. Alternatively, note that each pair is greater than 2/( (n+i) + (2n -i +1) ) *2 by AM-HM inequality. Wait, AM-HM states that (a +b)/2 >= 2/(1/a +1/b). So 1/a +1/b >=4/(a +b). So each pair 1/(n+i) +1/(2n -i +1) >=4/(3n +1). But adding up n/2 such pairs gives total sum >=4*(n/2)/(3n +1)=2n/(3n +1). For n>=1, 2n/(3n +1)=2/(3 +1/n)>=2/(4)=0.5 for n=1, but for n>=2, 3 +1/n <=3.5, so 2/(3 +1/n)>=2/3.5≈0.571>0.5. Therefore, for n>=2, sum>=2n/(3n +1)>1/2. For n=1, sum=1/2=2*1/(3*1 +1)=2/4=0.5. Therefore, this shows that the sum is >=2n/(3n +1)>=1/2 for n>=1, with equality for n=1. Therefore, for n>=2, sum>1/2. Hence, the inequality holds for n>=2.Therefore, this provides a proof using the AM-HM inequality on paired terms. Therefore, combining this with the induction step, we can conclude the result.Alternatively, use the AM-HM inequality on all terms. The harmonic mean of the terms is n / (sum_{k=n+1}^{2n} 1/k ). The arithmetic mean of the terms is (sum)/n. By AM-HM inequality, (sum)/n >= n / (sum_{k=n+1}^{2n}1/k ). But this seems circular.Alternatively, use the convexity of the function f(x)=1/x. By Jensen's inequality, the average of 1/x is greater than 1/(average of x). But the average of x from n+1 to 2n is (n+1 +2n)/2=(3n+1)/2. Therefore, the average of 1/x is greater than 1/( (3n+1)/2 )=2/(3n+1). Therefore, sum= n*(average)>=n*2/(3n+1). As before, sum>=2n/(3n+1). As before, 2n/(3n+1)>=1/2 when 4n>=3n+1 =>n>=1. But for n=1, equality; for n>=2, 4n>=3n+1 =>n>=1. Therefore, 2n/(3n+1)>=1/2 for n>=1. Hence, sum>=1/2 for all n>=1, with equality iff n=1. Therefore, for n>=2, sum>1/2.Therefore, combining these results, the inequality holds for all natural numbers n>=1, with strict inequality for n>=2. Therefore, the problem's statement has a minor inaccuracy for n=1, but assuming n>=2, the proof holds.Given that, the final answer would acknowledge that for n=1, the sum equals 1/2, but for all natural numbers n>=2, the sum is strictly greater than 1/2. However, since the problem states "if n is a natural number", it's technically incorrect for n=1. But likely, the intended answer is to ignore n=1 and proceed with n>=2, providing the induction proof or the AM-HM paired terms proof.Alternatively, another approach using the concept of the harmonic series difference:As mentioned earlier, H_{2n} - H_n = sum_{k=n+1}^{2n}1/k. It's known that H_{2n} - H_n = ln(2) + frac{1}{4n} - frac{1}{8n^2} + ..., using the expansion of the harmonic numbers. Since ln(2)≈0.693>0.5, and the correction terms are positive for large n, this suggests that the difference is greater than 1/2 for all n>=1. However, this is an asymptotic expansion and might not hold for small n. For n=1, H_2 - H_1 =1 +1/2 -1=1/2. For n=2, H_4 - H_2=1+1/2+1/3+1/4 - (1+1/2)=1/3+1/4=7/12≈0.583>0.5. For n=3, H_6 - H_3=1+1/2+1/3+1/4+1/5+1/6 - (1+1/2+1/3)=1/4+1/5+1/6=37/60≈0.616>0.5. So empirically, for n>=2, the sum exceeds 1/2. For n=1, it's exactly 1/2. Therefore, combining with the earlier induction step, which shows that the sum increases with n, we can conclude that for all n>=2, the sum is greater than 1/2.Therefore, the proof can be structured as follows:1. Show that for n=2, the sum is greater than 1/2.2. Show that for any n>=2, if the sum for n=k is greater than 1/2, then the sum for n=k+1 is also greater than 1/2.Alternatively, use the AM-HM inequality paired terms to show the sum is at least 2n/(3n+1), which is greater than 1/2 for n>=2.Alternatively, use the integral test to bound the sum below by ln((2n+1)/(n+1)) and show that this exceeds 1/2 for n>=2.Given that multiple approaches confirm the inequality for n>=2, and the problem's statement likely intended n>=2, we can proceed to write the proof using induction or the AM-HM approach.I think the induction approach is more straightforward for a beginner. Let's formalize that.Base case: n=2. Sum is1/3 +1/4=7/12>1/2. So base case holds.Inductive step: Assume for n=k, the sum S(k)=1/(k+1)+...+1/(2k) >1/2. Then for n=k+1, the sum S(k+1)=1/(k+2)+...+1/(2k+2). As shown earlier, S(k+1)=S(k) -1/(k+1) +1/(2k+1)+1/(2k+2). We computed the difference S(k+1)-S(k)=1/[(2k+1)(2k+2)]. Therefore, S(k+1)=S(k) + positive term. Since S(k)>1/2, then S(k+1)=S(k)+positive>1/2. Hence, by induction, the inequality holds for all n>=2.Therefore, the final proof is:For all natural numbers n>=2, the sum from k=n+1 to 2n of 1/k is greater than 1/2. This is proven by induction, with the base case n=2 verified directly, and the inductive step showing that if the statement holds for n=k, it holds for n=k+1 due to the addition of a positive difference. The case n=1 yields equality, which is an exception not covered by the strict inequality.However, since the problem states "if n is a natural number", which includes n=1, there's a discrepancy. But given the standard solutions and the problem's likely intent, we proceed with n>=2.

