💡First, I need to determine the cost of one packet of pumpkin seeds. Let's denote this cost as ( P ) dollars.Harry purchased:- 3 packets of pumpkin seeds, which cost ( 3P ) dollars,- 4 packets of tomato seeds at 1.50 each, totaling ( 4 times 1.50 = 6 ),- 5 packets of chili pepper seeds at 0.90 each, totaling ( 5 times 0.90 = 4.50 ).The total amount Harry spent is the sum of these amounts:[3P + 6 + 4.50 = 18]Next, I'll solve for ( P ):[3P = 18 - 6 - 4.50][3P = 7.50][P = frac{7.50}{3} = 2.50]Therefore, the cost of one packet of pumpkin seeds is 2.50.

💡First, determine the total number of books Laura has initially, which is (17n).After donating one book, the remaining number of books is (17n - 1).We need (17n - 1) to be a multiple of 11, which can be expressed as:[17n equiv 1 pmod{11}]Simplify the congruence:[17n equiv 1 pmod{11}][6n equiv 1 pmod{11}]Find the multiplicative inverse of 6 modulo 11. Testing values, we find:[6 times 2 = 12 equiv 1 pmod{11}]So, the inverse is 2.Thus, the congruence becomes:[n equiv 2 pmod{11}]The smallest positive integer (n) that satisfies this is 2.Verify with (n = 2):[17 times 2 = 34]After donating one book:[34 - 1 = 33][33 div 11 = 3]Therefore, the smallest possible value of (n) is 2.

💡First, I need to determine how many gallons Roman used. Since the total water used by both Roman and Remy is 33 gallons and Remy used 25 gallons, I can subtract Remy's usage from the total to find Roman's usage.Next, I know that Remy used 1 more gallon than a certain multiple of Roman's usage. Let's denote this multiple as ( M ). This relationship can be expressed with the equation ( 25 = M times 8 + 1 ).To find ( M ), I'll subtract 1 from 25 to get 24, and then divide by 8. This calculation will give me the value of ( M ).Finally, by performing the division, I determine that ( M ) is 3.

💡To translate the graph of the function ( y = sin 2x ) to the left by ( dfrac{pi}{3} ) units, we need to adjust the argument of the sine function accordingly. 1. **Understand the Translation Direction**: - Translating a graph to the left by ( c ) units involves replacing ( x ) with ( x + c ) in the function's equation.2. **Apply the Translation**: - Start with the original function: ( y = sin 2x ). - To shift it left by ( dfrac{pi}{3} ) units, substitute ( x ) with ( x + dfrac{pi}{3} ): [ y = sin [2(x + dfrac{pi}{3})] ]3. **Simplify the Expression**: - Distribute the 2 inside the sine function: [ y = sin (2x + dfrac{2pi}{3}) ]4. **Conclusion**: - The resulting equation after the translation is ( y = sin (2x + dfrac{2pi}{3}) ), which corresponds to option C.

💡First, I need to solve the congruence equation (5x + 9 equiv 3 pmod{20}). To isolate (x), I'll subtract 9 from both sides:[5x equiv -6 pmod{20}]Since (-6) modulo 20 is equivalent to 14, the equation becomes:[5x equiv 14 pmod{20}]Next, I need to find the modular inverse of 5 modulo 20. The inverse of 5 is a number (y) such that (5y equiv 1 pmod{20}). Testing values, I find that (5 times 4 = 20 equiv 0 pmod{20}), which doesn't satisfy the condition. Therefore, 5 does not have an inverse modulo 20, meaning the equation has no solution.Since there is no solution for (x), it's impossible to determine the value of (3x + 14) modulo 20.

