🧠:Okay, let's try to figure out this problem step by step. So, we have two sequences here: {a_n} and {b_n}. The problem gives us some information about each and asks for the sum of the first n terms of {b_n}. Let me start by writing down what I know.First, for the sequence {a_n}, the sum of the first n terms is given by S_n = 2a_n - 1. That's interesting because usually, the sum S_n is related to the terms a_1, a_2, ..., a_n. But here, it's expressed in terms of a_n itself. Hmm. Also, the sequence {b_n} has b_1 = 3, and each subsequent term is defined by b_{k+1} = a_k + b_k. So, each term of {b_n} after the first is the previous term plus the corresponding term from {a_n}. Our goal is to find the sum of the first n terms of {b_n}. Let's denote this sum as T_n. So, T_n = b_1 + b_2 + ... + b_n. To find T_n, we might need to express each b_k in terms of the a_k's and then sum them up. But before that, maybe we should figure out what the general term a_n is, since the sum S_n is given in terms of a_n.Let's start by analyzing the sequence {a_n}. The sum of the first n terms is S_n = 2a_n - 1. Let's recall that S_n = a_1 + a_2 + ... + a_n. Also, S_{n} can be related to S_{n-1} by S_n = S_{n-1} + a_n. So, maybe we can use that recursive relationship to find a formula for a_n.Given S_n = 2a_n - 1 and S_{n-1} = 2a_{n-1} - 1 (for n ≥ 2), then S_n = S_{n-1} + a_n. Substituting the expressions for S_n and S_{n-1}:2a_n - 1 = (2a_{n-1} - 1) + a_nLet me write that down:2a_n - 1 = 2a_{n-1} - 1 + a_nSimplify both sides. Let's subtract 2a_{n-1} - 1 from both sides:2a_n - 1 - (2a_{n-1} - 1) = a_nLeft side becomes 2a_n - 1 - 2a_{n-1} + 1 = 2a_n - 2a_{n-1}So, 2a_n - 2a_{n-1} = a_nSubtract a_n from both sides:2a_n - 2a_{n-1} - a_n = 0 => a_n - 2a_{n-1} = 0So, a_n = 2a_{n-1}Wait, that's a recurrence relation. So, the sequence {a_n} is a geometric sequence with common ratio 2. But we need to check the initial term.Let's find a_1. For n = 1, S_1 = a_1 = 2a_1 - 1. So:a_1 = 2a_1 - 1Subtract 2a_1 from both sides:-a_1 = -1 => a_1 = 1So, the first term is 1, and each subsequent term is double the previous. Therefore, a_n = 1 * 2^{n-1} = 2^{n-1}Wait, let's check this for n=2. S_2 = a_1 + a_2 = 1 + a_2. According to the formula, S_2 = 2a_2 - 1. So:1 + a_2 = 2a_2 - 1 => 1 + a_2 = 2a_2 -1 => 1 +1 = 2a_2 - a_2 => 2 = a_2. Which is correct, since a_2 = 2^{2-1} = 2^1 = 2.Similarly, for n=3: S_3 = 1 + 2 + 4 = 7. According to the formula, S_3 = 2a_3 -1 = 2*4 -1 = 8 -1 =7. Correct.So, the general term for {a_n} is a_n = 2^{n-1}.Okay, that's settled. Now, moving on to {b_n}. The sequence is defined by b_1 = 3 and b_{k+1} = a_k + b_k. So, each term is the previous term plus a_k. Let's write out the first few terms to see the pattern.b_1 = 3b_2 = a_1 + b_1 = 1 + 3 = 4b_3 = a_2 + b_2 = 2 + 4 = 6b_4 = a_3 + b_3 = 4 + 6 = 10b_5 = a_4 + b_4 = 8 + 10 = 18Hmm, so the terms of {b_n} are 3, 4, 6, 10, 18,...Looking at the differences between terms: from 3 to 4 is +1, 4 to 6 is +2, 6 to 10 is +4, 10 to 18 is +8... Wait a second, those differences are 1, 2, 4, 8, which are the terms of {a_n} starting from a_1=1. That makes sense because each b_{k+1} - b_k = a_k.So, the difference between consecutive terms of {b_n} is exactly the term a_{k}. So, the sequence {b_n} is a cumulative sum of the sequence {a_n} starting from 3. Wait, but how exactly?Wait, let me think. The recursive formula is b_{k+1} = b_k + a_k. So, starting from b_1 = 3, then b_2 = b_1 + a_1, b_3 = b_2 + a_2, etc. So, for term b_n, when n ≥ 2, b_n = b_1 + a_1 + a_2 + ... + a_{n-1}Because each step adds a previous a term. So, in general, for n ≥ 1,b_{n} = b_1 + Σ_{k=1}^{n-1} a_kBut Σ_{k=1}^{n-1} a_k is the sum of the first n-1 terms of {a_n}, which is S_{n-1} = 2a_{n-1} -1, as given in the problem.So, substituting that, b_n = 3 + S_{n-1} = 3 + 2a_{n-1} -1 = 2 + 2a_{n-1}But we know that a_{n-1} = 2^{(n-1)-1} = 2^{n-2}Therefore, b_n = 2 + 2*(2^{n-2}) = 2 + 2^{n-1}Wait, let's check that formula with the earlier terms.For n=1: b_1 = 3. According to the formula, plug n=1: 2 + 2^{1-1} = 2 + 1 = 3. Correct.n=2: b_2 = 4. Formula: 2 + 2^{2-1} = 2 + 2 =4. Correct.n=3: b_3=6. Formula: 2 + 2^{3-1}=2 +4=6. Correct.n=4: b_4=10. Formula: 2 + 2^{4-1}=2 +8=10. Correct. Good, so the general term for {b_n} is b_n = 2 + 2^{n-1}Therefore, each term b_n is equal to 2^{n-1} + 2. Now, we need to find the sum of the first n terms of {b_n}, which is T_n = Σ_{k=1}^n b_k = Σ_{k=1}^n (2 + 2^{k-1}) = Σ_{k=1}^n 2 + Σ_{k=1}^n 2^{k-1}The first sum is straightforward: Σ_{k=1}^n 2 = 2nThe second sum is a geometric series: Σ_{k=1}^n 2^{k-1} = 2^0 + 2^1 + ... + 2^{n-1} = (2^n -1)/(2 -1) ) = 2^n -1Therefore, T_n = 2n + (2^n -1) = 2^n + 2n -1Let me verify this with the first few terms.For n=1: T_1 = b_1 =3. According to formula: 2^1 +2*1 -1=2 +2 -1=3. Correct.For n=2: T_2 =3 +4=7. Formula:2^2 +2*2 -1=4 +4 -1=7. Correct.n=3: T_3=3+4+6=13. Formula:2^3 +6 -1=8 +6 -1=13. Correct.n=4: T_4=3+4+6+10=23. Formula:2^4 +8 -1=16 +8 -1=23. Correct.n=5: T_5=3+4+6+10+18=41. Formula:2^5 +10 -1=32 +10 -1=41. Correct.Looks like the formula is working. Therefore, the sum of the first n terms of {b_n} is 2^n + 2n -1.But let me just go through the steps again to ensure no mistakes.First, we found that {a_n} is a geometric sequence with a_1=1 and ratio 2, so a_n=2^{n-1}.Then, for {b_n}, since each term is the previous term plus a_{k-1}, starting from b_1=3, we can express b_n as 3 + sum_{k=1}^{n-1} a_k. Since sum_{k=1}^{n-1} a_k = S_{n-1} = 2a_{n-1} -1, substituting that gives b_n =3 + 2a_{n-1} -1 =2 +2a_{n-1}. Then since a_{n-1}=2^{n-2}, so b_n=2 +2*2^{n-2}=2 +2^{n-1}Therefore, each term b_n=2^{n-1} +2. Then summing from 1 to n:sum_{k=1}^n b_k = sum_{k=1}^n (2 +2^{k-1}) =2n + sum_{k=1}^n 2^{k-1}=2n + (2^n -1)Yes, that's correct. So, T_n=2^n +2n -1Therefore, the final answer should be boxed{2^n + 2n -1}Final Answerboxed{2^n + 2n - 1}

✅:Given the sequence {a_n} with the sum of its first n terms S_n = 2a_n - 1 for n = 1, 2, ldots, we start by analyzing the recurrence relation. We know that S_n = S_{n-1} + a_n. Substituting the given expressions for S_n and S_{n-1}:[2a_n - 1 = (2a_{n-1} - 1) + a_n]Simplifying this, we get:[2a_n - 1 = 2a_{n-1} - 1 + a_n implies a_n = 2a_{n-1}]This indicates that {a_n} is a geometric sequence with the first term a_1 = 1 (found by substituting n=1 into S_n = 2a_n - 1) and common ratio 2. Thus, a_n = 2^{n-1}.Next, for the sequence {b_n} defined by b_1 = 3 and b_{k+1} = a_k + b_k, we observe that each term b_n can be expressed as:[b_n = b_1 + sum_{k=1}^{n-1} a_k]The sum sum_{k=1}^{n-1} a_k is the sum of the first n-1 terms of {a_n}, which is S_{n-1} = 2a_{n-1} - 1. Substituting this into the expression for b_n:[b_n = 3 + 2a_{n-1} - 1 = 2 + 2a_{n-1}]Since a_{n-1} = 2^{n-2}, we have:[b_n = 2 + 2 cdot 2^{n-2} = 2 + 2^{n-1}]To find the sum of the first n terms of {b_n}, denoted T_n, we use:[T_n = sum_{k=1}^n b_k = sum_{k=1}^n (2 + 2^{k-1}) = sum_{k=1}^n 2 + sum_{k=1}^n 2^{k-1}]The first sum is straightforward:[sum_{k=1}^n 2 = 2n]The second sum is a geometric series:[sum_{k=1}^n 2^{k-1} = 2^n - 1]Combining these results, we get:[T_n = 2n + (2^n - 1) = 2^n + 2n - 1]Thus, the sum of the first n terms of the sequence {b_n} is:[boxed{2^n + 2n - 1}]

🔑:1. Given the sequence {a_n} with the partial sum S_n = 2a_n - 1 for ( n = 1, 2, cdots ), and the sequence {b_n} satisfying: - ( b_1 = 3 ) - ( b_{k+1} = a_k + b_k ) for ( k = 1, 2, cdots )2. Let's find the general term of the sequence {a_n}. We are given the partial sum ( S_n = 2a_n - 1 ). For ( n geq 2 ), we have the previous sum as ( S_{n-1} = 2a_{n-1} - 1 ). Therefore, by subtracting these two consecutive sums: [ S_n - S_{n-1} = (2a_n - 1) - (2a_{n-1} - 1) ] Simplifying this, we have: [ 2a_n - 2a_{n-1} = 0 implies a_n = a_{n-1} ] However, in the subtraction, we didn't arrive at a useful recurrence; instead, let's assume S_n propto 2^{n-1}. Note that if we understand the given conditions properly and revisit the assumption for a_n approx 2^{n-1}: [ a_n = 2 a_{n-1} implies a_n = 2^{n-1} quad text{(confirmed after trial and fitting boundary conditions)} ]3. Given ( b_1 = 3 ) and the recursive relation ( b_{k+1} = a_k + b_k ), we substitute a_k = 2^{k-1}: [ b_{k+1} = 2^{k-1} + b_k ] For instance: [ b_2 = 2^0 + b_1 = 1 + 3 = 4 ] [ b_3 = 2^1 + b_2 = 2 + 4 = 6 ] And so on. By induction or continued observation, we find: [ b_n = 2^{n-1} + 2 ]4. Determine the sum T_n = sum_{k=1}^n b_k using the general term b_k = 2^{k-1} + 2: [ T_n = sum_{k=1}^n (2^{k-1} + 2) ] Splitting this sum: [ T_n = sum_{k=1}^n 2^{k-1} + sum_{k=1}^n 2 ] For the first sum, which is a geometric series: [ sum_{k=1}^n 2^{k-1} = 2^0 + 2^1 + cdots + 2^{n-1} = 2^n - 1 ] For the second sum, which is simply an arithmetic sum of 2 added n times: [ sum_{k=1}^n 2 = 2n ]5. Combining these results, we have: [ T_n = (2^n - 1) + 2n = 2^n + 2n -1 ]Conclusion:[boxed{2^n + 2n - 1}]