✅:To prove that for any natural number ( n geq 2 ), the sum ( sum_{k=n+1}^{2n} frac{1}{k} > frac{1}{2} ), we use mathematical induction:Base Case (( n = 2 )):[frac{1}{3} + frac{1}{4} = frac{7}{12} approx 0.583 > frac{1}{2}.]The base case holds.Inductive Step:Assume the inequality holds for ( n = k ), i.e.,[sum_{m=k+1}^{2k} frac{1}{m} > frac{1}{2}.]For ( n = k+1 ), the sum becomes:[sum_{m=k+2}^{2(k+1)} frac{1}{m} = left( sum_{m=k+1}^{2k} frac{1}{m} right) - frac{1}{k+1} + frac{1}{2k+1} + frac{1}{2k+2}.]The difference between consecutive sums is:[left( -frac{1}{k+1} + frac{1}{2k+1} + frac{1}{2k+2} right) = frac{1}{(2k+1)(2k+2)} > 0.]Since the inductive hypothesis gives ( sum_{m=k+1}^{2k} frac{1}{m} > frac{1}{2} ), adding a positive term preserves the inequality:[sum_{m=k+2}^{2(k+1)} frac{1}{m} > frac{1}{2}.]Thus, by induction, the inequality holds for all ( n geq 2 ).Conclusion:For all natural numbers ( n geq 2 ),[frac{1}{n+1} + frac{1}{n+2} + ldots + frac{1}{2n} > frac{1}{2}.]The case ( n = 1 ) yields equality, but the strict inequality holds for ( n geq 2 ).(boxed{frac{1}{n+1} + frac{1}{n+2} + ldots + frac{1}{2n} > frac{1}{2} text{ for all natural numbers } n geq 2})

🔑:The goal is to show that for any natural number ( n ), the sum[frac{1}{n+1} + frac{1}{n+2} + ldots + frac{1}{2n}]is greater than ( frac{1}{2} ).# I. Solution:1. Consider the harmonic series up to ( 2n ): [ 1 + frac{1}{2} + ldots + frac{1}{n} + frac{1}{n+1} + ldots + frac{1}{2n} = 1 + frac{1}{2} + ldots + frac{1}{2n} ]2. Subtract the sum of the first ( n ) terms from both sides: [ left( frac{1}{n+1} + frac{1}{n+2} + ldots + frac{1}{2n} right) = left( 1 + frac{1}{2} + ldots + frac{1}{2n} right) - left( 1 + frac{1}{2} + ldots + frac{1}{n} right) ]3. Rewrite and regroup the terms on the right: [ frac{1}{n+1} + frac{1}{n+2} + ldots + frac{1}{2n} = 1 + left(frac{1}{2} - 1right) + frac{1}{3} + left(frac{1}{4} - frac{1}{2}right) + ldots + frac{1}{2n-1} + left(frac{1}{2n} - frac{1}{n}right) ]4. Simplify the right side: [ 1 - frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - frac{1}{6} + ldots + frac{1}{2n-1} - frac{1}{2n} ]5. Combine pairs of terms: [ left(1 - frac{1}{2}right) + left(frac{1}{3} - frac{1}{4}right) + left(frac{1}{5} - frac{1}{6}right) + ldots + left(frac{1}{2n-1} - frac{1}{2n}right) ]6. Note that the first term ( 1 - frac{1}{2} = frac{1}{2} ), and all other pairs are positive because each term is of the form ( frac{1}{k} - frac{1}{k+1} ): [ frac{1}{2} + left(frac{1}{3} - frac{1}{4}right) + left(frac{1}{5} - frac{1}{6}right) + ldots + left(frac{1}{2n-1} - frac{1}{2n}right) ]7. Hence, the entire sum is greater than ( frac{1}{2} ).Therefore, the assertion is proved.# II. Solution: (Proof by Induction)1. Start with the base case ( k = 2 ): [ s_2 = frac{1}{3} + frac{1}{4} > frac{1}{2} ]2. Assume that the statement holds for ( k ): [ s_k = frac{1}{k+1} + frac{1}{k+2} + ldots + frac{1}{2k} > frac{1}{2} ]3. Show it holds for ( k + 1 ): [ s_{k + 1} = frac{1}{k+2} + frac{1}{k+3} + ldots + frac{1}{2k} + frac{1}{2k+1} + frac{1}{2k+2} ]4. Break the sum for ( s_{k+1} ): [ s_{k+1} = s_k - frac{1}{k+1} + frac{1}{2k+1} + frac{1}{2k+2} ]5. Note that ( frac{1}{2k+1} + frac{1}{2k+2} > frac{1}{k+1} ): [ frac{1}{2k+1} + frac{1}{2k+2} = frac{1times (2k+2) + 1times (2k+1)}{(2k+1)(2k+2)} = frac{4k+3}{(2k+1)(2k+2)} > frac{2}{2k+2} = frac{1}{k+1} ]6. Therefore, [ s_{k+1} = s_k + frac{1}{(2k+1)(2k+2)} > s_k > frac{1}{2} ]7. By induction, the statement is true for all natural numbers ( n ).# III. Solution:1. Consider the sum ( s_n = frac{1}{n+1} + frac{1}{n+2} + ldots + frac{1}{2n} ).2. Substitute every term with ( frac{1}{2n} ), the smallest term in the range: [ s_n > frac{1}{2n} + frac{1}{2n} + ldots + frac{1}{2n} = n cdot frac{1}{2n} = frac{1}{2} ]Therefore, in all solutions, we demonstrate that[boxed{frac{1}{n+1} + frac{1}{n+2} + ldots + frac{1}{2n} > frac{1}{2}}]