💡Okay, so I have this problem where I need to calculate the sum of g(n) from n=3 to infinity. The function g(n) is defined as the sum starting from 1/3^n, then 1/4^n, 1/5^n, and so on. So, g(n) = 1/3^n + 1/4^n + 1/5^n + ... First, I need to understand what exactly g(n) represents. It looks like it's a series where each term is the reciprocal of an integer raised to the power of n, starting from 3. So, for each n, g(n) is the sum of 1/k^n where k starts at 3 and goes to infinity. So, the problem is asking for the sum of these g(n) from n=3 to infinity. That is, I need to compute:Sum_{n=3}^∞ [Sum_{k=3}^∞ 1/k^n]Hmm, that's a double sum. Maybe I can switch the order of summation? I remember that sometimes swapping the order of summation can make things easier. Let me try that.So, if I switch the order, I would have:Sum_{k=3}^∞ [Sum_{n=3}^∞ 1/k^n]That seems manageable. For each k starting from 3, I sum over n from 3 to infinity of 1/k^n. That inner sum is a geometric series, right?Yes, for each fixed k, the inner sum is Sum_{n=3}^∞ (1/k)^n. The formula for the sum of a geometric series starting at n=0 is a/(1 - r), where a is the first term and r is the common ratio. But here, the series starts at n=3, so I need to adjust for that.The sum from n=3 to infinity of r^n is equal to r^3 / (1 - r), provided that |r| < 1. In this case, r = 1/k, and since k starts at 3, 1/k is less than 1, so it converges.Therefore, the inner sum becomes (1/k)^3 / (1 - 1/k). Let's compute that:(1/k^3) / (1 - 1/k) = (1/k^3) / ((k - 1)/k) = (1/k^3) * (k / (k - 1)) = 1 / (k^2 (k - 1))So, the entire expression becomes:Sum_{k=3}^∞ [1 / (k^2 (k - 1))]Now, I need to compute this sum. Let's see if I can simplify the term 1 / (k^2 (k - 1)). Maybe partial fractions can help here.Let me try to express 1 / (k^2 (k - 1)) as A/(k - 1) + B/k + C/k^2. Let's find A, B, and C.So,1 = A k^2 + B (k - 1)k + C (k - 1)Let me expand the right-hand side:A k^2 + B (k^2 - k) + C (k - 1) = A k^2 + B k^2 - B k + C k - CCombine like terms:(A + B) k^2 + (-B + C) k + (-C)This must equal 1, which is a constant. Therefore, the coefficients of k^2 and k must be zero, and the constant term must be 1.So, setting up the equations:1. A + B = 0 (coefficient of k^2)2. -B + C = 0 (coefficient of k)3. -C = 1 (constant term)From equation 3: -C = 1 => C = -1From equation 2: -B + C = 0 => -B -1 = 0 => B = -1From equation 1: A + B = 0 => A -1 = 0 => A = 1So, the partial fractions decomposition is:1 / (k^2 (k - 1)) = 1/(k - 1) - 1/k - 1/k^2Therefore, the sum becomes:Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2]Let me split this into three separate sums:Sum_{k=3}^∞ 1/(k - 1) - Sum_{k=3}^∞ 1/k - Sum_{k=3}^∞ 1/k^2Let me handle each sum separately.First sum: Sum_{k=3}^∞ 1/(k - 1). Let's make a substitution: let m = k - 1. Then when k=3, m=2, and as k approaches infinity, m approaches infinity. So, this sum becomes Sum_{m=2}^∞ 1/m.Second sum: Sum_{k=3}^∞ 1/k. That's straightforward, it's just the harmonic series starting from k=3.Third sum: Sum_{k=3}^∞ 1/k^2. That's the sum of reciprocals of squares starting from k=3.So, putting it all together:Sum_{m=2}^∞ 1/m - Sum_{k=3}^∞ 1/k - Sum_{k=3}^∞ 1/k^2Notice that the first two sums are similar. Let's write them out:Sum_{m=2}^∞ 1/m = 1/2 + 1/3 + 1/4 + 1/5 + ...Sum_{k=3}^∞ 1/k = 1/3 + 1/4 + 1/5 + ...So, subtracting these, we get:[1/2 + 1/3 + 1/4 + ...] - [1/3 + 1/4 + 1/5 + ...] = 1/2That's nice, the first two sums cancel out all terms except 1/2.So, now we have:1/2 - Sum_{k=3}^∞ 1/k^2So, the entire expression simplifies to 1/2 minus the sum from k=3 to infinity of 1/k^2.Now, I need to compute Sum_{k=3}^∞ 1/k^2. I know that the sum from k=1 to infinity of 1/k^2 is π^2/6, which is a famous result known as the Basel problem. So, if I subtract the first two terms, I can get the sum starting from k=3.Sum_{k=3}^∞ 1/k^2 = Sum_{k=1}^∞ 1/k^2 - 1/1^2 - 1/2^2 = π^2/6 - 1 - 1/4 = π^2/6 - 5/4Therefore, the expression becomes:1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4Let me compute this:1/2 is 2/4, 5/4 is 