🧠:Okay, so I need to find the minimum constant C such that the sum from k=1 to n of 1/(a_k a_{k+2}) is less than or equal to C for any positive integer n. The sequence a_n is defined as [(2 + sqrt(5))^n + 1/(2^n)], where [x] is the floor function, the greatest integer not exceeding x. Hmm. Let me try to break this down step by step.First, let's understand the sequence a_n. Each term is the floor of (2 + sqrt(5))^n plus 1/(2^n). Since (2 + sqrt(5)) is a number greater than 1, raising it to the power of n will make it grow exponentially. The term 1/(2^n) is a small positive number that gets added before taking the floor. So essentially, a_n is just the integer part of (2 + sqrt(5))^n + something very small. But since we take the floor, it's actually just floor((2 + sqrt(5))^n + 1/(2^n)). But because (2 + sqrt(5))^n is not an integer, adding a small positive number (1/(2^n)) and then taking the floor might just give floor((2 + sqrt(5))^n) + 0 or maybe sometimes +1? Wait, but if (2 + sqrt(5))^n is irrational, adding 1/(2^n) which is rational, maybe the sum is not an integer, so the floor would just be the integer part. Let me check for small n.Let’s compute a_1: (2 + sqrt(5))^1 + 1/2 = 2 + sqrt(5) + 0.5. The numerical value of sqrt(5) is approximately 2.236, so 2 + 2.236 is about 4.236, plus 0.5 gives 4.736. Then floor(4.736) is 4. So a_1 = 4.Similarly, a_2: (2 + sqrt(5))^2 + 1/4. Let's compute (2 + sqrt(5))^2: expand it. (a + b)^2 = a^2 + 2ab + b^2. So 2^2 + 2*2*sqrt(5) + (sqrt(5))^2 = 4 + 4 sqrt(5) + 5 = 9 + 4 sqrt(5). Numerically, that's 9 + 4*2.236 ≈ 9 + 8.944 ≈ 17.944, plus 0.25 is 18.194, so floor is 18. So a_2 = 18.a_3: (2 + sqrt(5))^3 + 1/8. Let's compute (2 + sqrt(5))^3. Let me use recursion or expansion. Alternatively, note that (2 + sqrt(5))^n + (2 - sqrt(5))^n is an integer, because when expanded, the terms with sqrt(5) cancel out. For example, (2 + sqrt(5)) + (2 - sqrt(5)) = 4, (2 + sqrt(5))^2 + (2 - sqrt(5))^2 = (9 + 4 sqrt(5)) + (9 - 4 sqrt(5)) = 18, etc. In general, (2 + sqrt(5))^n + (2 - sqrt(5))^n is an integer. Since 2 - sqrt(5) is approximately -0.236, its absolute value is less than 1, so (2 - sqrt(5))^n approaches 0 as n increases. Therefore, (2 + sqrt(5))^n is very close to an integer when n is large, specifically, it's equal to that integer minus (2 - sqrt(5))^n. Wait, but (2 + sqrt(5))^n + (2 - sqrt(5))^n is integer, so (2 + sqrt(5))^n = integer - (2 - sqrt(5))^n. Therefore, (2 + sqrt(5))^n is equal to that integer minus a small positive number (since (2 - sqrt(5)) is negative but raised to even power would be positive, odd power negative. Wait, actually, 2 - sqrt(5) is negative (since sqrt(5) ≈ 2.236 > 2), so 2 - sqrt(5) ≈ -0.236. Therefore, (2 - sqrt(5))^n alternates in sign: for even n, it's positive, for odd n, negative. But its magnitude is (sqrt(5) - 2)^n, which is less than 1. Therefore, (2 + sqrt(5))^n + (2 - sqrt(5))^n is integer, and (2 + sqrt(5))^n is that integer minus (2 - sqrt(5))^n. Therefore, depending on the parity of n, (2 + sqrt(5))^n is either slightly less than an integer (if n is even) or slightly more than an integer (if n is odd)? Wait, let's test with n=1: (2 + sqrt(5)) + (2 - sqrt(5)) = 4, which is integer. So (2 + sqrt(5)) = 4 - (2 - sqrt(5)). Since (2 - sqrt(5)) is negative, (2 + sqrt(5)) = 4 - negative ≈ 4 + 0.236 ≈ 4.236. So (2 + sqrt(5))^1 ≈ 4.236, so floor((2 + sqrt(5))^1) is 4. Similarly, (2 + sqrt(5))^2 = 9 + 4 sqrt(5) ≈ 17.944, and (2 - sqrt(5))^2 = 9 - 4 sqrt(5) ≈ 9 - 8.944 ≈ 0.056. Therefore, (2 + sqrt(5))^2 ≈ 17.944, and (2 + sqrt(5))^2 + (2 - sqrt(5))^2 = 18, so (2 + sqrt(5))^2 = 18 - (2 - sqrt(5))^2 ≈ 18 - 0.056 ≈ 17.944. So here, (2 + sqrt(5))^2 ≈ 17.944, so floor is 17. Wait, but earlier when we calculated a_2, we added 1/4 and then took the floor. Wait, the sequence a_n is [(2 + sqrt(5))^n + 1/(2^n)]. So in the case of n=2, (2 + sqrt(5))^2 + 1/4 ≈ 17.944 + 0.25 ≈ 18.194, floor is 18. So a_2 = 18. But (2 + sqrt(5))^2 ≈ 17.944, so floor((2 + sqrt(5))^2) = 17, but with the added 1/4, the floor becomes 18. So actually, the 1/(2^n) term is critical here because it might tip the value over to the next integer. So for each n, (2 + sqrt(5))^n is just less than an integer (if n is even) or just above an integer (if n is odd)? Wait, let me check n=1,2,3:n=1: (2 + sqrt(5))^1 ≈ 4.236, add 1/2 ≈ 0.5, total ≈ 4.736, floor 4. So a_1=4. But (2 + sqrt(5))^1 + (2 - sqrt(5))^1 = 4, so (2 + sqrt(5))^1 = 4 - (2 - sqrt(5))^1. Since (2 - sqrt(5)) ≈ -0.236, (2 - sqrt(5))^1 ≈ -0.236. Therefore, (2 + sqrt(5))^1 ≈ 4 - (-0.236) = 4.236. Then adding 0.5 gives 4.736, floor is 4. So a_1 is 4.n=2: (2 + sqrt(5))^2 ≈ 17.944, add 1/4 ≈ 0.25, total ≈ 18.194, floor is 18. But (2 + sqrt(5))^2 + (2 - sqrt(5))^2 = 18, so (2 + sqrt(5))^2 = 18 - (2 - sqrt(5))^2 ≈ 18 - 0.056 ≈ 17.944. So adding 0.25 gives 18.194, floor 18.n=3: (2 + sqrt(5))^3. Let's compute it. Let's note that (2 + sqrt(5))^3 = (2 + sqrt(5))*(9 + 4 sqrt(5)) = 2*(9 + 4 sqrt(5)) + sqrt(5)*(9 + 4 sqrt(5)) = 18 + 8 sqrt(5) + 9 sqrt(5) + 4*5 = 18 + 17 sqrt(5) + 20 = 38 + 17 sqrt(5). Numerically, that's 38 + 17*2.236 ≈ 38 + 38.012 ≈ 76.012. Adding 1/8 ≈ 0.125 gives ≈76.137, floor is 76. So a_3=76. But (2 + sqrt(5))^3 + (2 - sqrt(5))^3 is integer. Let's compute (2 + sqrt(5))^3 + (2 - sqrt(5))^3. Let’s compute using the identity for a^3 + b^3 = (a + b)(a^2 - ab + b^2). Here, a = 2 + sqrt(5), b = 2 - sqrt(5). Then a + b = 4, ab = (2)^2 - (sqrt(5))^2 = 4 - 5 = -1, a^2 + b^2 = (a + b)^2 - 2ab = 16 - 2*(-1) = 18. So a^3 + b^3 = 4*(18 - (-1)) = 4*19 = 76. Therefore, (2 + sqrt(5))^3 + (2 - sqrt(5))^3 =76. Therefore, (2 + sqrt(5))^3 =76 - (2 - sqrt(5))^3. Since (2 - sqrt(5))≈-0.236, (2 - sqrt(5))^3≈-0.236^3≈-0.013. So (2 + sqrt(5))^3≈76 - (-0.013)=76.013. Wait, but earlier when I calculated numerically, (2 + sqrt(5))^3≈76.012. Hmm, maybe approximation errors. Anyway, so (2 + sqrt(5))^3≈76.013, then adding 1/8=0.125 gives≈76.138, which still floors to 76. So a_3=76.Wait, but (2 + sqrt(5))^3 is≈76.013, then adding 1/8≈0.125 gives≈76.138, floor is 76. But if (2 + sqrt(5))^n is slightly more than an integer when n is odd? Wait, for n=1: (2 + sqrt(5))≈4.236, which is 4 + 0.236. For n=2:≈17.944, which is 17 +0.944. For n=3:≈76.013, which is 76 +0.013. Hmm, so n=1 and n=3, (2 + sqrt(5))^n is slightly above an integer, but for n=2, it's slightly below. Wait, that contradicts my earlier thought. Wait, but given that (2 + sqrt(5))^n + (2 - sqrt(5))^n is integer, and (2 - sqrt(5)) is negative with absolute value less than 1. Therefore, when n is even, (2 - sqrt(5))^n is positive, so (2 + sqrt(5))^n = integer - positive small number, so (2 + sqrt(5))^n is slightly less than integer. When n is odd, (2 - sqrt(5))^n is negative, so (2 + sqrt(5))^n = integer - negative small number = integer + positive small number, so slightly more than integer. Therefore, for even n, (2 + sqrt(5))^n is just below an integer, and for odd n, just above. Therefore, when we add 1/(2^n) to (2 + sqrt(5))^n and take the floor:For even n: (2 + sqrt(5))^n is just below integer m, so adding 1/(2^n) might push it to m or stay below m. For example, n=2: (2 + sqrt(5))^2≈17.944, add 0.25≈18.194, floor is 18. So even though (2 + sqrt(5))^2 is≈17.944, adding 1/4 gets it over 18, so floor is 18. For n=4: let's compute (2 + sqrt(5))^4. Let's use recursion. Let’s note that (2 + sqrt(5))^{n+1} = (2 + sqrt(5))*(2 + sqrt(5))^n. But maybe a better approach is to use the recurrence relation. Since (2 + sqrt(5)) and (2 - sqrt(5)) are roots of the quadratic equation x^2 -4x -1=0. Therefore, the sequence (2 + sqrt(5))^n + (2 - sqrt(5))^n satisfies the recurrence relation: S_n = 4S_{n-1} + S_{n-2} with S_0 = 2, S_1 = 4. Let me check: S_2 = 4*S_1 + S_0 = 4*4 + 2 = 18, which matches. S_3 =4*18 +4=76, S_4=4*76 +18= 322. Therefore, (2 + sqrt(5))^4 + (2 - sqrt(5))^4=322. So (2 + sqrt(5))^4=322 - (2 - sqrt(5))^4. Since (2 - sqrt(5))^4 is positive (even exponent), so (2 + sqrt(5))^4≈322 - small positive≈321.999... Hmm, but (2 + sqrt(5))^4 is actually equal to 322 - (something less than 1). Let's compute (2 - sqrt(5))^4. Since 2 - sqrt(5)≈-0.236, so (-0.236)^4≈0.236^4≈(0.236^2)^2≈(0.0556)^2≈0.00309. So (2 + sqrt(5))^4≈322 -0.00309≈321.9969. Adding 1/16≈0.0625 gives≈322.0594, floor is 322. So a_4=322.So in this case, for even n, (2 + sqrt(5))^n is slightly less than an integer, but adding 1/(2^n) can sometimes push it over to the next integer. For example, n=2: added 0.25 to≈17.944 gives≈18.194. For n=4: added≈0.0625 to≈321.9969 gives≈322.0594. So floor is 322. For odd n, like n=1,3: (2 + sqrt(5))^n is slightly more than integer. Adding 1/(2^n) will make it slightly more, but the floor will still be the integer part. For example, n=1: (2 + sqrt(5))≈4.236, add 0.5≈4.736, floor 4. n=3:≈76.013 + 0.125≈76.138, floor 76. So in all cases, a_n = floor((2 + sqrt(5))^n + 1/(2^n)) equals the integer S_n - 1 when n is even? Wait, S_n is (2 + sqrt(5))^n + (2 - sqrt(5))^n. For even n: (2 + sqrt(5))^n = S_n - (2 - sqrt(5))^n. Since (2 - sqrt(5))^n is positive, (2 + sqrt(5))^n = S_n - small positive. So (2 + sqrt(5))^n + 1/(2^n) = S_n - small positive + 1/(2^n). Depending on whether 1/(2^n) is bigger than the small positive, which is (2 - sqrt(5))^n. Let's see: For even n, (2 - sqrt(5))^n = (sqrt(5)-2)^n, which is approximately (0.236)^n. So 1/(2^n) versus (0.236)^n. Since 0.236 < 0.5, so (0.236)^n decreases faster than 1/(2^n). For example, n=2: (0.236)^2≈0.055, 1/4=0.25, so 0.25>0.055. So 1/(2^n) > (sqrt(5)-2)^n for even n. Therefore, for even n: (2 + sqrt(5))^n + 1/(2^n) = S_n - (sqrt(5)-2)^n + 1/(2^n) > S_n, since 1/(2^n) - (sqrt(5)-2)^n >0. Therefore, floor((2 + sqrt(5))^n +1/(2^n))=S_n -1 +1= S_n? Wait, because S_n is integer, and (2 + sqrt(5))^n +1/(2^n) is S_n - (sqrt(5)-2)^n +1/(2^n). Since 1/(2^n) > (sqrt(5)-2)^n, then S_n - (sqrt(5)-2)^n +1/(2^n) is S_n + (1/(2^n) - (sqrt(5)-2)^n). Since both terms are positive, so this is greater than S_n. But since (1/(2^n) - (sqrt(5)-2)^n) is less than 1 (because 1/(2^n) <1 and (sqrt(5)-2)^n <1), so the total is S_n + something less than 1, so floor would be S_n. Wait, but (2 + sqrt(5))^n +1/(2^n) is S_n - (sqrt(5)-2)^n +1/(2^n). For even n:Let me take n=2: S_2=18, (sqrt(5)-2)^2≈0.055, 1/4=0.25. So (2 + sqrt(5))^2 +1/4≈17.944 +0.25≈18.194, which is S_2 + (0.25 -0.055)=18 +0.195≈18.195. So floor is 18, which is S_2. Similarly, n=4: S_4=322, (sqrt(5)-2)^4≈0.00309, 1/16≈0.0625. So (2 + sqrt(5))^4 +1/16≈321.9969 +0.0625≈322.0594, floor is 322=S_4.For odd n: (2 + sqrt(5))^n = S_n - (2 - sqrt(5))^n. But since n is odd, (2 - sqrt(5))^n is negative. So (2 + sqrt(5))^n = S_n - negative = S_n + positive. So (2 + sqrt(5))^n is S_n + |(2 - sqrt(5))^n|. Then adding 1/(2^n): (2 + sqrt(5))^n +1/(2^n) = S_n + |(2 - sqrt(5))^n| +1/(2^n). Since |(2 - sqrt(5))^n| = (sqrt(5)-2)^n, which is less than 1/(2^n) for n >=1? Let's check n=1: (sqrt(5)-2)^1≈0.236, 1/2=0.5. So 0.236 <0.5. For n=3: (0.236)^3≈0.013, 1/8≈0.125. Still 0.013 <0.125. So for odd n: |(2 - sqrt(5))^n| +1/(2^n) = (sqrt(5)-2)^n +1/(2^n). Since (sqrt(5)-2)≈0.236 <0.5, so (sqrt(5)-2)^n < (0.5)^n. Therefore, (sqrt(5)-2)^n +1/(2^n) < (0.5)^n + (0.5)^n = 2*(0.5)^n =1/(2^{n-1}) <1 for n>=1. Therefore, (2 + sqrt(5))^n +1/(2^n) = S_n + something less than 1, so floor is S_n. Wait, but in the case of n=1: (2 + sqrt(5)) +1/2≈4.236 +0.5≈4.736, floor is 4, but S_1=4. Wait, S_1=(2 + sqrt(5)) + (2 - sqrt(5))=4. So (2 + sqrt(5))^1 +1/2=4.736, floor is 4, which is S_1 -0.736. Wait, but according to the previous logic, S_n + something less than 1, but S_n is 4, and the total is 4.736. Wait, that contradicts. Wait, for odd n: (2 + sqrt(5))^n = S_n - (2 - sqrt(5))^n. Since n is odd, (2 - sqrt(5))^n is negative. Therefore, (2 + sqrt(5))^n = S_n - negative = S_n + |(2 - sqrt(5))^n|. Then adding 1/(2^n): total is S_n + |(2 - sqrt(5))^n| +1/(2^n). Since |(2 - sqrt(5))^n| +1/(2^n) is positive. So for example, n=1: S_1=4, |(2 - sqrt(5))^1|=sqrt(5)-2≈0.236, 1/2=0.5. So total is 4 +0.236 +0.5=4.736. So floor is 4. So floor(S_n + something less than 1) = S_n. But here, floor(4.736)=4, which is S_1=4. Similarly, n=3: S_3=76, |(2 - sqrt(5))^3|≈0.013, 1/8≈0.125. So total is 76 +0.013 +0.125≈76.138, floor is 76=S_3. Therefore, in all cases, a_n = floor((2 + sqrt(5))^n +1/(2^n))=S_n. Wait, because when n is even, (2 + sqrt(5))^n +1/(2^n) is slightly above S_n - (sqrt(5)-2)^n +1/(2^n), which is S_n + (1/(2^n) - (sqrt(5)-2)^n). Since 1/(2^n) > (sqrt(5)-2)^n for even n, as we saw, so total is S_n + positive amount less than 1, so floor is S_n. For odd n, (2 + sqrt(5))^n +1/(2^n) is S_n + |(2 - sqrt(5))^n| +1/(2^n), which is S_n + something less than 1, so floor is S_n. Therefore, in general, a_n = S_n. So the sequence a_n is actually equal to S_n, which is (2 + sqrt(5))^n + (2 - sqrt(5))^n. Therefore, a_n = (2 + sqrt(5))^n + (2 - sqrt(5))^n. Because floor((2 + sqrt(5))^n +1/(2^n))=S_n. Because (2 + sqrt(5))^n +1/(2^n) is less than S_n +1, and greater than S_n (for even n, since adding 1/(2^n) > (sqrt(5)-2)^n), and for odd n, (2 + sqrt(5))^n is S_n - (2 - sqrt(5))^n, but (2 - sqrt(5))^n is negative, so (2 + sqrt(5))^n = S_n + |(2 - sqrt(5))^n|. Adding 1/(2^n) gives S_n + |(2 - sqrt(5))^n| +1/(2^n), which is less than S_n +1, because |(2 - sqrt(5))^n| +1/(2^n) <1. Therefore, in both cases, floor of that is S_n. Therefore, a_n = S_n = (2 + sqrt(5))^n + (2 - sqrt(5))^n.Wait, let me verify with n=1: a_1=4, which is (2 + sqrt(5)) + (2 - sqrt(5))=4, which matches. n=2:18, which is (2 + sqrt(5))^2 + (2 - sqrt(5))^2=18. n=3:76, same as above. So indeed, a_n = S_n. Therefore, the sequence a_n is actually the integer sequence defined by a_n = (2 + sqrt(5))^n + (2 - sqrt(5))^n. That's a key insight. Therefore, even though the problem defines a_n as floor((2 + sqrt(5))^n +1/(2^n)), it turns out that this floor is exactly equal to S_n. Therefore, we can treat a_n as S_n, which is an integer sequence.So, given that, the problem reduces to summing 1/(a_k a_{k+2}) from k=1 to n, and finding the minimal C such that the sum is <=C for all n.So, let's first try to analyze the sum S = sum_{k=1}^n 1/(a_k a_{k+2}).Given that a_n satisfies a recurrence relation. Earlier, I noted that S_n =4 S_{n-1} + S_{n-2}. Let's confirm that.Given that a_n = (2 + sqrt(5))^n + (2 - sqrt(5))^n.Let’s check if this satisfies a recurrence. Let’s suppose that a_n = p a_{n-1} + q a_{n-2}. Let's compute for n=2: a_2=18, a_1=4, a_0=2 (if we define a_0). Wait, let's compute a_0: (2 + sqrt(5))^0 + (2 - sqrt(5))^0=1 +1=2. So a_0=2. Let's see:a_2 = 18, 4 a_1 + a_0 =4*4 +2=18. So that works. a_3=76, 4 a_2 +a_1=4*18 +4=76. Yes. Therefore, the recurrence is a_n =4 a_{n-1} +a_{n-2} with a_0=2, a_1=4. Therefore, the sequence a_n satisfies a linear recurrence relation of order 2.Given this recurrence, perhaps we can express 1/(a_k a_{k+2}) in terms that telescope. Let's see.Let me consider the term 1/(a_k a_{k+2}). Maybe we can find constants A and B such that 1/(a_k a_{k+2}) = A (1/a_k - 1/a_{k+2}) or something similar. Alternatively, using the recurrence relation to relate a_{k+2} to a_{k+1} and a_k. From the recurrence, a_{k+2}=4 a_{k+1} +a_k. Therefore, perhaps we can write 1/(a_k a_{k+2}) =1/(a_k (4 a_{k+1} +a_k)). But not sure if that helps. Alternatively, maybe we can find a telescoping expression.Alternatively, consider that since a_{k+2} =4 a_{k+1} +a_k, then maybe we can relate reciprocal terms. Let me consider the difference between terms. Let's see:Suppose we take 1/(a_k a_{k+1}) and see how it relates to 1/(a_{k+1} a_{k+2}). Maybe using the recurrence to express a relation between these terms. Let's compute 1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2}) = [a_{k+2} -a_k]/(a_k a_{k+1} a_{k+2}). But a_{k+2}=4a_{k+1} +a_k, so a_{k+2} -a_k=4 a_{k+1}. Therefore, the difference becomes [4 a_{k+1}]/(a_k a_{k+1} a_{k+2})=4/(a_k a_{k+2}). Therefore, 4/(a_k a_{k+2})= [1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2})]. Therefore, 1/(a_k a_{k+2})= (1/4)[1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2})].That's a crucial step! Therefore, each term 1/(a_k a_{k+2}) can be written as (1/4) times the difference between 1/(a_k a_{k+1}) and 1/(a_{k+1} a_{k+2}}). Therefore, when we sum over k from 1 to n, we get a telescoping sum.Let me verify this. Let's take the expression:1/(a_k a_{k+2}) = (1/4)[1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2})]Multiply both sides by 4:4/(a_k a_{k+2}) =1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2})Therefore, the sum from k=1 to n of 1/(a_k a_{k+2}) equals (1/4) times the sum from k=1 to n of [1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2})].This is a telescoping series, so when we expand the sum:Sum_{k=1}^n [1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2})] = [1/(a_1 a_2) -1/(a_2 a_3)] + [1/(a_2 a_3) -1/(a_3 a_4)] + ... + [1/(a_n a_{n+1}) -1/(a_{n+1} a_{n+2})].All the intermediate terms cancel out, leaving:Sum = 1/(a_1 a_2) -1/(a_{n+1} a_{n+2}).Therefore, the original sum Sum_{k=1}^n 1/(a_k a_{k+2}) = (1/4)[1/(a_1 a_2) -1/(a_{n+1} a_{n+2})].Since all terms a_n are positive, 1/(a_{n+1} a_{n+2}) is positive, so the sum is less than (1/4)(1/(a_1 a_2)). But let's compute a_1 and a_2. From earlier:a_1=4, a_2=18. Therefore, 1/(a_1 a_2)=1/(4*18)=1/72≈0.0139.Therefore, the sum is (1/4)(1/72 -1/(a_{n+1} a_{n+2})).But since 1/(a_{n+1} a_{n+2}) is positive, the entire sum is less than (1/4)(1/72)=1/288≈0.00347. Wait, but that contradicts the earlier examples. Wait, wait, perhaps I messed up the coefficients.Wait, according to the previous step:Sum_{k=1}^n 1/(a_k a_{k+2}) = (1/4)[1/(a_1 a_2) -1/(a_{n+1} a_{n+2})].Therefore, as n approaches infinity, the sum approaches (1/4)(1/(a_1 a_2)) because 1/(a_{n+1} a_{n+2}) approaches zero (since a_n grows exponentially). Therefore, the total sum is bounded by 1/(4 a_1 a_2). But a_1=4, a_2=18, so 1/(4*4*18)=1/(4*72)=1/288≈0.00347. But that seems too small. Wait, let me check the algebra again.Wait, we had:4/(a_k a_{k+2})=1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2}}).Therefore, 1/(a_k a_{k+2})= (1/4)[1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2}}).Therefore, summing over k=1 to n:Sum_{k=1}^n 1/(a_k a_{k+2})= (1/4) Sum_{k=1}^n [1/(a_k a_{k+1}) -1/(a_{k+1} a_{k+2})].The sum on the right telescopes to (1/4)[1/(a_1 a_2) -1/(a_{n+1} a_{n+2})].Therefore, the total sum is (1/4)(1/(a_1 a_2) -1/(a_{n+1} a_{n+2})).Given that a_1=4, a_2=18, then 1/(a_1 a_2)=1/72≈0.013888. Then multiplying by 1/4 gives≈0.003472. But the sum from k=1 to n of 1/(a_k a_{k+2}) is equal to 1/4*(1/72 -1/(a_{n+1} a_{n+2})).Wait, but then as n increases, the term 1/(a_{n+1} a_{n+2}) becomes very small because a_n grows exponentially. Therefore, the sum approaches 1/(4*72)=1/288≈0.003472. But that would mean the sum is bounded by 1/288. But this contradicts my earlier calculation for n=1: sum from k=1 to 1 of 1/(a_1 a_3)=1/(4*76)=1/304≈0.003289, which is less than 1/288≈0.003472. Wait, but according to the formula, when n=1, the sum is (1/4)(1/72 -1/(a_2 a_3)). Let's compute:For n=1: (1/4)(1/(4*18) -1/(18*76))= (1/4)(1/72 -1/(1368))≈(1/4)(0.013888 -0.000731)≈(1/4)(0.013157)=≈0.003289, which matches 1/(4*76)=1/304≈0.003289. Similarly, for n=2: sum from k=1 to 2 of 1/(a_k a_{k+2})=1/(4*76) +1/(18*322)=≈0.003289 +1/(5796)≈0.003289 +0.000172≈0.003461. According to the formula: (1/4)(1/72 -1/(a_3 a_4))= (1/4)(1/72 -1/(76*322))≈(1/4)(0.013888 -1/24472)≈(1/4)(0.013888 -0.0000408)≈(1/4)(0.013847)=≈0.0034618, which matches. So as n increases, the sum approaches 1/(4*72)=1/288≈0.003472. Therefore, the sum is always less than 1/288 and approaches it as n approaches infinity. Therefore, the minimal constant C is 1/288. But wait, 1/288 is approximately 0.003472, but when n=2, the sum is≈0.003461, which is slightly less than 1/288. Wait, but 1/288≈0.003472222..., so the sum approaches 1/288 from below. Therefore, the sum is strictly increasing and approaching 1/288. Therefore, the supremum of the sum over all n is 1/288, so the minimal C is 1/288. Therefore, the answer should be boxed{dfrac{1}{288}}.But let me verify once more. Since the sum is (1/4)(1/(a_1 a_2) -1/(a_{n+1} a_{n+2})). As n approaches infinity, 1/(a_{n+1} a_{n+2}) approaches zero, so the sum approaches (1/4)(1/(a_1 a_2))=1/(4*4*18)=1/288. Therefore, since the sum is always less than 1/288 and approaches it, the minimal C is 1/288. Therefore, the answer is 1/288.Final Answerboxed{dfrac{1}{288}}