🧠:Okay, let's see. I need to evaluate the expression (2 sin 2α - 3 cos 2α)/(4 sin 2α + 5 cos 2α) given that tan α = 3. Hmm, where do I start? Well, since we know tan α, maybe we can find sin α and cos α first. Then use those to compute sin 2α and cos 2α. Alternatively, maybe there's a way to express the entire expression in terms of tan α directly. Let me think.First approach: Find sin α and cos α. Since tan α = 3, which is opposite over adjacent in a right triangle, so if we imagine a right triangle where the opposite side is 3 and the adjacent is 1, then the hypotenuse would be sqrt(3² + 1²) = sqrt(10). So, sin α = 3/sqrt(10) and cos α = 1/sqrt(10). But wait, is α in a specific quadrant? The problem doesn't specify, but since tan α is positive, α is either in the first or third quadrant. However, typically, unless specified, we can assume it's in the first quadrant. So, sin α and cos α are positive. So, sin α = 3/sqrt(10), cos α = 1/sqrt(10).Then, compute sin 2α and cos 2α. Using double-angle formulas:sin 2α = 2 sin α cos α = 2*(3/sqrt(10))*(1/sqrt(10)) = 6/10 = 3/5.cos 2α = cos²α - sin²α = (1/10) - (9/10) = -8/10 = -4/5.Alternatively, cos 2α can also be computed as 1 - 2 sin²α or 2 cos²α - 1, but let's check:1 - 2 sin²α = 1 - 2*(9/10) = 1 - 18/10 = (10/10 - 18/10) = -8/10 = -4/5. Same result. Good.So, sin 2α = 3/5, cos 2α = -4/5.Now plug these into the expression:Numerator: 2 sin 2α - 3 cos 2α = 2*(3/5) - 3*(-4/5) = 6/5 + 12/5 = 18/5.Denominator: 4 sin 2α + 5 cos 2α = 4*(3/5) + 5*(-4/5) = 12/5 - 20/5 = -8/5.Therefore, the expression is (18/5) / (-8/5) = (18/5)*(-5/8) = -18/8 = -9/4. So the answer is -9/4. Hmm, that seems straightforward, but let me check if there's another method to confirm.Alternative approach: Express the entire expression in terms of tan α. Since we have sin 2α and cos 2α in the expression, maybe we can write it as a fraction involving tan 2α. Let's see.First, note that sin 2α = 2 tan α / (1 + tan²α) and cos 2α = (1 - tan²α)/(1 + tan²α). Wait, but those are formulas for sin 2α and cos 2α in terms of tan α? Wait, actually, let me recall:sin 2α = 2 tan α / (1 + tan²α)cos 2α = (1 - tan²α)/(1 + tan²α)Yes, that's correct. These are derived from the double-angle identities. So, using these, maybe we can express the entire expression in terms of tan α.Given tan α = 3, let's compute sin 2α and cos 2α using these formulas:sin 2α = 2*3 / (1 + 3²) = 6 / (1 + 9) = 6/10 = 3/5. Same as before.cos 2α = (1 - 3²)/(1 + 3²) = (1 - 9)/10 = -8/10 = -4/5. Also same as before. So same result. Therefore, same numerator and denominator.Alternatively, maybe we can divide numerator and denominator by cos 2α. Let me see. Let's try that.Original expression: [2 sin 2α - 3 cos 2α]/[4 sin 2α + 5 cos 2α]Divide numerator and denominator by cos 2α:[2 tan 2α - 3]/[4 tan 2α + 5]But wait, because sin 2α / cos 2α is tan 2α. So yes, this would give us:Numerator: 2 tan 2α - 3Denominator: 4 tan 2α + 5But then we need to compute tan 2α. Since tan α = 3, tan 2α = 2 tan α / (1 - tan²α) = 2*3 / (1 - 9) = 6 / (-8) = -3/4.So tan 2α = -3/4. Therefore, substitute back:Numerator: 2*(-3/4) - 3 = (-6/4) - 3 = (-3/2) - 3 = (-3/2 - 6/2) = -9/2Denominator: 4*(-3/4) + 5 = (-12/4) + 5 = (-3) + 5 = 2Therefore, the expression is (-9/2)/2 = -9/4. Same result. So this method also gives -9/4.Another way: Let's verify if both methods indeed give the same answer, which they do. Therefore, the answer is -9/4.But wait, maybe there's a different approach? Let's think. For instance, using complex numbers or phasors? Probably overcomplicating. Or maybe parametrize alpha. Since tan α = 3, then alpha is arctan(3). So, perhaps compute 2α first, then compute sin and cos of 2α. But that's essentially the same as the first method. Alternatively, use the original expression and write it in terms of tan alpha. Wait, but we already did that by expressing it in terms of tan 2α, which relates to tan alpha.Alternatively, consider writing the entire expression as a fraction in terms of tan α. Let's see:Express numerator and denominator in terms of sin 2α and cos 2α. Wait, but since tan 2α is known, and we've already leveraged that. So perhaps the second method is more efficient because it reduces the expression in terms of tan 2α directly, avoiding the need to compute sin 2α and cos 2α individually.But regardless, both methods lead to the same result. Therefore, the answer is -9/4.But let me check once again for computational errors. Let's recompute using the first method:Given tan α = 3, so sin α = 3/sqrt(10), cos α = 1/sqrt(10).sin 2α = 2 sin α cos α = 2*(3/sqrt(10))*(1/sqrt(10)) = 6/10 = 3/5. Correct.cos 2α = cos²α - sin²α = (1/10) - (9/10) = -8/10 = -4/5. Correct.Numerator: 2*(3/5) - 3*(-4/5) = 6/5 + 12/5 = 18/5. Correct.Denominator: 4*(3/5) + 5*(-4/5) = 12/5 - 20/5 = -8/5. Correct.18/5 divided by -8/5 is indeed (18/5)*(5/-8) = 18/-8 = -9/4. Correct.Second method:tan 2α = 2*3/(1 - 9) = 6/-8 = -3/4. Correct.Numerator: 2*(-3/4) - 3 = -6/4 - 3 = -3/2 - 3 = -9/2. Correct.Denominator: 4*(-3/4) + 5 = -3 + 5 = 2. Correct.So (-9/2)/2 = -9/4. Correct.Therefore, regardless of the method, the result is -9/4.Wait, but let me think if there's any alternative approach that might give a different answer, just to be thorough.Alternatively, suppose we let t = tan α. Then, perhaps express the original expression in terms of t. Let's try that.Given tan α = t = 3.We can use the expressions for sin 2α and cos 2α in terms of t.As earlier:sin 2α = 2t/(1 + t²) = 2*3/(1 + 9) = 6/10 = 3/5.cos 2α = (1 - t²)/(1 + t²) = (1 - 9)/10 = -8/10 = -4/5.So substituting into the original expression:Numerator: 2*(3/5) - 3*(-4/5) = same as before.Denominator: 4*(3/5) + 5*(-4/5) = same as before.So same result. Therefore, this is consistent.Alternatively, express the entire fraction as a function of t. Let's see:Original expression: [2 sin 2α - 3 cos 2α]/[4 sin 2α + 5 cos 2α]Express sin 2α and cos 2α in terms of t:[2*(2t/(1 + t²)) - 3*((1 - t²)/(1 + t²))]/[4*(2t/(1 + t²)) + 5*((1 - t²)/(1 + t²))]Simplify numerator and denominator:Numerator:[ (4t)/(1 + t²) - 3(1 - t²)/(1 + t²) ] = [4t - 3(1 - t²)] / (1 + t²)= [4t - 3 + 3t²] / (1 + t²)Denominator:[ (8t)/(1 + t²) + 5(1 - t²)/(1 + t²) ] = [8t + 5(1 - t²)] / (1 + t²)= [8t + 5 - 5t²] / (1 + t²)Therefore, the entire expression becomes:[4t - 3 + 3t²] / [8t + 5 - 5t²]Now plug t = 3 into this expression:Numerator: 4*3 - 3 + 3*(9) = 12 - 3 + 27 = 36Denominator: 8*3 + 5 -5*(9) = 24 + 5 -45 = -16Wait, wait, this gives 36 / (-16) = -9/4. Wait, but according to this calculation:Numerator: 4*3 =12; 12 -3 =9; 9 + 27 =36. Correct.Denominator:8*3=24;24 +5=29;29 -45= -16. Correct.So 36/-16 simplifies to -9/4. Same result. So even this method gives the same answer.So this is another method, expressing everything in terms of t = tan α, and simplifying the expression algebraically before substituting t =3. Which also leads to -9/4.Therefore, regardless of the method used, the answer is -9/4. Therefore, the solution is consistent across different approaches. Hence, confident that the answer is correct.But just to be absolutely thorough, let me check once more with another angle. Suppose we compute α numerically. Since tan α =3, α is arctan(3). Let's compute α in degrees or radians, compute 2α, then compute sin 2α and cos 2α numerically, then compute the expression.Let's try that.Compute α = arctan(3). Let's use a calculator. arctan(3) ≈ 71.565 degrees. Then 2α ≈143.13 degrees.Compute sin(143.13 degrees): Let's compute sin(143.13°). Since sin(180° - x) = sin x. So sin(143.13°) = sin(36.87°). 36.87 degrees is a 3-4-5 triangle. sin(36.87°) = 3/5 =0.6. So sin(143.13°) ≈0.6. Similarly, cos(143.13°)= -cos(36.87°)= -4/5= -0.8.So sin 2α ≈0.6, which is 3/5. cos 2α≈ -0.8, which is -4/5. Then numerator:2*0.6 -3*(-0.8)=1.2 +2.4=3.6. Denominator:4*0.6 +5*(-0.8)=2.4 -4= -1.6. Then 3.6 / (-1.6) = -2.25, which is -9/4. So numerically, this also checks out.Therefore, all methods confirm that the answer is -9/4. So I think that's solid.Conclusion: The value of the expression is -9/4.Final Answerboxed{-dfrac{9}{4}}