✅:Given the sequence {a_n} defined bya_n = leftlfloor (2 + sqrt{5})^n + frac{1}{2^n} rightrfloor quad (n in mathbf{Z}_+),where lfloor x rfloor denotes the greatest integer not exceeding x, we need to determine the minimum value of the constant C such that for any positive integer n, the inequalitysum_{k=1}^{n} frac{1}{a_k a_{k+2}} leqslant Cholds.First, we recognize that the sequence a_n is actually the integer sequence defined by a_n = (2 + sqrt{5})^n + (2 - sqrt{5})^n. This sequence satisfies the recurrence relation a_n = 4a_{n-1} + a_{n-2} with initial terms a_0 = 2 and a_1 = 4.Next, we express the term frac{1}{a_k a_{k+2}} using the recurrence relation. We find that:frac{1}{a_k a_{k+2}} = frac{1}{4} left( frac{1}{a_k a_{k+1}} - frac{1}{a_{k+1} a_{k+2}} right).Summing this from k=1 to n results in a telescoping series:sum_{k=1}^{n} frac{1}{a_k a_{k+2}} = frac{1}{4} left( frac{1}{a_1 a_2} - frac{1}{a_{n+1} a_{n+2}} right).Since a_1 = 4 and a_2 = 18, we have frac{1}{a_1 a_2} = frac{1}{72}. As n approaches infinity, the term frac{1}{a_{n+1} a_{n+2}} approaches zero. Therefore, the sum converges to:frac{1}{4} cdot frac{1}{72} = frac{1}{288}.Thus, the minimum value of the constant C such that the inequality holds for all positive integers n is:boxed{dfrac{1}{288}}.

🔑:1. Define constants and initial analysis: Let ( x_1 = 2 + sqrt{5} ) and ( x_2 = 2 - sqrt{5} ). Then we have: [ a_n = left[ x_1^n + frac{1}{2^n} right] ] Also, define ( T_n = x_1^n + x_2^n ).2. Recurrence relation and initial values: Using the properties of ( x_1 ) and ( x_2 ), we can derive the following recurrence relation: [ T_{n+2} = (x_1 + x_2) T_{n+1} - x_1 x_2 T_n ] Here, ( x_1 + x_2 = 4 ) and ( x_1 x_2 = (2 + sqrt{5})(2 - sqrt{5}) = -1 ). Therefore, the recurrence becomes: [ T_{n+2} = 4T_{n+1} + T_n ] The initial values can be calculated as: [ T_1 = x_1 + x_2 = 4 ] [ T_2 = x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2 x_1 x_2 = 4^2 - 2(-1) = 16 + 2 = 18 ]3. Integral properties and approximation: Notice that ( -frac{1}{2} < 2 - sqrt{5} < 0 ), hence ( |x_2| < frac{1}{2} ). Therefore, for any positive integer ( n ), we have: [ -frac{1}{2^n} < x_2^n < frac{1}{2^n} ] Hence, [ x_1^n + x_2^n - 1 < x_1^n + frac{1}{2^n} - 1 < a_n leq x_1^n + frac{1}{2^n} = x_1^n + (-1 + frac{1}{2^n}) + 1 < x_1^n + x_2^n + 1 ] Since (x_1^n + x_2^n - 1), ( a_n ), and ( x_1^n + x_2^n + 1) are integers, we have: [ a_n = x_1^n + x_2^n = T_n ] Thus, ( a_n ) satisfies the recurrence relation (a_{n+2} = 4a_{n+1} + a_n ) with ( a_1 = 4 ) and ( a_2 = 18 ).4. Evaluate series sum expression: [ frac{1}{a_k a_{k+2}} = frac{1}{4} cdot frac{4a_{k+1}}{a_k a_{k+1} a_{k+2}} = frac{1}{4} cdot frac{a_{k+2} - a_k}{a_k a_{k+1} a_{k+2}} = frac{1}{4} left( frac{1}{a_k a_{k+1}} - frac{1}{a_{k+1} a_{k+2}} right) ] Therefore, [ sum_{k=1}^n frac{1}{a_k a_{k+2}} = frac{1}{4} sum_{k=1}^n left( frac{1}{a_k a_{k+1}} - frac{1}{a_{k+1} a_{k+2}} right) = frac{1}{4} left( frac{1}{a_1 a_2} - frac{1}{a_{n+1} a_{n+2}} right) ] Given ( a_1 = 4 ) and ( a_2 = 18 ), we find: [ frac{1}{a_1 a_2} = frac{1}{4 cdot 18} = frac{1}{72} ] Hence, [ sum_{k=1}^n frac{1}{a_k a_{k+2}} = frac{1}{4} left( frac{1}{72} - frac{1}{a_{n+1} a_{n+2}} right) = frac{1}{288} - frac{1}{4 a_{n+1} a_{n+2}} ]5. Bounding and limits: Notice ( a_n > 0 ) and ( a_{n+2} > 4a_{n+1} ). Thus, as ( n to infty ): [ a_n geq 4^{n-2} a_2 text{ for } n geq 2 ] Thus, [ lim_{n to infty} frac{1}{a_{n+1} a_{n+2}} = 0 ] Therefore, [ sum_{k=1}^n frac{1}{a_k a_{k+2}} < frac{1}{288} ] and [ lim_{n to infty} sum_{k=1}^n frac{1}{a_k a_{k+2}} = frac{1}{288} ]6. Conclusion: The smallest value of ( C ) satisfying the given inequality is: [ boxed{frac{1}{288}} ]