✅:Given the expression (frac{2 sin 2alpha - 3 cos 2alpha}{4 sin 2alpha + 5 cos 2alpha}) and (tan alpha = 3), we need to evaluate it.First, we find (sin alpha) and (cos alpha). Since (tan alpha = 3), we can consider a right triangle with opposite side 3, adjacent side 1, and hypotenuse (sqrt{10}). Thus, (sin alpha = frac{3}{sqrt{10}}) and (cos alpha = frac{1}{sqrt{10}}).Using the double-angle formulas:[sin 2alpha = 2 sin alpha cos alpha = 2 left(frac{3}{sqrt{10}}right) left(frac{1}{sqrt{10}}right) = frac{6}{10} = frac{3}{5}][cos 2alpha = cos^2 alpha - sin^2 alpha = left(frac{1}{sqrt{10}}right)^2 - left(frac{3}{sqrt{10}}right)^2 = frac{1}{10} - frac{9}{10} = -frac{8}{10} = -frac{4}{5}]Substituting these into the expression:[text{Numerator: } 2 sin 2alpha - 3 cos 2alpha = 2 left(frac{3}{5}right) - 3 left(-frac{4}{5}right) = frac{6}{5} + frac{12}{5} = frac{18}{5}][text{Denominator: } 4 sin 2alpha + 5 cos 2alpha = 4 left(frac{3}{5}right) + 5 left(-frac{4}{5}right) = frac{12}{5} - frac{20}{5} = -frac{8}{5}]Thus, the expression simplifies to:[frac{frac{18}{5}}{-frac{8}{5}} = frac{18}{5} times frac{5}{-8} = frac{18}{-8} = -frac{9}{4}]We verified this result using alternative methods, including expressing the entire expression in terms of (tan alpha) and numerical computation, all leading to the same result.[boxed{-dfrac{9}{4}}]

🔑:Given the expression (frac{2 sin 2alpha - 3 cos 2alpha}{4 sin 2alpha + 5 cos 2alpha}) and the value (operatorname{tg} alpha = 3),1. Express (sin 2alpha) and (cos 2alpha) in terms of (tg alpha):Using the double-angle identities for sine and cosine:[sin 2alpha = frac{2 tg alpha}{1 + tg^2 alpha}][cos 2alpha = frac{1 - tg^2 alpha}{1 + tg^2 alpha}]2. Substitute these identities into the given expression:[frac{2 sin 2alpha - 3 cos 2alpha}{4 sin 2alpha + 5 cos 2alpha} = frac{2 left(frac{2 tg alpha}{1 + tg^2 alpha}right) - 3 left(frac{1 - tg^2 alpha}{1 + tg^2 alpha}right)}{4 left(frac{2 tg alpha}{1 + tg^2 alpha}right) + 5 left(frac{1 - tg^2 alpha}{1 + tg^2 alpha}right)}]3. Combine the terms in the numerator and the denominator:[= frac{frac{4 tg alpha}{1 + tg^2 alpha} - frac{3 - 3 tg^2 alpha}{1 + tg^2 alpha}}{frac{8 tg alpha}{1 + tg^2 alpha} + frac{5 - 5 tg^2 alpha}{1 + tg^2 alpha}}][= frac{frac{4 tg alpha - (3 - 3 tg^2 alpha)}{1 + tg^2 alpha}}{frac{8 tg alpha + (5 - 5 tg^2 alpha)}{1 + tg^2 alpha}}]4. Simplify the fraction:[= frac{4 tg alpha - 3 + 3 tg^2 alpha}{8 tg alpha + 5 - 5 tg^2 alpha}]5. Substitute (tg alpha = 3) into the simplified expression:[tg alpha = 3 implies tg^2 alpha = 9][= frac{4 cdot 3 - 3 + 3 cdot 9}{8 cdot 3 + 5 - 5 cdot 9}]6. Calculate the values in the numerator and denominator:[= frac{12 - 3 + 27}{24 + 5 - 45}][= frac{36}{-16}]7. Simplify the fraction:[= -frac{36}{16} = -frac{9}{4}] Conclusion:[boxed{-frac{9}{4}}]

🧠:Okay, let's see. I need to prove that two lines are parallel if and only if any plane that intersects one of them also intersects the other, and only at one point. Then, derive the transitivity of parallelism in space. Hmm. First, let me recall some basic concepts. In three-dimensional space, two lines can be parallel, intersecting, or skew. Parallel lines are coplanar and don't intersect. Skew lines aren't coplanar and don't intersect. So the problem is about establishing a condition for parallelism based on plane intersections. The statement says that for two lines to be parallel, it's necessary and sufficient that any plane intersecting one also intersects the other (only once). So I need to prove both directions: if two lines are parallel, then any plane intersecting one must intersect the other, and conversely, if any plane intersecting one line intersects the other, then the lines are parallel. Starting with the necessity: assume two lines are parallel. Then, they are coplanar. If a plane intersects one of them, does it have to intersect the other? Well, if the two lines are in their own plane, say plane α. If another plane β intersects one line, say line l1, at a point P, then plane β must intersect plane α along a line. Since l1 and l2 are both in α, the line of intersection of α and β would intersect l1 at P. If l2 is parallel to l1, then unless the line of intersection is parallel to both, which would mean it doesn't intersect either, but since β already intersects l1 at P, the line of intersection can't be parallel to l1. Therefore, the line of intersection must intersect both l1 and l2. Wait, but l2 is parallel to l1. If two lines are parallel, a third line intersecting one must intersect the other unless it's parallel. Wait, but in this case, the line of intersection of α and β is not parallel to l1 because it intersects l1. So, since l2 is parallel to l1, the line of intersection must also intersect l2. Hence, plane β would intersect l2 as well. So that point of intersection would be a single point, because if the plane contained l2, then it would contain both lines, but since they are parallel and in plane α, plane β would have to be plane α, but plane β could be different. Wait, no. If the lines are parallel and in plane α, then any plane intersecting one of them either contains both or intersects them each at a single point. If the plane is not α, then it can't contain both, because two parallel lines determine a unique plane. So if plane β is different from α and intersects l1 at a point, then β intersects α along a line that intersects both l1 and l2. Therefore, plane β must intersect l2 at another point. So yes, if lines are parallel, any plane intersecting one intersects the other. That seems to work. Now the sufficiency: if any plane that intersects one line intersects the other, then the lines must be parallel. So suppose lines l1 and l2 are such that every plane intersecting l1 also intersects l2, and vice versa. Need to show they are parallel. Assume for contradiction they are not parallel. Then they could either intersect or be skew. If they intersect, then they are coplanar, but not parallel. In that case, there exists a plane containing both. But if we take a plane that intersects one line at a point not on the other line, does that plane intersect the other line? Wait, if the lines intersect at a point, say P, then any plane through l1 would contain P. But if we take a plane that intersects l1 at a point Q different from P, then if the plane is not the plane containing both lines, it might not intersect l2. Wait, but if two lines intersect at P, then any plane containing a point Q on l1 (other than P) and not containing l2 would intersect l2 only if it passes through P. Hmm, this is getting confusing. Maybe another approach. Suppose lines are not parallel. Then either they are skew or intersecting. If they are skew, then by definition, they don't lie in the same plane. So, there exists a plane containing one line and not the other. Wait, but if they are skew, can we find a plane that intersects one but not the other? Wait, the original condition is that any plane intersecting one must intersect the other. So if they are skew, there should exist a plane that intersects one but not the other. How? For skew lines, there exist planes that contain one line and are parallel to the other. For example, take a line l1 and a skew line l2. Then, we can construct a plane containing l1 and parallel to l2. Since l2 is skew, it doesn't lie in that plane, and since the plane is parallel to l2, the plane doesn't intersect l2. Hence, such a plane would intersect l1 (as it contains l1) but not l2, violating the condition. Therefore, if lines are skew, the condition fails. Similarly, if lines are intersecting but not parallel, then they lie in a plane. So take a plane that intersects one line at a point not shared by the other line. Wait, but if they intersect at a point P, then any plane through l1 must contain P, and if it's a different plane, maybe it doesn't contain l2. Wait, no. If two lines intersect at P, they determine a unique plane. Any plane containing l1 must either be that unique plane or intersect l1 at a line. Wait, no, planes can intersect a line at a single point. Wait, actually, a plane can't intersect a line at a single point unless it's not containing the line. Wait, if a plane intersects a line, the intersection is either the whole line (if the line is in the plane) or a single point. So if we have two intersecting lines l1 and l2, which meet at P. If I take a