🧠:Okay, so I need to show that ( n^{7} - n^{3} ) is divisible by 120 for every natural number n. Hmm, let's start by understanding what the problem is asking. Divisible by 120 means that when you divide ( n^{7} - n^{3} ) by 120, there's no remainder. 120 factors into prime factors as ( 2^3 times 3 times 5 ). So, if I can show that ( n^{7} - n^{3} ) is divisible by 8, 3, and 5 for any natural number n, then it would be divisible by 120. That seems like a good approach because if a number is divisible by each of the prime power components of a composite number, then it's divisible by the composite itself. First, let me verify that 120's prime factors are indeed 8, 3, and 5. 8 is ( 2^3 ), and 3 and 5 are primes. Multiplying them together: 8*3=24, 24*5=120. Yep, that's correct. So the plan is to check divisibility by 8, 3, and 5 separately.Starting with divisibility by 8. Let's look at ( n^{7} - n^{3} ). Maybe factor out ( n^3 ): that gives ( n^3(n^4 - 1) ). Then, factor ( n^4 - 1 ) as a difference of squares: ( (n^2)^2 - 1^2 = (n^2 - 1)(n^2 + 1) ). Then, ( n^2 - 1 ) factors further into (n - 1)(n + 1). So altogether, the expression becomes ( n^3(n - 1)(n + 1)(n^2 + 1) ).Hmm, so for divisibility by 8, we need to ensure that the product of these terms has at least three factors of 2. Let's think about the factors of 2 in each part.First, consider n^3. Depending on whether n is even or odd, n^3 will have different factors of 2. If n is even, say n = 2k, then n^3 = 8k^3, which contributes three factors of 2. Then, even if the rest of the terms don't contribute any, we already have 8 as a factor. If n is odd, then n^3 is odd, so we need the other terms (n - 1)(n + 1)(n^2 + 1) to contribute the necessary factors of 2.If n is odd, then both (n - 1) and (n + 1) are even numbers. Because they are consecutive even numbers (since n is odd, subtracting and adding 1 makes even numbers two apart), one of them is divisible by 4. So, (n - 1)(n + 1) would be divisible by 2*4 = 8. But wait, n is odd, so let me check:Take n = 2k + 1, then n - 1 = 2k, and n + 1 = 2k + 2 = 2(k + 1). So, their product is 2k * 2(k + 1) = 4k(k + 1). Now, k and k + 1 are consecutive integers, so one of them is even. Therefore, 4k(k + 1) is divisible by 8. Therefore, (n - 1)(n + 1) is divisible by 8 when n is odd. Then, multiplying by n^2 + 1: but n is odd, so n^2 is 1 mod 4, so n^2 + 1 is 2 mod 4, which is 2*(something). Therefore, the entire term (n - 1)(n + 1)(n^2 + 1) would be divisible by 8*2 = 16 when n is odd. So even though n^3 is odd, the rest contributes 16, so total is 16, which is divisible by 8. Wait, but actually, in the expression, we have n^3 multiplied by that. Wait, but if n is odd, n^3 is odd, and the other terms (n - 1)(n + 1)(n^2 + 1) contribute 16, so total factors of 2 would be 16, which is more than enough for 8. If n is even, n^3 gives 8, and the other terms (n - 1)(n + 1)(n^2 + 1) would be odd*odd*odd if n is even. Wait, if n is even, then n - 1 and n + 1 are odd, and n^2 + 1 is also odd (since even squared is even, plus 1 is odd). So, when n is even, the other terms contribute nothing, but n^3 already has 8. So in either case, the entire expression is divisible by 8. Therefore, divisibility by 8 is proven.Next, divisibility by 3. To show that ( n^7 - n^3 ) is divisible by 3. Let's consider modulo 3. By Fermat's little theorem, for any integer n, ( n^3 equiv n mod 3 ). Therefore, ( n^3 equiv n mod 3 ), so ( n^7 = n^{3 + 4} = n^3 cdot n^4 equiv n cdot n^4 = n^5 mod 3 ). Hmm, but that's not immediately helpful. Maybe better to use Fermat's little theorem again. Since 3 is prime, ( n^3 equiv n mod 3 ), so ( n^7 = n^{3 times 2 +1} = (n^3)^2 cdot n equiv n^2 cdot n = n^3 equiv n mod 3 ). Therefore, ( n^7 equiv n mod 3 ), so ( n^7 - n equiv 0 mod 3 ). But our expression is ( n^7 - n^3 ). Wait, so maybe compute ( n^7 - n^3 ) modulo 3. Let's see:( n^7 - n^3 = n^3(n^4 - 1) ). Maybe factor ( n^4 -1 = (n^2 -1)(n^2 +1) ). But modulo 3, possible residues are 0,1,2.Alternatively, let's test n modulo 3. For n ≡ 0 mod 3: then n^7 - n^3 ≡ 0 - 0 = 0 mod 3.For n ≡ 1 mod 3: 1^7 -1^3 =1 -1=0 mod3.For n ≡2 mod3: 2^7=128, 128 mod3: 128 divided by3 is 42*3=126, 128-126=2, so 2 mod3. Similarly, 2^3=8 mod3=2. Therefore, 2 -2=0 mod3. Therefore, in all cases, n^7 -n^3 ≡0 mod3. So divisible by3.Okay, that works. So divisibility by3 is established.Now, divisibility by5. Let's check modulo5. We need to show that ( n^7 -n^3 equiv0 mod5 ). Again, perhaps using Fermat's little theorem. Since 5 is prime, Fermat's little theorem tells us that ( n^4 equiv1 mod5 ) if n is not divisible by5. If n is divisible by5, then n^7 -n^3 ≡0 -0=0 mod5.For n not divisible by5, ( n^4 ≡1 mod5 ). Then, n^7 =n^{4 +3}=n^4 *n^3 ≡1 *n^3 =n^3 mod5. Therefore, n^7 -n^3 ≡n^3 -n^3=0 mod5. Therefore, in both cases, whether n is divisible by5 or not, ( n^7 -n^3 ≡0 mod5 ). Therefore, divisible by5.So putting it all together: since ( n^7 -n^3 ) is divisible by8,3, and5, and these are pairwise coprime (8 is 2^3, and 3 and5 are primes different from2), then by the Chinese remainder theorem, the expression is divisible by8*3*5=120. Hence, proven.But wait, just to make sure, let me check with some example numbers.Take n=1: 1^7 -1^3=1-1=0, which is divisible by120.n=2: 128 -8=120. 120/120=1. So yes.n=3:3^7=2187, 3^3=27. 2187-27=2160. 2160/120=18. So yes.n=4:4^7=16384, 4^3=64. 16384-64=16320. 16320/120=136. So divisible.n=5:5^7=78125, 5^3=125. 78125-125=78000. 78000/120=650. Yes.n=6:6^7=279936, 6^3=216. 279936-216=279720. 279720/120=2331. So yes.Okay, so these examples work. Therefore, the reasoning seems correct. So the key steps were factoring the expression, applying Fermat's little theorem for primes3 and5, and checking divisibility by8 through case analysis based on parity ofn. Therefore, the original statement is true.Final AnswerThe expression ( n^{7} - n^{3} ) is divisible by 120 for every natural number ( n ), as shown by the reasoning above. The final answer is boxed{120}.

✅:To show that ( n^{7} - n^{3} ) is divisible by 120 for every natural number ( n ), we need to demonstrate that it is divisible by the prime power components of 120, which are ( 8 ) (i.e., ( 2^3 )), ( 3 ), and ( 5 ). Divisibility by 81. Case 1: ( n ) is even If ( n ) is even, ( n^3 ) is divisible by ( 8 ). Hence, ( n^3(n^4 - 1) ) is divisible by ( 8 ).2. Case 2: ( n ) is odd If ( n ) is odd, then ( n - 1 ) and ( n + 1 ) are consecutive even numbers, one of which is divisible by 4. Thus, their product ( (n - 1)(n + 1) ) is divisible by ( 8 ). Additionally, ( n^2 + 1 ) is even, contributing another factor of 2. Therefore, the product ( n^3(n - 1)(n + 1)(n^2 + 1) ) is divisible by ( 8 ). Divisibility by 3Using Fermat's Little Theorem:- For any integer ( n ), ( n^3 equiv n mod 3 ).- ( n^7 equiv n^3 mod 3 ).- Therefore, ( n^7 - n^3 equiv n^3 - n^3 equiv 0 mod 3 ). Divisibility by 5Using Fermat's Little Theorem:- For any integer ( n ), ( n^4 equiv 1 mod 5 ) if ( n ) is not divisible by 5.- If ( n ) is divisible by 5, ( n^7 - n^3 equiv 0 mod 5 ).- If ( n ) is not divisible by 5, ( n^7 equiv n^3 mod 5 ), so ( n^7 - n^3 equiv 0 mod 5 ). ConclusionSince ( n^{7} - n^{3} ) is divisible by ( 8 ), ( 3 ), and ( 5 ), and these divisors are pairwise coprime, it follows by the Chinese Remainder Theorem that ( n^{7} - n^{3} ) is divisible by ( 8 times 3 times 5 = 120 ).[boxed{120}]

🔑:To show that ( n^{7} - n^{3} ) is divisible by 120 for all natural numbers ( n ), we need to show that it is divisible by 3, 5, and 8. This is because 120 can be factored as ( 3 times 5 times 8 ) and these numbers are pairwise relatively prime.Let's rewrite ( n^{7} - n^{3} ) in a factored form. [n^{7} - n^{3} = n^{3}(n^{4} - 1)]Noting that ( n^4 - 1 ) is a difference of squares, we can further factor it:[n^4 - 1 = (n^2 - 1)(n^2 + 1)]And then ( n^2 - 1 ) can be factored again as another difference of squares:[n^2 - 1 = (n - 1)(n + 1)]Thus,[n^{7} - n^{3} = n^{3}(n^2 - 1)(n^2 + 1) = n^3 (n - 1)(n + 1)(n^2 + 1)]Breaking it down:[n^{7} - n^{3} = (n - 1) cdot n cdot (n + 1) cdot n^2 cdot (n^2 + 1)]Now, we show divisibility by 3, 5, and 8.1. Divisibility by 3: The numbers ((n - 1)), (n), and ((n + 1)) are three consecutive integers. Thus, one of them must be divisible by 3. Therefore, the product ((n - 1)n(n + 1)) is divisible by 3. Since (n^3) and (n^2 + 1) do not affect the overall divisibility by 3, (n^{7} - n^{3}) is divisible by 3. 2. Divisibility by 8: - If ( n ) is even, write ( n = 2k ). Then, [ n^3 = (2k)^3 = 8k^3 ] which is clearly divisible by 8. Therefore, ( n^{7} - n^{3} ) is divisible by 8. - If ( n ) is odd, ( n^2 ) is odd, so both ( (n^2 + 1) ) and ( (n - 1) cdot n cdot (n + 1) ) will contain even factors: ( n ) is odd, and both ( n-1 ) and ( n+1 ) are even. In fact, one of ( n-1 ) or ( n+1 ) is divisible by 4 while the other is divisible by 2, given their overall evenness. This means ( (n-1) cdot (n+1) ) contributes ( 8 ). Thus, ((n-1)cdot ncdot (n+1)) is divisible by 8 for odd ( n ).3. Divisibility by 5: - If ( n equiv 0 text{ or } pm 1 (text{mod } 5) ), one of ( n, n-1, n+1 ) is divisible by 5. Hence, the product ((n-1)n(n+1)) is divisible by 5. - If ( n equiv 2 text{ or } 3 (text{mod } 5) ): For ( n = 5k + 2 ), we calculate: [ n^2 + 1 = (5k + 2)^2 + 1 = 25k^2 + 20k + 4 + 1 = 25k^2 + 20k + 5 = 5(5k^2 + 4k + 1) ] so ( n^2 + 1 ) is divisible by 5. For ( n = 5k + 3 ), we similarly calculate: [ n^2 + 1 = (5k + 3)^2 + 1 = 25k^2 + 30k + 9 + 1 = 25k^2 + 30k + 10 = 5(5k^2 + 6k + 2) ] so ( n^2 + 1 ) is divisible by 5.Since ( n^{7} - n^{3} ) is divisible by 3, 5, and 8 for any natural number ( n ), it must be divisible by ( 120 = 3 times 5 times 8 ).[boxed{text{120 is a divisor of } n^{7} - n^{3} text{ for all natural } n}]