plane that intersects l1 at a point Q ≠ P, then does this plane necessarily intersect l2? Not necessarily. Because if the plane is arranged such that it contains Q and is arranged in space such that it doesn't pass through P or any other part of l2. Wait, but l2 passes through P. If the plane contains Q and doesn't contain P, then it can't contain l2. But can such a plane exist? Let me think. Let’s say l1 and l2 intersect at P. Take a point Q on l1 different from P. Can we construct a plane that contains Q but does not contain P? Yes, infinitely many planes pass through Q and do not contain P. But such a plane would intersect l1 at Q. However, does it intersect l2? Since l2 passes through P, and the plane doesn't contain P, unless the plane intersects l2 at some other point. But l2 is a straight line. If the plane doesn't contain P, then the intersection with l2 would be a single point if the plane intersects l2 somewhere else. But if the plane is constructed such that it doesn't intersect l2 at all, then that would violate the condition. But can a plane passing through Q (on l1) avoid intersecting l2? If the lines are intersecting at P, then l2 is the line passing through P and some other point. If we take a plane through Q that doesn't contain P and is not parallel to l2, then it should intersect l2 somewhere. Wait, but in 3D space, two lines that are not coplanar are skew, but here l1 and l2 are coplanar. So any plane that intersects l1 at Q (not P) must intersect l2 unless it's parallel to l2. Wait, but if the plane is not parallel to l2, it should intersect l2. But if the plane is parallel to l2, then it can contain Q and be parallel to l2. If such a plane exists, then it would intersect l1 at Q but not intersect l2, since it's parallel to l2. Therefore, if lines are intersecting (and hence not parallel), then we can construct a plane that intersects one but not the other, which would violate the condition. Therefore, if the condition holds, the lines cannot be intersecting or skew; hence, they must be parallel. So that proves the necessary and sufficient condition. Now, deriving the theorem about transitivity of parallelism in space. The transitivity would state that if line a is parallel to line b, and line b is parallel to line c, then line a is parallel to line c. How can we derive this from the previous theorem? Let's see. Assume a || b and b || c. We need to show a || c. Suppose not. Then a and c are either intersecting or skew. If they are intersecting, then they lie in a plane. If they are skew, they don't. Let's see. If a and c are not parallel, then there exists a plane that intersects one but not the other. Wait, according to the previous theorem, if two lines are not parallel, then there exists a plane that intersects one but not the other. So if a and c are not parallel, there's a plane intersecting, say, a but not c. But since a is parallel to b, by the theorem, any plane intersecting a must intersect b. Similarly, since b is parallel to c, any plane intersecting b must intersect c. Therefore, the plane that intersects a must also intersect b, and then since the plane intersects b, it must intersect c. Therefore, the plane that intersects a must also intersect c. Contradiction. Hence, such a plane cannot exist, so a and c must be parallel. Therefore, parallelism is transitive. So that would be the transitivity theorem. Wait, let me check that again. Suppose a || b and b || c. If a is not parallel to c, then there exists a plane that intersects a but not c. But because a || b, any plane intersecting a must intersect b. And since b || c, any plane intersecting b must intersect c. Therefore, the plane that intersects a must intersect b and then must intersect c. Therefore, no such plane can exist that intersects a but not c. Therefore, a and c must be parallel. Yes, that seems correct. So the transitivity follows from the initial theorem. So putting it all together: first, the equivalence between parallelism and the plane intersection property, then using that to show transitivity.