🧠:Alright, let me try to work through this problem step by step. It's a bit complex, but I'll take it slowly.First, let me restate the problem to make sure I understand it correctly. We have 1991 numbers: x₁, x₂, ..., x₁₉₉₁. These numbers satisfy the condition that the sum of the absolute differences between consecutive terms is 1991. In other words:|x₁ - x₂| + |x₂ - x₃| + ... + |x₁₉₉₀ - x₁₉₉₁| = 1991.Then, for each k from 1 to 1991, we define yₖ as the average of the first k x's: yₖ = (x₁ + x₂ + ... + xₖ)/k.The task is to find the maximum possible value of the sum:|y₁ - y₂| + |y₂ - y₃| + ... + |y₁₉₉₀ - y₁₉₉₁|.So, we need to maximize the sum of the absolute differences between consecutive y's, given that the sum of the absolute differences between consecutive x's is 1991.Hmm, okay. Let me think about how to approach this.First, maybe I should express each |yₖ - yₖ₊₁| in terms of the x's. Since yₖ is the average of the first k x's, and yₖ₊₁ is the average of the first k+1 x's, their difference can be written as:yₖ₊₁ - yₖ = [ (x₁ + ... + xₖ + xₖ₊₁)/(k+1) ] - [ (x₁ + ... + xₖ)/k ]Let me compute this difference:= [ (Sₖ + xₖ₊₁)/(k+1) ] - [ Sₖ / k ], where Sₖ = x₁ + ... + xₖ.Let me combine these terms:= (k(Sₖ + xₖ₊₁) - (k+1)Sₖ) / [k(k+1)]= [kSₖ + kxₖ₊₁ - kSₖ - Sₖ] / [k(k+1)]= [kxₖ₊₁ - Sₖ] / [k(k+1)]But Sₖ is the sum of the first k x's, so Sₖ = k * yₖ. Therefore:= [k xₖ₊₁ - k yₖ] / [k(k+1)]= [xₖ₊₁ - yₖ] / (k+1)Therefore, the difference yₖ₊₁ - yₖ is equal to (xₖ₊₁ - yₖ)/(k+1).Therefore, the absolute value |yₖ₊₁ - yₖ| = |xₖ₊₁ - yₖ| / (k+1).So, the entire sum we need to maximize is:Σₖ=1 to 1990 |xₖ₊₁ - yₖ| / (k+1).Our goal is to maximize this sum given that the sum of |xᵢ - xᵢ₊₁| from i=1 to 1990 is 1991.Hmm. So, we need to choose x₁, ..., x₁₉₉₁ such that the sum of their consecutive absolute differences is 1991, and such that the sum of |xₖ₊₁ - yₖ|/(k+1) is maximized.This seems a bit abstract. Maybe I can think of some strategies. For example, maybe arranging the x's in a particular way, such as alternating between high and low values, or having a staircase pattern, or some other configuration that would maximize each term |xₖ₊₁ - yₖ|/(k+1).Alternatively, perhaps we can apply some inequality here, like the triangle inequality or Cauchy-Schwarz, but given that we're dealing with absolute values and maximizing a sum subject to another sum constraint, maybe Hölder's inequality could be useful? Let's think.But first, let me try to understand the structure of the problem better.First, note that each term |xₖ₊₁ - yₖ|/(k+1) is part of the sum we're trying to maximize. So, to maximize the entire sum, we need each individual term |xₖ₊₁ - yₖ|/(k+1) to be as large as possible. However, the variables x₁, ..., x₁₉₉₁ are interdependent because each yₖ is the average of the first k x's, so changing xᵢ affects all yⱼ for j ≥ i. Additionally, the sum of |xᵢ - xᵢ₊₁| is fixed at 1991, which also constrains the x's.This seems like an optimization problem with constraints. Perhaps using Lagrange multipliers? But given the high dimensionality (1991 variables), that might be intractable. Maybe we can find a pattern or recursive relation.Alternatively, perhaps we can model the sequence x₁, x₂, ..., x₁₉₉₁ such that each step maximizes |xₖ₊₁ - yₖ|/(k+1) without violating the total sum constraint. But this is a dynamic problem where each choice affects future terms.Alternatively, let's consider that yₖ is the average of the first k x's. So, yₖ = (x₁ + ... + xₖ)/k. Then, yₖ₊₁ = (k yₖ + xₖ₊₁)/(k+1). Therefore, xₖ₊₁ = (k+1)yₖ₊₁ - k yₖ.Wait, that might be useful. Let me write that down:xₖ₊₁ = (k+1)yₖ₊₁ - k yₖ.So, xₖ₊₁ is expressed in terms of yₖ and yₖ₊₁. Then, perhaps we can model the problem in terms of the y's instead of the x's. Let's try that.Given that xₖ₊₁ = (k+1)yₖ₊₁ - k yₖ, we can write xₖ₊₁ - xₖ = [(k+1)yₖ₊₁ - k yₖ] - [k yₖ - (k-1)yₖ₋₁] (if k ≥ 2). Wait, but xₖ = k yₖ - (k-1)yₖ₋₁ for k ≥ 2? Let's check.For k=1: y₁ = x₁, so x₁ = 1*y₁.For k=2: y₂ = (x₁ + x₂)/2 ⇒ x₂ = 2y₂ - x₁ = 2y₂ - y₁.Similarly, for k=3: y₃ = (x₁ + x₂ + x₃)/3 ⇒ x₃ = 3y₃ - x₁ - x₂ = 3y₃ - (y₁ + (2y₂ - y₁)) ) = 3y₃ - 2y₂.So in general, xₖ = k yₖ - (k-1)yₖ₋₁ for k ≥ 1, with y₀ undefined, but for k=1, x₁ = 1*y₁ - 0*y₀ = y₁. That seems okay.Therefore, xₖ = k yₖ - (k-1)yₖ₋₁ for k ≥ 1. Then, the difference between xₖ and xₖ₊₁ would be:xₖ₊₁ - xₖ = [ (k+1)yₖ₊₁ - k yₖ ] - [ k yₖ - (k-1)yₖ₋₁ ]= (k+1)yₖ₊₁ - k yₖ - k yₖ + (k-1)yₖ₋₁= (k+1)yₖ₊₁ - 2k yₖ + (k-1)yₖ₋₁.But this seems complicated. Maybe instead, let's consider that the sum of |xₖ - xₖ₊₁| is 1991, and each |xₖ - xₖ₊₁| is equal to | [ (k+1)yₖ₊₁ - k yₖ ] - [ k yₖ - (k-1)yₖ₋₁ ] |.Wait, perhaps this approach is getting too convoluted. Let me step back.Alternatively, since we have expressions for xₖ in terms of y's, maybe we can express the original constraint (sum of |xₖ - xₖ₊₁| = 1991) in terms of the y's. Then, our target expression to maximize is the sum of |yₖ - yₖ₊₁|, but actually, our target is sum of |yₖ - yₖ₊₁|, but in the problem statement, the target sum is sum of |y₁ - y₂| + ... + |y₁₉₉₀ - y₁₉₉₁|, which is the same as sum_{k=1}^{1990} |yₖ - yₖ₊₁|. Wait, but earlier, we derived that |yₖ - yₖ₊₁| = |xₖ₊₁ - yₖ| / (k+1). So, the sum we need to maximize is sum_{k=1}^{1990} |xₖ₊₁ - yₖ| / (k+1).But if we can express the original constraint in terms of the y's, maybe there's a relationship between these two sums.Alternatively, maybe using the triangle inequality. Let's think about the relationship between the two sums.We have sum_{k=1}^{1990} |xₖ₊₁ - xₖ| = 1991, and we need to relate this to sum_{k=1}^{1990} |xₖ₊₁ - yₖ| / (k+1).Is there a way to bound the second sum in terms of the first?Alternatively, maybe consider that |xₖ₊₁ - yₖ| ≤ |xₖ₊₁ - xₖ| + |xₖ - yₖ|, but I'm not sure if that helps.Wait, but yₖ is the average of the first k x's. So, |xₖ - yₖ| is the absolute difference between xₖ and the average of the first k x's. Maybe this can be related to the variance or something, but that might be complicated.Alternatively, perhaps arrange the x's in such a way that each xₖ₊₁ is as far as possible from yₖ, given the constraints.But since yₖ depends on all previous x's, this seems recursive. Let's consider small cases first to see if we can find a pattern.Let's try with a small number of variables, say n=2, n=3, and see what happens.Case 1: n=2.We have x₁ and x₂. The condition is |x₁ - x₂| = 2 (since 1991 is replaced by 2 here? Wait, original problem has 1991 terms with sum 1991. If n=2, then there is 1 term: |x₁ - x₂| = 2? Wait, maybe not. Wait, in the original problem, there are 1991 numbers, so there are 1990 differences. So, if we have n numbers, the sum of n-1 differences is equal to n-1. So, scaling down, for n=2, sum is 1, which equals |x₁ - x₂|.But maybe let's try n=3.Wait, perhaps this is confusing. Let me adjust. Let's take the original problem but with n=3 variables instead of 1991. Then the sum |x₁ - x₂| + |x₂ - x₃| = 3 (since original sum is 1991 for 1990 differences; so for n variables, the sum is n-1). Wait, perhaps not. Wait, original problem: 1991 variables, 1990 differences, sum 1991. So, the average difference is 1991 / 1990 ≈ 1.0005. So, each |xₖ - xₖ₊₁| is on average about 1.But maybe in the case of n=3 variables (so two differences), the sum is 3, so average 1.5 per difference.But perhaps this scaling complicates things. Maybe instead, for simplicity, let's take the original problem but with a small n, like n=2.Wait, for n=2, variables x₁, x₂. The sum |x₁ - x₂| = 1 (since original problem has sum 1991 for 1990 differences; if we scale down to n=2, there is 1 difference, so sum 1). Then y₁ = x₁, y₂ = (x₁ + x₂)/2.The expression to maximize is |y₁ - y₂|.Compute |y₁ - y₂| = |x₁ - (x₁ + x₂)/2| = |(x₁ - x₂)/2| = |x₁ - x₂| / 2.But the sum |x₁ - x₂| = 1, so |y₁ - y₂| = 1/2. Therefore, the maximum is 1/2.But in this case, the maximum is achieved when |x₁ - x₂| = 1, so the maximum of |y₁ - y₂| is 1/2.So, in this case, the maximum is 1/2.Similarly, for n=3, variables x₁, x₂, x₃. The sum |x₁ - x₂| + |x₂ - x₃| = 2 (since original problem has 1991 variables with 1990 differences summing to 1991; so scaling down, 3 variables have 2 differences summing to 2). Then, y₁ = x₁, y₂ = (x₁ + x₂)/2, y₃ = (x₁ + x₂ + x₃)/3. The expression to maximize is |y₁ - y₂| + |y₂ - y₃|.Compute each term:|y₁ - y₂| = |x₁ - (x₁ + x₂)/2| = |(x₁ - x₂)/2|.|y₂ - y₃| = |(x₁ + x₂)/2 - (x₁ + x₂ + x₃)/3| = |(3(x₁ + x₂) - 2(x₁ + x₂ + x₃))/6| = |(x₁ + x₂ - 2x₃)/6|.So the total sum is |(x₁ - x₂)/2| + |(x₁ + x₂ - 2x₃)/6|.Given that |x₁ - x₂| + |x₂ - x₃| = 2.We need to maximize the sum S = |(x₁ - x₂)/2| + |(x₁ + x₂ - 2x₃)/6|.Hmm. Let's see. Let me denote a = x₁ - x₂ and b = x₂ - x₃. Then |a| + |b| = 2. Then x₁ = x₂ + a, x₃ = x₂ - b. Then, substitute into S:First term: |a| / 2.Second term: |( (x₂ + a) + x₂ - 2(x₂ - b) ) / 6 | = |(2x₂ + a - 2x₂ + 2b)/6| = |(a + 2b)/6|.Therefore, S = |a|/2 + |a + 2b|/6.Subject to |a| + |b| = 2.We need to maximize S.Let me assume a and b are real numbers with |a| + |b| = 2. Let's consider different cases based on the signs of a and b.Case 1: a ≥ 0, b ≥ 0. Then |a| = a, |b| = b, so a + b = 2. Then, S = a/2 + |a + 2b|/6.But since a and b are non-negative, a + 2b is also non-negative. Therefore:S = a/2 + (a + 2b)/6 = (3a + a + 2b)/6 = (4a + 2b)/6 = (2a + b)/3.But since a + b = 2, we can write b = 2 - a. Then,S = (2a + (2 - a))/3 = (a + 2)/3. To maximize this, since a can be up to 2 (when b=0), so maximum is (2 + 2)/3 = 4/3 ≈ 1.333.Case 2: a ≥ 0, b ≤ 0. Then |a| = a, |b| = -b, so a - b = 2.Compute S:First term: a/2.Second term: |a + 2b|/6. Since b is negative, let's write b = -c, c ≥ 0. Then a + 2b = a - 2c.But since a - c = 2 (from |a| + |b| = a + c = 2, wait, no: original sum is |a| + |b| = a + |b| = a + c = 2, where c = |b|.But in this case, if b ≤ 0, then |b| = -b = c, so a + c = 2. Then, S = a/2 + |a - 2c|/6.But since c = 2 - a (from a + c = 2), substitute:|a - 2(2 - a)| = |a - 4 + 2a| = |3a - 4|.Therefore, S = a/2 + |3a - 4|/6.We need to maximize this for a ∈ [0,2].Let's consider two subcases:Subcase 1: 3a - 4 ≥ 0 ⇒ a ≥ 4/3.Then, S = a/2 + (3a - 4)/6 = (3a + 3a - 4)/6 = (6a -4)/6 = a - 2/3.Since a ∈ [4/3, 2], the maximum at a=2 is 2 - 2/3 = 4/3 ≈ 1.333.Subcase 2: 3a - 4 < 0 ⇒ a < 4/3.Then, S = a/2 + (4 - 3a)/6 = (3a + 4 - 3a)/6 = 4/6 = 2/3 ≈ 0.666.So in this case, maximum is 4/3 when a=2, which would mean c=0, so b=0. But wait, a=2, c=0, then b=0. But original assumption was b ≤ 0, but here b=0. So it's on the boundary. So in this case, when a=2, b=0, we have S=4/3.Case 3: a ≤ 0, b ≥ 0. Then |a| = -a, |b| = b, so -a + b = 2.Then, S = |a|/2 + |a + 2b|/6 = (-a)/2 + |a + 2b|/6.Let me write a = -c, where c ≥ 0. Then, |a| = c, and the condition becomes c + b = 2.So, S = c/2 + | -c + 2b | /6.But since a = -c, and b = 2 - c (from c + b = 2).Then, | -c + 2b | = | -c + 2(2 - c) | = | -c + 4 - 2c | = |4 - 3c|.Thus, S = c/2 + |4 - 3c|/6.Again, consider two subcases:Subcase 1: 4 - 3c ≥ 0 ⇒ c ≤ 4/3.Then, S = c/2 + (4 - 3c)/6 = (3c + 4 - 3c)/6 = 4/6 = 2/3.Subcase 2: 4 - 3c < 0 ⇒ c > 4/3.Then, S = c/2 + (3c -4)/6 = (3c + 3c -4)/6 = (6c -4)/6 = c - 2/3.Since c ∈ [0,2], and here c >4/3 ≈1.333. Then maximum at c=2 is 2 - 2/3 = 4/3.So again, maximum S=4/3 when c=2, which corresponds to a=-2, b=0. But b=0 is allowed (non-negative). Thus, in this case, S=4/3.Case 4: a ≤ 0, b ≤ 0. Then |a| + |b| = -a -b =2.Then, S = (-a)/2 + |a + 2b| /6.Let a = -c, b = -d, where c, d ≥0. Then, c + d =2.Then, S = c/2 + | -c -2d | /6 = c/2 + | - (c + 2d) | /6 = c/2 + (c + 2d)/6.So, S = (3c + c + 2d)/6 = (4c + 2d)/6 = (2c + d)/3.But c + d =2 ⇒ d=2 -c.Therefore, S=(2c + 2 -c)/3=(c +2)/3.To maximize, c can be up to 2 (when d=0). Then S=(2 +2)/3=4/3.So, again maximum S=4/3 when c=2, d=0. So a=-2, b=0.So, in all cases, the maximum sum S is 4/3. So for n=3 variables, the maximum is 4/3.Comparing to the original problem scaled down, where sum of differences is n-1. Here, n=3, sum of differences is 2, and the maximum S is 4/3.Hmm. Interesting. So, for n=2, maximum S=1/2; for n=3, maximum S=4/3. Let's see if we can find a pattern.Wait, 1/2 for n=2: (2-1)/2 = 0.5.4/3 for n=3: (3-1)/something? Not sure. Alternatively, maybe for n=2, the maximum is (n-1)/2, but 1/2 ≠ (2-1)/2 = 0.5. Wait, 1/2 is indeed (2-1)/2.For n=3, 4/3 ≈1.333. Is there a relation? (3-1)*2/3=4/3. Yes! For n=3, maximum S= (n-1)*2/3. For n=2, it's (n-1)*1/2=1/2. Maybe the general formula is (n-1)* something.Wait, maybe the maximum sum S is (n-1)*c, where c is a constant depending on n?Alternatively, perhaps the maximum sum is 2*(n-1)/n.Wait, for n=2: 2*(1)/2=1. But we have 1/2. Doesn't fit.Alternatively, maybe the maximum S for n variables is (n-1)/2.Wait, for n=2: 1/2, which matches. For n=3: 4/3, which is more than (3-1)/2=1. So that doesn't fit.Alternatively, perhaps we need to consider the harmonic series or something else.Alternatively, looking at the n=2 and n=3 cases:For n=2: S_max = 1/2.For n=3: S_max = 4/3.Let me compute the ratio between S_max and the sum of |x differences|:For n=2: 1/2 over 1 = 0.5.For n=3: 4/3 over 2 ≈ 0.666.So, the ratio increases as n increases. Maybe for the original problem, the ratio approaches 1 as n becomes large? But 1991 is quite large.Alternatively, in the n=2 case, S_max = 1/2, which is the sum of |x differences| multiplied by 1/2.In n=3, S_max = 4/3, which is the sum of |x differences| (which is 2) multiplied by 2/3.Wait, 4/3 = (2) * (2/3).So, in n=3, the multiplier is 2/3.Hmm, so 1/2, 2/3... Maybe for n, the multiplier is (n-1)/n?Wait, for n=2: (2-1)/2 = 1/2, which matches.For n=3: (3-1)/3 = 2/3, which matches.So, if this pattern holds, then for the original problem with n=1991 variables, the maximum sum S would be (1991 - 1)/1991 multiplied by the sum of |x differences|, which is 1991. Hence:S_max = (1990/1991) * 1991 = 1990.Wait, that would be the case if the multiplier is (n-1)/n. But in the n=3 case, we had S_max = 4/3, which is (2/3)*2. So, 2/3 is (n-1)/n * sum of |x differences|.Wait, sum of |x differences| for n variables is n-1 (in the scaled-down problem). For the original problem, the sum is 1991, which is equal to the number of differences (1990). So in the original problem, the sum is equal to the number of differences. So if we follow the pattern, the maximum S would be (n-1)/n multiplied by the sum of |x differences|, which is (1990/1991)*1991 = 1990.But in the n=3 case, the sum of |x differences| is 2 (which is n-1), and the maximum S is 4/3 = (2/3)*2. So, the same formula applies.Therefore, the conjecture is that for n variables, where the sum of |x_i - x_{i+1}| from i=1 to n-1 is equal to n-1, the maximum sum S of |y_i - y_{i+1}| is (n-1)/n * (n-1) = (n-1)^2 /n.Wait, but in the n=2 case, (2-1)^2 /2 = 1/2, which matches.In the n=3 case, (3-1)^2 /3 = 4/3, which also matches.Therefore, the formula seems to hold. Therefore, for the original problem with n=1991 variables, the maximum sum S would be (1991 -1)^2 /1991 = 1990² /1991.But wait, in our scaled-down examples, the sum of |x differences| was n-1, and S_max was (n-1)^2 /n. However, in the original problem, the sum of |x differences| is 1991, which is equal to n-1 where n=1991. Therefore, substituting into the formula, S_max = (1991 -1)^2 /1991 = 1990² /1991 ≈ (1990)^2 /1991 ≈ roughly 1990*(1990/1991) ≈ 1990*(1 - 1/1991) ≈ 1990 - 1990/1991 ≈ 1989.0005... but the exact value is 1990² /1991.But wait, in the n=2 case, sum of |x differences| is 1 (n-1=1), and S_max=1/2=(1)^2 /2=1/2.Similarly, for n=3, sum of |x differences|=2, and S_max=4/3=(2)^2 /3=4/3.Hence, generalizing, if the sum of |x_i - x_{i+1}| from i=1 to n-1 is S, then the maximum sum of |y_i - y_{i+1}| is S^2 /n. But in our case, S =n-1, so it's (n-1)^2 /n.But in our original problem, S =1991, and n=1991. Therefore, S_max=1991² /1991=1991.Wait, that contradicts the previous thought. Wait, maybe I made a mistake here.Wait, in the scaled-down examples, S (sum of |x differences|) is n-1, and the maximum sum of |y differences| is (n-1)^2 /n.But in the original problem, the sum of |x differences| is 1991, which is equal to n-1 where n=1991. Therefore, substituting into the formula, we get (n-1)^2 /n =1990² /1991. However, if instead, the formula is S^2 /n, where S is the sum of |x differences|, then since S=1991, then S_max=1991² /1991=1991. But in our scaled examples, when S=n-1, it gives (n-1)^2 /n, which is different from S^2 /n when S=n-1.Wait, maybe there's confusion here. Let me clarify.Suppose in general, for a sequence of n numbers x₁,...,x_n with sum_{i=1}^{n-1} |x_i - x_{i+1}| = T, then the maximum sum of |y_i - y_{i+1}| is T^2 /n.But in our scaled examples, T was n-1, leading to (n-1)^2 /n. But if T is arbitrary, then S_max=T^2 /n.But in our original problem, T=1991, n=1991, so S_max=1991² /1991=1991.But in the n=2 case, when T=1, S_max=1/2=1² /2. Similarly, for n=3, T=2, S_max=4/3=2² /3.So this seems to hold. Therefore, if the formula is S_max = T² /n, then in our original problem, it's 1991² /1991=1991. But that contradicts the n=2 and n=3 cases where T=n-1, and S_max=(n-1)^2 /n.But wait, maybe the formula is S_max = T² / (4(n-1)) or something else. Wait, let's re-examine the n=2 case.In n=2, sum |x₁ -x₂|=1=T. Then S_max=1/2= T² / (4*1)=1/4, which is not matching. So maybe not.Alternatively, in the n=2 case, S_max= T/2. For n=3, S_max= T² / (something). Wait, T=2 for n=3, S_max=4/3. Then, 4/3= (2)^2 /3=4/3. So for n=3, S_max= T² /n.But for n=2, S_max= T/2=1/2. Which is also T² /n when T=1: 1² /2=1/2. So yes, if T is the sum of |x differences|, then S_max= T² /n.Therefore, for the original problem, where T=1991 and n=1991, S_max=1991² /1991=1991.But that seems too straightforward. However, according to the examples:- For n=2, T=1: S_max=1² /2=0.5.- For n=3, T=2: S_max=2² /3=4/3≈1.333.Thus, in the original problem, S_max=1991² /1991=1991.But that seems surprising. Wait, if we set all the differences |x_i -x_{i+1}|=1 for each i, then the sum is 1991, which satisfies the condition. Then, what would the sum of |y_i - y_{i+1}| be?Wait, let's test this with n=3. If |x₁ -x₂|=1 and |x₂ -x₃|=1, then sum is 2. Then, let's set x₁=0, x₂=1, x₃=0. Then:y₁=0y₂=(0+1)/2=0.5y₃=(0+1+0)/3≈0.333Thus, |y₁ - y₂|=0.5, |y₂ - y₃|≈0.166, sum≈0.666, which is less than 4/3≈1.333. So in this case, the sum is lower.But in the previous analysis for n=3, the maximum was achieved when two differences are 2 and 0. For example, x₁=-2, x₂=0, x₃=0. Then:|x₁ -x₂|=2, |x₂ -x₃|=0, sum=2.Then, y₁=-2, y₂=(-2+0)/2=-1, y₃=(-2+0+0)/3≈-0.666.Thus, |y₁ - y₂|=1, |y₂ - y₃|≈0.333, sum≈1.333=4/3.So in this case, the sum is higher. Therefore, arranging the x's to have a large difference followed by zeros gives a higher sum.Therefore, the maximum occurs when we have one large jump and the rest zero, which corresponds to setting one difference to T and others to zero.But in the original problem, if we set one |x_i -x_{i+1}|=1991 and the rest zero, then the sum is 1991.In that case, let's compute the sum of |y_k - y_{k+1}|.Suppose we have x₁, x₂, ..., x_{1991}, where x₁ = a, x₂ = a + 1991, and x₃ = x₂, ..., x_{1991}=x₂. Then, the only non-zero difference is |x₁ -x₂|=1991, and the rest are zero.Then, compute y_k:For k=1: y₁ =x₁ =a.For k=2: y₂=(x₁ +x₂)/2=(a + a +1991)/2=a + 1991/2.For k=3: y₃=(x₁ +x₂ +x₃)/3=(a + (a +1991) + (a +1991))/3=(3a + 3982)/3=a + 1991*2/3.Similarly, for k=4: y₄=(4a + 1991*3)/4=a + 1991*3/4.Continuing this pattern, for k=m: y_m =a + 1991*(m-1)/m.Therefore, the difference |y_{m} - y_{m+1}|= |a +1991*(m-1)/m - [a +1991*m/(m+1)]| =1991*|(m-1)/m - m/(m+1)|.Compute this:=1991*| [ (m-1)(m+1) - m² ] / [m(m+1)] |=1991*| [ (m² -1) -m² ] / [m(m+1)] |=1991*| -1 / [m(m+1)] | =1991 / [m(m+1)].Therefore, each term |y_{m} - y_{m+1}| for m from 1 to 1990 is 1991 / [m(m+1)].Therefore, the total sum would be Σ_{m=1}^{1990} 1991 / [m(m+1)].This is a telescoping series:Σ_{m=1}^{1990} 1991*(1/m - 1/(m+1)) )=1991*(1 - 1/1991)=1991*(1990/1991)=1990.Therefore, the sum is 1990.Wait, this matches the previous conjecture. Therefore, if we set one difference to 1991 and the rest to zero, the sum of |y_k - y_{k+1}| is 1990.But is this the maximum? Earlier examples with smaller n suggest that arranging the differences to have one large jump and the rest zero gives the maximum sum. Let's verify with n=3.For n=3, if we set |x₁ -x₂|=2 and |x₂ -x₃|=0, then the sum S=4/3≈1.333. Alternatively, if we spread the differences as 1 and 1, the sum S≈0.666. So indeed, concentrating the difference gives a higher sum.Similarly, for n=2, setting the single difference to 1 gives S=0.5, which is the maximum.Therefore, it's plausible that for the original problem, arranging the x's such that one difference is 1991 and the rest are zero gives the maximum sum of 1990.But is there a way to confirm that this is indeed the maximum? Let's think.Suppose we have multiple non-zero differences. For example, suppose we have two differences: |x₁ -x₂|=a and |x₂ -x₃|=b, with a + b=1991. Then, compute the sum S. Would this sum be less than 1990?But in the case of two differences, with a and b, maybe we can compute S and see.However, given the complexity of the problem with 1991 variables, it's more efficient to consider that the telescoping sum gives 1990, and any other configuration would lead to a smaller sum.Alternatively, we can think of the sum S as:Σ_{k=1}^{1990} |x_{k+1} - y_k| / (k+1).If we use the Cauchy-Schwarz inequality, maybe we can bound this sum.But first, note that each term is |x_{k+1} - y_k| / (k+1). To apply Cauchy-Schwarz, we can write:Σ |a_k b_k| ≤ sqrt(Σ a_k²) * sqrt(Σ b_k²).But I'm not sure if that would directly help here.Alternatively, let's consider that:Σ_{k=1}^{1990} |x_{k+1} - y_k| / (k+1) ≤ Σ_{k=1}^{1990} |x_{k+1} - y_k| / 2.But this is a rough bound.Alternatively, note that in the telescoping case, we have each |x_{k+1} - y_k| = 1991 / (k+1), which gives us the sum 1990. So perhaps this is the maximum.Alternatively, suppose we try to maximize each term |x_{k+1} - y_k| / (k+1). To maximize the sum, each term should be as large as possible. However, the variables are dependent.In the case where we have one large jump at the beginning, then x₂ =x₁ ±1991, and the rest of the x's are equal to x₂. Then, for each k ≥1, y_k is the average of the first k x's. If x₁, x₂, x₃,... are such that x₂ =x₁ +1991 and x₃=...=x₁₉₉₁=x₂, then:For k=1: y₁=x₁.For k=2: y₂=(x₁ +x₂)/2=x₁ +1991/2.For k=3: y₃=(x₁ +x₂ +x₃)/3=x₁ + (2*1991)/3.Similarly, yₖ=x₁ +1991*(k-1)/k.Therefore, the difference yₖ₊₁ -yₖ=1991*(k/(k+1) - (k-1)/k)=1991*( (k² - (k² -1))/k(k+1))=1991*(1)/(k(k+1)).Therefore, the absolute difference is 1991/(k(k+1)), and summing from k=1 to 1990 gives a telescoping sum:Σ1991/(k(k+1))=1991Σ(1/k -1/(k+1))=1991*(1 -1/1991)=1990.Therefore, in this case, the sum is 1990.Now, suppose we have a different configuration where there are multiple jumps. Let's say two jumps: |x₁ -x₂|=a, |x₂ -x₃|=b, and the rest are zero, with a + b=1991.Then, compute the sum S.But this will require more complex calculations. Let me try with a small example.Let n=3, as before, with two jumps a and b such that a + b=1991. Then:x₁, x₂=x₁ ±a, x₃=x₂ ±b, and x₄=...=x₁₉₉₁=x₃.But this becomes complicated. Alternatively, take n=4 for simplicity.But maybe instead, consider that distributing the total difference into multiple jumps would lead to smaller contributions to the sum S. Because each jump affects multiple y's, but the total contribution might be less due to averaging.In the case where we have a single jump, the entire difference is concentrated at the beginning, maximizing the impact on all subsequent y's. Whereas, if the jumps are spread out, each jump only affects a limited number of y's, leading to smaller contributions.Therefore, it's plausible that concentrating the entire difference into one jump maximizes the sum S. Therefore, the maximum possible value is 1990.Hence, after this lengthy exploration and checking with smaller cases, I believe the maximum possible value of the given expression is 1990.Final Answerboxed{1990}

✅:Given numbers ( x_1, x_2, ldots, x_{1991} ) satisfy the condition:[|x_1 - x_2| + |x_2 - x_3| + cdots + |x_{1990} - x_{1991}| = 1991,]where ( y_k = frac{1}{k}(x_1 + x_2 + cdots + x_k) ) for ( k = 1, 2, ldots, 1991 ). We need to find the maximum possible value of the expression:[|y_1 - y_2| + |y_2 - y_3| + cdots + |y_{1990} - y_{1991}|.]First, we express the difference ( y_{k+1} - y_k ) in terms of ( x_{k+1} ) and ( y_k ):[y_{k+1} - y_k = frac{x_{k+1} - y_k}{k+1}.]Thus, the absolute value ( |y_{k+1} - y_k| ) becomes:[|y_{k+1} - y_k| = frac{|x_{k+1} - y_k|}{k+1}.]To maximize the sum ( sum_{k=1}^{1990} |y_{k+1} - y_k| ), we need to consider the structure of the sequence ( x_1, x_2, ldots, x_{1991} ). By analyzing smaller cases and generalizing, we found that concentrating the total difference into one large jump maximizes the sum of absolute differences between consecutive averages ( y_k ).Setting one large jump at the beginning, for example, ( x_1 = a ) and ( x_2 = a + 1991 ), with all subsequent ( x_i = x_2 ), we compute the averages ( y_k ). This configuration leads to a telescoping series when summing the absolute differences between consecutive ( y_k ):[sum_{k=1}^{1990} frac{1991}{k(k+1)} = 1991 sum_{k=1}^{1990} left( frac{1}{k} - frac{1}{k+1} right) = 1991 left( 1 - frac{1}{1991} right) = 1990.]Thus, the maximum possible value of the given expression is:[boxed{1990}]

🔑:1. Given the sequences x_1, x_2, ldots, x_{1991} satisfying: [ left|x_{1}-x_{2}right| + left|x_{2}-x_{3}right| + cdots + left|x_{1990}-x_{1991}right| = 1991 ]2. Define y_k = frac{1}{k}(x_1 + x_2 + cdots + x_k) for k = 1, 2, ldots, 1991.3. To find the maximum value of: [ left|y_{1}-y_{2}right| + left|y_{2}-y_{3}right| + cdots + left|y_{1990}-y_{1991}right| ]4. For k = 1, 2, ldots, 1990, we have: [ left|y_{k} - y_{k+1}right| = left|frac{1}{k}(x_{1} + x_{2} + cdots + x_k) - frac{1}{k+1}(x_{1} + x_{2} + cdots + x_{k+1})right| ]5. Rewrite the expression inside the absolute value: [ left|y_{k} - y_{k+1}right| = left|frac{1}{k}(x_{1} + x_{2} + cdots + x_k) - frac{1}{k+1}(x_{1} + x_{2} + cdots + x_k + x_{k+1})right| ]6. Simplify the right-hand side: [ = left|frac{k+1}{k(k+1)}(x_{1} + x_{2} + cdots + x_k) - frac{1}{k+1} x_{k+1}right| = left|frac{1}{k(k+1)} (x_{1} + x_{2} + cdots + x_k - k x_{k+1})right| ]7. Use the inequality property of absolute values: [ leq frac{1}{k(k+1)} left(|x_{1} - x_{2}| + |x_{2} - x_{3}| + cdots + |k cdot x_{k} - k cdot x_{k+1}|right) ]8. Summing this expression from k=1 to k=1990, we get: [ begin{aligned} left|y_{1} - y_{2}right| + left|y_{2} - y_{3}right| + cdots + left|y_{1990} - y_{1991}right| &leq sum_{k=1}^{1990} frac{1}{k(k+1)} left(|x_{1} - x_{2}| + |x_{2} - x_{3}| + cdots + |k cdot x_{k} - k cdot x_{k+1}|right) end{aligned} ]9. We can rewrite the sum using the property of telescoping series: [ = sum_{k=1}^{1990} frac{1}{k(k+1)} (|x_{1} - x_{2}| + 2 |x_{2} - x_{3}| + cdots + k |x_{k} - x_{k+1}|) ]10. Using the series summation properties: [ sum_{k=1}^{1990} frac{1}{k(k+1)} = (1 - frac{1}{1991}) + (1 - frac{2}{1991}) + cdots + (1 - frac{1990}{1991}) ]11. Summing these terms: [ left|y_{1}-y_{2}right| + left|y_{2}-y_{3}right| + cdots + left|y_{1990}-y_{1991}right| leq |x_{1} - x_{2}| sum_{k=1}^{1990} frac{1}{k(k+1)} + cdots + |x_{1990} - x_{1991}| cdot frac{1990}{1990 cdot 1991} ]12. Simplifying: [ leq left|x_{1}-x_{2}right| left(1 - frac{1}{1991}right) + cdots + left|x_{1990} - x_{1991}right| left(1 - frac{1990}{1991}right) ]13. Given that the sum of the absolute values of the differences is 1991, we have: [ leq 1991 left(1 - frac{1}{1991}right) = 1990 ]14. Hence, the maximum value is achieved when:[ x_1 = 1991, x_2 = cdots = x_{1991} = 0 ]Thus, the maximum value of the sought expression is:[boxed{1990}]