✅:To establish the necessary and sufficient condition for parallelism and derive the transitivity theorem, we proceed as follows:--- Theorem: Two lines ( ell_1 ) and ( ell_2 ) in space are parallel if and only if every plane intersecting one line also intersects the other (at exactly one point).---# Proof:1. Necessity (Parallel lines ⇒ Planes intersect both): Assume ( ell_1 parallel ell_2 ). They lie in a unique plane ( alpha ). Let ( beta ) be any plane intersecting ( ell_1 ) at point ( P ). Since ( beta neq alpha ), ( beta ) intersects ( alpha ) along a line ( m ). Because ( ell_1 ) and ( ell_2 ) are parallel, ( m ) (which intersects ( ell_1 ) at ( P )) cannot be parallel to ( ell_1 ). Thus, ( m ) must intersect ( ell_2 ) at a distinct point ( Q ). Hence, ( beta ) intersects ( ell_2 ) at ( Q ). Since ( beta ) was arbitrary, every plane intersecting ( ell_1 ) also intersects ( ell_2 ). Symmetrically, the same holds for planes intersecting ( ell_2 ).2. Sufficiency (Planes intersect both ⇒ Parallel lines): Suppose every plane intersecting ( ell_1 ) also intersects ( ell_2 ). Assume for contradiction ( ell_1 ) and ( ell_2 ) are not parallel. There are two cases:- Case 1: ( ell_1 ) and ( ell_2 ) intersect at a point ( P ). Let ( Q neq P ) be a point on ( ell_1 ). Construct a plane ( beta ) containing ( Q ) but not ( P ). Since ( beta ) intersects ( ell_1 ) at ( Q ), it must intersect ( ell_2 ). However, ( ell_2 ) passes through ( P ), and ( beta notni P ), so ( beta ) cannot intersect ( ell_2 ). Contradiction.- Case 2: ( ell_1 ) and ( ell_2 ) are skew. There exists a plane ( beta ) containing ( ell_1 ) and parallel to ( ell_2 ) (by the parallel plane construction for skew lines). This plane intersects ( ell_1 ) but not ( ell_2 ), violating the condition. Contradiction.Thus, ( ell_1 parallel ell_2 ).--- Transitivity of Parallelism:Theorem: If ( ell_1 parallel ell_2 ) and ( ell_2 parallel ell_3 ), then ( ell_1 parallel ell_3 ).# Proof:Assume ( ell_1 parallel ell_2 ) and ( ell_2 parallel ell_3 ). Suppose ( ell_1 notparallel ell_3 ). Then there exists a plane ( beta ) intersecting ( ell_1 ) but not ( ell_3 ) (by the contrapositive of the earlier theorem). However:- Since ( ell_1 parallel ell_2 ), ( beta ) must intersect ( ell_2 ).- Since ( ell_2 parallel ell_3 ), ( beta ) must intersect ( ell_3 ).This contradicts the existence of ( beta ). Hence, ( ell_1 parallel ell_3 ).--- Conclusion:Parallelism in space is transitive, and the intersection property with planes fully characterizes parallel lines.

🔑:1. Establishing the Necessary Condition: - Suppose ( a parallel b ), meaning lines (a) and (b) are parallel. - Let ( a ) and ( b ) both lie in the same plane ( gamma ), so ( a subset gamma ) and ( b subset gamma ). - Consider a plane ( alpha ) that intersects line ( a ). - Since ( alpha ) intersects ( a ), it must also intersect plane ( gamma ). Let the intersection line be ( l = alpha cap gamma ). - The line ( l ), intersecting the parallel lines (a) and (b), must intersect both (a) and (b) at unique points because (a) and (b) are in the same plane and are parallel. - Thus, ( l ) intersects both (a) and (b) at unique points, ensuring any plane that intersects one of the parallel lines (a), must also intersect the other parallel line (b).2. Necessary Condition Proven: - The condition is necessary: if lines ( a parallel b ), then any plane ( alpha ) that intersects line ( a ) will also intersect line ( b ).3. Establishing the Sufficient Condition: - Suppose that any plane ( alpha ) that intersects line ( a ) also intersects line ( b ). - To argue the sufficiency, consider the possibilities for the spatial relationship between (a) and (b): - If (a) and (b) were intersecting, there would exist a plane containing ( a ) but not containing ( b ), contradicting the given condition. - If (a) and (b) were skew (i.e., neither parallel nor intersecting), a plane intersecting ( a ) would not necessarily intersect ( b ) as the two lines don't share a common plane. - Since both intersecting and skew lines are ruled out and they must intersect when introduced by the same plane, (a) and (b) must be parallel.4. Sufficient Condition Proven: - The condition is sufficient: if any plane that intersects line ( a ) also intersects line ( b ), then lines ( a parallel b ).5. Proving Transitivity of Parallelism: - Given ( a parallel b ) and ( b parallel c ), we must show ( a parallel c ). - Consider an arbitrary plane ( alpha ) that intersects line ( a ). - By our previous argument, if (a parallel b), then ( alpha ) intersects ( b ) as well. - Similarly, if ( b parallel c ), then ( alpha ), intersecting ( b ), must also intersect ( c ). - Therefore, ( alpha ) intersects both ( a ) and ( c ), establishing that ( a parallel c ).Conclusively, the transitivity of parallelism is proven by demonstrating the necessary and sufficient conditions for two lines to be parallel. [boxed{a parallel c}]