🧠:Okay, let me try to figure out this problem step by step. So, the sports meeting lasted for n days, where n is greater than 1. A total of m medals were awarded. The way they awarded medals each day is a bit specific: on the first day, they gave out 1 medal and then 1/7 of the remaining medals. On the second day, 2 medals and then 1/7 of the remaining, and so on until the nth day, where they gave out exactly n medals and there were none left. We need to find n and m.Hmm, let's break this down. Let's start with the first day. On day 1, they give out 1 medal. Then, they give out 1/7 of the remaining medals. So after giving 1 medal, there are m - 1 medals left. Then they give away 1/7 of those, so the remaining after day 1 would be (m - 1) - 1/7*(m - 1) = 6/7*(m - 1). Right?Wait, but the problem says "on the first day, 1 medal was awarded, and 1/7 of the remaining m - 1 medals were also awarded." Wait, maybe after giving 1 medal, they give 1/7 of the remaining (which is m - 1) medals. Wait, but m - 1 is the remaining after the first medal. So maybe on day 1, total medals awarded are 1 + (1/7)*(m - 1). Then the remaining medals would be (m - 1) - (1/7)*(m - 1) = 6/7*(m - 1). So that seems correct.Similarly, on day 2, they start by awarding 2 medals, then 1/7 of the remaining. So after day 1, the remaining medals are 6/7*(m - 1). Then on day 2, they give out 2 medals first, so remaining after that is 6/7*(m - 1) - 2. Then they give out 1/7 of that remaining amount, so the remaining after day 2 would be 6/7*( [6/7*(m - 1) - 2] ).This seems like a recursive process where each day, after awarding k medals (on day k), they give out 1/7 of the remaining, leaving 6/7 of that remaining. Then this repeats until day n, where they just give out n medals and there are none left. So on day n, they don't have to give out 1/7 of the remaining because all medals are exhausted.So maybe we can model this with equations. Let me try to formalize this.Let’s denote R_k as the remaining medals at the start of day k. Then:On day 1:- Award 1 medal. Remaining: R_1 - 1- Then award 1/7 of the remaining: (1/7)(R_1 - 1)- Total awarded on day 1: 1 + (1/7)(R_1 - 1)- Remaining after day 1: (R_1 - 1) - (1/7)(R_1 - 1) = (6/7)(R_1 - 1)So R_2 = (6/7)(R_1 - 1)Similarly, on day 2:- Award 2 medals. Remaining: R_2 - 2- Then award 1/7 of that remaining: (1/7)(R_2 - 2)- Total awarded on day 2: 2 + (1/7)(R_2 - 2)- Remaining after day 2: (R_2 - 2) - (1/7)(R_2 - 2) = (6/7)(R_2 - 2)Thus, R_3 = (6/7)(R_2 - 2)Continuing this pattern, on day k:- Award k medals. Remaining: R_k - k- Then award 1/7 of the remaining: (1/7)(R_k - k)- Total awarded on day k: k + (1/7)(R_k - k)- Remaining after day k: (6/7)(R_k - k) = R_{k+1}And this continues until day n, where after awarding n medals, there are no medals left. So on day n:- Award n medals. Remaining: R_n - n = 0 => R_n = nBut since on day n, they don't award 1/7 of the remaining, because all medals are exhausted. So the last day is different. So the recurrence relation is:For k from 1 to n-1:R_{k+1} = (6/7)(R_k - k)But on day n, R_n = n.So starting from R_1 = m (total medals). Then R_2 = (6/7)(m - 1), R_3 = (6/7)(R_2 - 2), and so on until R_n = (6/7)(R_{n-1} - (n - 1)) ), and then on day n, R_n = n. Therefore, we can write:R_n = (6/7)(R_{n-1} - (n - 1)) ) = nSo we can work backwards from R_n = n to find R_{n-1}, then R_{n-2}, etc., until we reach R_1 = m. Maybe this reverse approach is easier.Let me try this. Let's denote R_n = n. Then:R_n = (6/7)(R_{n-1} - (n - 1)) ) = nSo solving for R_{n-1}:(6/7)(R_{n-1} - (n - 1)) ) = nMultiply both sides by 7/6:R_{n-1} - (n - 1) = (7/6) nTherefore:R_{n-1} = (7/6) n + (n - 1) = (7n/6 + n - 1) = (7n/6 + 6n/6 - 6/6) = (13n - 6)/6But R_{n-1} must be an integer because the number of medals is a whole number. Therefore, (13n - 6) must be divisible by 6. Let's note that.Similarly, moving up to R_{n-2}:R_{n-1} = (6/7)(R_{n-2} - (n - 2)) ) = (13n - 6)/6So:(6/7)(R_{n-2} - (n - 2)) ) = (13n - 6)/6Multiply both sides by 7/6:R_{n-2} - (n - 2) = (7/6)*(13n - 6)/6Simplify the right side:(7*(13n - 6))/36Therefore,R_{n-2} = (7*(13n - 6))/36 + (n - 2)Hmm, this seems complicated. Let me compute that step by step.First, R_{n-2} = (7*(13n - 6))/36 + (n - 2)Convert n - 2 to 36 denominator:(n - 2) = 36(n - 2)/36So,R_{n-2} = [7*(13n - 6) + 36(n - 2)] / 36Compute numerator:7*(13n - 6) = 91n - 4236*(n - 2) = 36n - 72Total numerator: 91n - 42 + 36n - 72 = 127n - 114Therefore,R_{n-2} = (127n - 114)/36Again, R_{n-2} must be an integer, so (127n - 114) must be divisible by 36.This is getting messy. Maybe there's a pattern here. Let me check for small n. Since n >1, let's try n=2 first.Wait, if n=2, then on day 2, they give out 2 medals and that's all. Let's check:For n=2:R_2 = 2. Then R_1 = (7/6)*(R_2) +1 = (7/6)*2 +1 = 14/6 +1 = 14/6 + 6/6 = 20/6 = 10/3 ≈ 3.333. But m must be integer. So n=2 is invalid.n=3:Start with R_3 =3.Then R_2 = (7/6)*3 +2 = 21/6 +12/6=33/6=11/2=5.5 Not integer. So invalid.n=4:R_4=4R_3=(7/6)*4 +3=28/6 +18/6=46/6=23/3 ≈7.666 Not integer.n=5:R_5=5R_4=(7/6)*5 +4=35/6 +24/6=59/6≈9.833 Not integer.n=6:R_6=6R_5=(7/6)*6 +5=7 +5=12So R_5=12, which is integer. Then R_4=(7/6)*(12) +4=14 +4=18Wait, wait, no. Wait, let me recast.Wait, when n=6, R_6=6.Then R_5 = (7/6)*6 +5=7 +5=12. Wait, but how did we get that?Wait, the formula for R_{n-1} is (13n -6)/6. For n=6, that would be (13*6 -6)/6 = (78 -6)/6=72/6=12. So R_5=12.Then moving up, R_4 would be computed as:R_5 = (6/7)(R_4 -4) =12So (6/7)(R_4 -4)=12 => R_4 -4=12*(7/6)=14 => R_4=14 +4=18Similarly, R_4=18.Then R_3:R_4=(6/7)(R_3 -3)=18So R_3 -3=18*(7/6)=21 => R_3=21 +3=24R_3=24.Then R_2:R_3=(6/7)(R_2 -2)=24Thus, R_2 -2=24*(7/6)=28 => R_2=28 +2=30R_2=30.Then R_1:R_2=(6/7)(R_1 -1)=30So R_1 -1=30*(7/6)=35 => R_1=35 +1=36Therefore, m=R_1=36.So n=6 and m=36. Let's verify if this works.Starting with m=36.Day 1:Award 1 medal. Remaining:36-1=35. Then 1/7 of 35 is 5. So total awarded on day1:1+5=6. Remaining:35-5=30.Day2:Award2 medals. Remaining:30-2=28. 1/7 of 28 is4. Awarded:2+4=6. Remaining:28-4=24.Day3:Award3 medals. Remaining:24-3=21. 1/7 of21=3. Awarded:3+3=6. Remaining:21-3=18.Day4:Award4 medals. Remaining:18-4=14. 1/7 of14=2. Awarded:4+2=6. Remaining:14-2=12.Day5:Award5 medals. Remaining:12-5=7. 1/7 of7=1. Awarded:5+1=6. Remaining:7-1=6.Day6:Award6 medals. Remaining:6-6=0. All medals exhausted.Yes, that works! Each day, after awarding k medals, they award 1/7 of the remaining, and each day the remaining reduces by 6/7 of the previous remaining minus k. And on day 6, exactly 6 medals are awarded with none left. So the answer is n=6 days and m=36 medals.But let me check if there are other possible solutions. For example, when n=6 works, maybe higher n could also work? Let's check n=7.For n=7:R_7=7R_6=(13*7 -6)/6=(91 -6)/6=85/6≈14.166. Not integer. So invalid.Similarly, n=8:R_8=8R_7=(13*8 -6)/6=(104-6)/6=98/6≈16.333 Not integer.n=9:R_9=9R_8=(13*9 -6)/6=(117 -6)/6=111/6=37/2=18.5 Not integer.n=10:R_10=10R_9=(13*10 -6)/6=124/6≈20.666 Not integer.So it seems only n=6 gives integer values going back. Therefore, the solution is n=6 and m=36.Final AnswerThe sports meeting lasted for boxed{6} days, and a total of boxed{36} medals were awarded.

✅:The sports meeting lasted for ( n ) days and awarded a total of ( m ) medals. Each day ( k ), ( k ) medals were awarded followed by ( frac{1}{7} ) of the remaining medals. On the ( n )-th day, exactly ( n ) medals were awarded, exhausting all medals. We need to find ( n ) and ( m ).1. Day 1: - Award 1 medal, remaining ( m - 1 ). - Award ( frac{1}{7} ) of remaining: ( frac{1}{7}(m - 1) ). - Remaining: ( frac{6}{7}(m - 1) ).2. Day 2: - Award 2 medals, remaining ( frac{6}{7}(m - 1) - 2 ). - Award ( frac{1}{7} ) of remaining: ( frac{1}{7} left( frac{6}{7}(m - 1) - 2 right) ). - Remaining: ( frac{6}{7} left( frac{6}{7}(m - 1) - 2 right) ).3. Pattern: - For each day ( k ), remaining medals after day ( k ) is ( frac{6}{7}(R_k - k) ). - On day ( n ), ( R_n = n ).4. Reverse Calculation: - Start from ( R_n = n ) and work backwards. - For ( n = 6 ): - ( R_6 = 6 ). - ( R_5 = frac{7}{6} times 6 + 5 = 12 ). - ( R_4 = frac{7}{6} times 12 + 4 = 18 ). - ( R_3 = frac{7}{6} times 18 + 3 = 24 ). - ( R_2 = frac{7}{6} times 24 + 2 = 30 ). - ( R_1 = frac{7}{6} times 30 + 1 = 36 ).5. Verification: - Day 1: 1 medal + 5 medals (1/7 of 35) = 6, remaining 30. - Day 2: 2 medals + 4 medals (1/7 of 28) = 6, remaining 24. - Day 3: 3 medals + 3 medals (1/7 of 21) = 6, remaining 18. - Day 4: 4 medals + 2 medals (1/7 of 14) = 6, remaining 12. - Day 5: 5 medals + 1 medal (1/7 of 7) = 6, remaining 6. - Day 6: 6 medals, remaining 0.Thus, the sports meeting lasted for (boxed{6}) days, and a total of (boxed{36}) medals were awarded.

🔑:Let ( a_k ) denote the number of prizes distributed on the ( k )-th day. 1. On the first day, the number of prizes distributed is given by: [ a_1 = 1 + frac{1}{7}(m - 1) = frac{1}{7}(m + 6) ]2. For the ( k )-th day, the number of prizes distributed can be expressed recursively as: [ a_k = k + frac{1}{7} left(m - a_1 - a_2 - cdots - a_{k-1} - kright) ]3. Simplifying the recursive formula, we have: [ a_k = k + frac{1}{7} left(m - sum_{i=1}^{k-1} a_i - kright) ]4. This recursive formula can be simplified further to: [ a_k - a_{k-1} = 1 + frac{1}{7}left(-a_{k-1} - 1right) ] which implies: [ a_k = frac{6}{7} a_{k-1} + frac{6}{7} ]5. Note that the fixed point of the function ( f(x) = frac{6}{7} x + frac{6}{7} ) is 6, since solving ( x = frac{6}{7} x + frac{6}{7} ) yields: [ x - frac{6}{7} x = frac{6}{7} ] [ frac{1}{7} x = frac{6}{7} ] [ x = 6 ]6. Thus, the solution to the recursive sequence can be written as: [ a_k = left(frac{6}{7}right)^{k-1} left(a_1 - 6right) + 6 ] Substituting ( a_1 = frac{1}{7}(m + 6) ), we have: [ a_k = left(frac{6}{7}right)^{k-1} left(frac{1}{7}(m + 6) - 6right) + 6 = frac{1}{7} left(frac{6}{7}right)^{k-1} (m - 36) + 6 ]7. Sum of the sequences from ( a_1 ) to ( a_n ): [ m = a_1 + a_2 + cdots + a_n ] Simplifying the sum: [ m = sum_{k=1}^n left[ frac{1}{7} left( frac{6}{7} right)^{k-1} (m - 36) + 6 right] ]8. Recognizing the geometric series sum: [ sum_{k=1}^n left( frac{6}{7} right)^{k-1} = frac{1 - left( frac{6}{7} right)^n}{1 - frac{6}{7}} = frac{7}{1} left( 1 - left( frac{6}{7} right)^n right) = 7 left( 1 - left( frac{6}{7} right)^n right) ]9. Thus, [ m = (m - 36) left[ 1 - left( frac{6}{7} right)^n right] + 6n ] which simplifies to: [ m = (m - 36) left[ 1 - left( frac{6}{7} right)^n right] + 6n ]10. Solving for ( m ): [ m left( 1 - left[ 1 - left( frac{6}{7} right)^n right] right) = 36 left[ 1 - left( frac{6}{7} right)^n right] + 6n ] [ m left( left( frac{6}{7} right)^n right) = 36 left[ left( frac{6}{7} right)^n right] + 6n - 36 ] [ m = frac{7^n}{6^{n-1}}(n-6) + 36 ]11. Since both ( m ) and ( n ) are integers and ( 7^n ) and ( 6^{n-1} ) are co-prime, ( (n-6) ) must be divisible by ( 6^{n-1} ).12. Considering ( n > 1 ), it is clear that ( |n-6| < 6^{n-1} ). This implies ( n = 6 ), yielding: [ m = 36 ]13. Conclusion: The event lasted ( 6 ) days, and a total of ( 36 ) prizes were distributed. [ boxed{6 text{ days, } 36 text{ prizes}} ]

🧠:Alright, let me try to figure out this problem step by step. So, we have triangle ABC with AB = 10, and the equation sin(3A) + 3 sin C = 4. We need to find the length of the altitude from point C to side AB. Hmm, okay. Let me start by recalling some trigonometric identities and triangle properties that might help here.First, in any triangle, the sum of the angles is 180 degrees (or π radians). So, A + B + C = π. Also, the Law of Sines relates the sides and angles: a/sin A = b/sin B = c/sin C, where a, b, c are the lengths opposite angles A, B, C respectively. But here, we have AB = 10, which is side c (opposite angle C) if we follow the standard notation where side a is BC, side b is AC, and side c is AB. Wait, actually, let me confirm the notation. If AB is opposite angle C, then yes, AB = c = 10. So, sides: BC = a, AC = b, AB = c = 10.We need to find the altitude from C to AB. Let's denote that altitude as h. The area of the triangle can be expressed as (1/2)*base*height. Here, base AB is 10, so the area is (1/2)*10*h = 5h. Alternatively, the area can also be expressed using sides and sines: (1/2)*ab*sin C, but I need to figure out which sides and angles are involved here. Maybe using the Law of Sines or Cosines could help relate the angles and sides.But the given equation is sin(3A) + 3 sin C = 4. That seems a bit complex. Let me recall the triple angle formula for sine: sin(3θ) = 3 sin θ - 4 sin³ θ. So, maybe I can expand sin(3A) using that identity. Let's try that.So, substituting the identity into the equation:sin(3A) + 3 sin C = 4 => 3 sin A - 4 sin³ A + 3 sin C = 4Hmm, now we have an equation involving sin A and sin C. Since in triangle ABC, angles A, B, C add up to π, we can express angle C as π - A - B. But we might need another relation here. Alternatively, using the Law of Sines, we can relate sin A and sin C.From the Law of Sines: a/sin A = c/sin C => a/ sin A = 10/sin C => a = (10 sin A)/sin C. Similarly, b = (10 sin B)/sin C. But I'm not sure if that's immediately helpful.Wait, perhaps we can find a relationship between angles A and C. Let me express angle C in terms of A and B: C = π - A - B. But unless we know something about angle B, that might not help. Alternatively, maybe there's a way to relate angles A and C through the sides. Since AB is 10, and we need the altitude from C to AB, which is h. Let's think about how h relates to the sides and angles.In triangle ABC, the altitude h from C to AB divides AB into two segments. Let's call the foot of the altitude D. Then AD = b cos A and DB = a cos B, right? Because in right triangles ADC and BDC, AD = AC cos A = b cos A, and DB = BC cos B = a cos B. Since AD + DB = AB = 10, we have b cos A + a cos B = 10. But again, this might be too many variables unless we can relate them.Alternatively, the altitude h can be expressed in terms of sides and angles. From triangle ADC, h = b sin A, and from triangle BDC, h = a sin B. So, h = b sin A = a sin B. Also, from the area formula, h = (2 * area)/10 = area/5.Hmm, perhaps we can express the area in terms of angles A and C. The area is (1/2)*a*b*sin C. But how does that help?Wait, let's get back to the given equation: sin(3A) + 3 sin C = 4. The maximum value of sine is 1, so sin(3A) ≤ 1 and 3 sin C ≤ 3. So, the maximum possible value of sin(3A) + 3 sin C is 1 + 3 = 4. So, the equation given is sin(3A) + 3 sin C = 4, which is equal to the maximum possible sum. Therefore, for this equation to hold, both terms must achieve their maximum values simultaneously. That is, sin(3A) = 1 and 3 sin C = 3, which implies sin C = 1. So, let's check if that's possible.If sin(3A) = 1, then 3A = π/2 + 2πk, for some integer k. But since A is an angle in a triangle, 0 < A < π, so 0 < 3A < 3π. The principal solution would be 3A = π/2, so A = π/6 (30 degrees). Alternatively, 3A = 5π/2, which would make A = 5π/6 (150 degrees), but then angle A is 150 degrees, which would leave only 30 degrees for angles B and C combined, but since sin C = 1 implies C = π/2 (90 degrees), which would not be possible. Therefore, the only feasible solution is A = π/6 (30 degrees), and sin C = 1, so C = π/2 (90 degrees).Therefore, angles A = 30°, C = 90°, so angle B = 180° - 30° - 90° = 60°. Therefore, triangle ABC is a right-angled triangle at C, with angle A = 30°, angle B = 60°, and hypotenuse AB = 10. Wait, but if angle C is 90°, then AB would be the hypotenuse. In a right-angled triangle with hypotenuse 10 and angle A = 30°, the sides would be as follows: side opposite 30° (BC) is half the hypotenuse, so BC = 5, and side AC (adjacent to 30°) is 5√3. Therefore, the altitude from C to AB... Wait, but in a right-angled triangle at C, the altitude from C to AB is actually the same as the legs? Wait, no. Wait, in a right-angled triangle, the legs are the altitudes. If the triangle is right-angled at C, then the altitude from C to AB is just the length of the leg CC, but since C is the right angle, the altitude is zero? Wait, no. Wait, that doesn't make sense. Wait, let's think again.If triangle ABC is right-angled at C, then sides AC and BC are the legs, and AB is the hypotenuse. The altitude from C to AB is the same as the legs only in the case of an isosceles right-angled triangle, but here, we have angles 30°, 60°, 90°, so sides are 5, 5√3, 10. The altitude from C to AB in a right-angled triangle can be calculated by the formula: (leg1 * leg2)/ hypotenuse. So, altitude h = (AC * BC)/AB = (5√3 * 5)/10 = (25√3)/10 = (5√3)/2 ≈ 4.33. So, h = (5√3)/2. But let's verify that.Alternatively, in a right-angled triangle at C, the area is (1/2)*AC*BC. Also, the area is (1/2)*AB*h. Therefore, (1/2)*AC*BC = (1/2)*AB*h => h = (AC*BC)/AB. Which matches the formula above. So, with AC = 5√3, BC = 5, AB = 10, h = (5√3 * 5)/10 = 25√3/10 = 5√3/2. So, h = (5√3)/2. Therefore, the altitude is 5√3/2. So, is this the answer?Wait, but let's verify that the given equation holds. If A = 30°, then 3A = 90°, so sin(3A) = sin 90° = 1. And sin C = sin 90° = 1. So, the equation becomes 1 + 3*1 = 4, which matches. Therefore, this solution is valid. Therefore, the altitude is indeed 5√3/2. So, the answer should be 5√3/2. Let me just make sure that there are no other possible solutions.Earlier, we considered that sin(3A) + 3 sin C = 4 is maximum when sin(3A) = 1 and sin C = 1. Are there other possibilities where the sum is 4? For example, could sin(3A) be greater than 1? No, because sine cannot exceed 1. Similarly, sin C cannot exceed 1, so 3 sin C cannot exceed 3. Hence, the only way the sum is 4 is if both terms achieve their maximums: sin(3A) = 1 and 3 sin C = 3 => sin C = 1. Therefore, that's the only solution. Therefore, angle C must be 90°, and angle A must be 30°, leading to the altitude h = 5√3/2.Therefore, the altitude from C to AB is 5√3/2. So, the answer is boxed{dfrac{5sqrt{3}}{2}}.Wait, but let me check again. If triangle ABC is right-angled at C, then AB is the hypotenuse of length 10. Then, with angle A = 30°, the sides opposite 30°, 60°, 90° are 5, 5√3, 10. Therefore, side AC (adjacent to angle A) is 5√3, and BC (opposite angle A) is 5. Then, the area is (1/2)*5*5√3 = (25√3)/2. Then, the altitude h from C to AB is (2*Area)/AB = (25√3)/2 / 10 = (25√3)/20 = 5√3/4. Wait, hold on, this contradicts the previous result. Which one is correct?Wait, confusion here. Wait, if the triangle is right-angled at C, then the legs are AC and BC, and AB is the hypotenuse. If angle A is 30°, then side BC (opposite angle A) is 5, and AC (adjacent to angle A) is 5√3. Then, the altitude from C to AB is different from the legs. Wait, but in a right-angled triangle, the altitude from the right angle to the hypotenuse can be calculated as (AC * BC)/AB. So, (5√3 * 5)/10 = 25√3 / 10 = 5√3 / 2. So that's 5√3 / 2. Alternatively, using area: Area is (1/2)*5√3*5 = (25√3)/2. Then, h = (2*Area)/AB = (25√3)/2 / 10 = (25√3)/20 = 5√3/4. Wait, wait, this is conflicting. Which is it?Hold on, there must be a mistake here. Let me clarify. In a right-angled triangle, the area is (1/2)*leg1*leg2. But if we consider the hypotenuse as the base, then the altitude h to the hypotenuse is given by h = (leg1*leg2)/hypotenuse. Let me verify this formula.Yes, in a right-angled triangle, the altitude h from the right angle to the hypotenuse is h = (leg1 * leg2)/ hypotenuse. So, here, legs are 5 and 5√3, hypotenuse is 10. So, h = (5 * 5√3)/10 = (25√3)/10 = 5√3/2. Therefore, h = 5√3/2. But when I calculated using area, I did:Area = (1/2)*5*5√3 = 25√3/2. Then, h = (2*Area)/AB = (25√3/2)*2 /10 = 25√3 /10 = 5√3/2. Wait, wait, no. Wait, area is (1/2)*base*height. If we take AB as the base (length 10), then area is also (1/2)*10*h. Therefore:(1/2)*10*h = (1/2)*5*5√3 => 5h = (25√3)/2 => h = (25√3)/10 = 5√3/2.Yes, so both methods give h = 5√3/2. Earlier mistake in the area calculation was probably due to a miscalculation. Therefore, h is indeed 5√3/2. So, the altitude is 5√3/2. Therefore, the answer is boxed{dfrac{5sqrt{3}}{2}}.But let me just confirm once more that angle C is 90°, angle A is 30°, which gives us this configuration. Since the given equation sin(3A) + 3 sin C = 4 is satisfied (sin 90° + 3 sin 90° = 1 + 3 = 4), this is valid. Therefore, the triangle is right-angled at C, with angles 30°, 60°, 90°, sides 5, 5√3, 10, and altitude 5√3/2. Therefore, the answer should be 5√3/2.Alternatively, is there another triangle configuration that satisfies the equation sin(3A) + 3 sin C = 4? For example, suppose angle A is something else, and angle C is something else, such that even if sin(3A) and sin C are less than 1, their sum could still be 4. But since sine functions have a maximum of 1, and 3 sin C has a maximum of 3, the only way their sum is 4 is if both terms are at their maximum. Therefore, there can't be another solution. Hence, the only possible solution is angle A = 30°, angle C = 90°, leading to altitude 5√3/2.Therefore, the answer is boxed{dfrac{5sqrt{3}}{2}}.Final Answerboxed{dfrac{5sqrt{3}}{2}}

✅:Given triangle ( triangle ABC ) with ( AB = 10 ) and the equation ( sin 3A + 3 sin C = 4 ), we need to find the length of the altitude from ( C ) to ( AB ).1. Using Trigonometric Identities: - The triple angle formula for sine: ( sin 3A = 3 sin A - 4 sin^3 A ). - Substituting this into the given equation: ( 3 sin A - 4 sin^3 A + 3 sin C = 4 ).2. Maximizing Sine Values: - The maximum value of ( sin theta ) is 1. For the equation ( sin 3A + 3 sin C = 4 ) to hold, both terms must be at their maximum values. - Therefore, ( sin 3A = 1 ) and ( 3 sin C = 3 ), leading to ( sin 3A = 1 ) and ( sin C = 1 ).3. Determining Angles: - ( sin 3A = 1 ) implies ( 3A = frac{pi}{2} ) (since ( A ) is an angle in a triangle), so ( A = frac{pi}{6} ) (30 degrees). - ( sin C = 1 ) implies ( C = frac{pi}{2} ) (90 degrees).4. Triangle Properties: - With ( A = 30^circ ) and ( C = 90^circ ), the remaining angle ( B = 60^circ ). - Triangle ( triangle ABC ) is a right-angled triangle at ( C ), with hypotenuse ( AB = 10 ).5. Calculating Sides: - In a 30-60-90 triangle, the sides are in the ratio ( 1 : sqrt{3} : 2 ). - The side opposite ( 30^circ ) (BC) is 5, and the side opposite ( 60^circ ) (AC) is ( 5sqrt{3} ).6. Finding the Altitude: - The altitude from ( C ) to ( AB ) in a right-angled triangle is given by ( frac{text{leg}_1 times text{leg}_2}{text{hypotenuse}} ). - Substituting the values: ( frac{5 times 5sqrt{3}}{10} = frac{25sqrt{3}}{10} = frac{5sqrt{3}}{2} ).Thus, the length of the altitude from ( C ) to ( AB ) is (boxed{dfrac{5sqrt{3}}{2}}).

🔑:1. Given ( triangle ABC ) with ( AB = 10 ) and the equation ( sin 3A + 3 sin C = 4 ), we start by leveraging the range of the sine function. For any angle ( x ), we have:[-1 leq sin x leq 1]2. Applying this to both ( sin 3A ) and ( sin C ):[-1 leq sin 3A leq 1 quad text{and} quad -1 leq sin C leq 1]3. Given the equation ( sin 3A + 3 sin C = 4 ), we recognize:[sin 3A leq 1 quad text{and} quad 3 sin C leq 3]4. The sum ( 1 + 3 = 4 ) is reached only if both ( sin 3A = 1 ) and ( sin C = 1 ). Hence:[sin 3A = 1 quad text{and} quad sin C = 1]5. Since ( 0^circ < C < 180^circ ) for a triangle, the condition ( sin C = 1 ) implies:[C = 90^circ]6. Consequently, ( triangle ABC ) is a right-angled triangle at ( C ).7. Substituting back for angle ( A ), we know in a right-angled triangle with ( C = 90^circ ), other angles should lie between ( 0^circ ) and ( 90^circ ). Hence ( 0^circ < A < 90^circ ) applies.8. For ( sin 3A = 1 ) and considering ( 0^circ < A < 90^circ ), ( 3A ) must lie in the first quadrant:[0^circ < 3A < 270^circ]9. With ( sin 3A = 1 ), ( 3A = 90^circ ) provides:[A = 30^circ]10. Accordingly, ( triangle ABC ) is identified as a ( 30^circ - 60^circ - 90^circ ) triangle, with hypotenuse ( AB = 10 ).11. The side ( BC ), opposite ( 30^circ ), is half the hypotenuse:[BC = frac{1}{2} times 10 = 5]12. The side ( AC ), opposite ( 60^circ ), calculates as:[AC = frac{sqrt{3}}{2} times 10 = 5sqrt{3}]13. The area of ( triangle ABC ) finds as:[text{Area} = frac{1}{2} times AC times BC = frac{1}{2} times 5 times 5sqrt{3} = frac{25sqrt{3}}{2}]14. Suppose ( h ) is the length of the altitude from ( C ) to ( AB ). The area is also expressed as:[text{Area} = frac{1}{2} times AB times h = 5h]15. Equate the two area expressions:[5h = frac{25sqrt{3}}{2}]16. Solve for ( h ):[h = frac{25sqrt{3}}{2} cdot frac{1}{5} = frac{5sqrt{3}}{2}]Conclusion:[boxed{frac{5sqrt{3}}{2}}]