5/4, so 2/4 + 5/4 = 7/4So, 7/4 - π^2/6Wait, but hold on. The original sum was Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2] which simplified to 1/2 - Sum_{k=3}^∞ 1/k^2But now, I have 1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4 = (1/2 + 5/4) - π^2/6 = (7/4) - π^2/6Hmm, but wait, that seems a bit complicated. Is there another way to approach this?Wait, maybe I made a mistake in the partial fractions decomposition. Let me double-check that.We had:1 / (k^2 (k - 1)) = A/(k - 1) + B/k + C/k^2Multiplying both sides by k^2 (k - 1):1 = A k^2 + B (k - 1)k + C (k - 1)Expanding:1 = A k^2 + B k^2 - B k + C k - CCombine like terms:(A + B) k^2 + (-B + C) k + (-C) = 1So, equating coefficients:A + B = 0-B + C = 0-C = 1So, from the last equation, C = -1Then from the second equation: -B + (-1) = 0 => -B -1 = 0 => B = -1From the first equation: A + (-1) = 0 => A = 1So, the partial fractions are correct: 1/(k - 1) - 1/k - 1/k^2So, that part is correct.Therefore, the sum is indeed 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6But wait, that can't be right because the original sum was supposed to be a finite value, but 7/4 - π^2/6 is approximately 1.75 - 1.6449 ≈ 0.105, which is a small positive number.But let me think again. The original problem was to compute Sum_{n=3}^∞ g(n), where g(n) = Sum_{k=3}^∞ 1/k^n.So, the sum is over n starting at 3, and for each n, sum over k starting at 3.After swapping the order, we had Sum_{k=3}^∞ Sum_{n=3}^∞ 1/k^n = Sum_{k=3}^∞ [1/k^3 + 1/k^4 + 1/k^5 + ...] = Sum_{k=3}^∞ [1/k^3 / (1 - 1/k)] = Sum_{k=3}^∞ 1/(k^2 (k - 1)).Which we decomposed into partial fractions and got 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6.But wait, 7/4 is 1.75 and π^2/6 is approximately 1.6449, so 1.75 - 1.6449 ≈ 0.105.But let me check if I can compute this sum another way.Alternatively, maybe I can express the sum as telescoping.Wait, the partial fractions decomposition was 1/(k - 1) - 1/k - 1/k^2.So, when we sum from k=3 to infinity, we have:Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2] = Sum_{k=3}^∞ [1/(k - 1) - 1/k] - Sum_{k=3}^∞ 1/k^2The first part, Sum_{k=3}^∞ [1/(k - 1) - 1/k], is a telescoping series.Let me write out the terms:When k=3: 1/2 - 1/3When k=4: 1/3 - 1/4When k=5: 1/4 - 1/5And so on.So, adding these up, most terms cancel:(1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ... = 1/2 - lim_{m→∞} 1/m = 1/2 - 0 = 1/2So, the first part is 1/2.Then, subtracting the sum of 1/k^2 from k=3 to infinity, which we found earlier as π^2/6 - 1 - 1/4 = π^2/6 - 5/4.So, the total sum is 1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4 = (1/2 + 5/4) - π^2/6 = 7/4 - π^2/6.So, that's approximately 1.75 - 1.6449 ≈ 0.105.But wait, is there a way to express this exactly? Because 7/4 - π^2/6 is an exact expression, but maybe it can be simplified further.Alternatively, perhaps I made a mistake in the partial fractions decomposition. Let me double-check.Wait, another approach: instead of partial fractions, maybe I can express 1/(k^2 (k - 1)) as something else.Let me consider that 1/(k^2 (k - 1)) can be written as [1/(k - 1) - 1/k] / k.Because:[1/(k - 1) - 1/k] = 1/(k - 1) - 1/k = (k - (k - 1)) / (k(k - 1)) = 1 / (k(k - 1))So, [1/(k - 1) - 1/k] / k = 1/(k^2 (k - 1))Which is exactly the term we have.Therefore, 1/(k^2 (k - 1)) = [1/(k - 1) - 1/k] / kSo, the sum becomes:Sum_{k=3}^∞ [1/(k - 1) - 1/k] / kWhich is equal to Sum_{k=3}^∞ [1/(k(k - 1)) - 1/k^2]Wait, that's the same as before, because 1/(k(k - 1)) = 1/(k - 1) - 1/k.So, that doesn't seem to help much.Alternatively, maybe I can write 1/(k^2 (k - 1)) as 1/(k - 1) - 1/k - 1/k^2, which is what we did earlier.So, I think the partial fractions approach is correct.Therefore, the sum is indeed 7/4 - π^2/6.But let me compute this exactly:7/4 is 1.75, π^2 is approximately 9.8696, so π^2/6 ≈ 1.6449.So, 1.75 - 1.6449 ≈ 0.1051.But wait, the problem is asking for an exact value, not an approximate one.So, 7/4 - π^2/6 is an exact expression, but perhaps it can be written differently.Alternatively, maybe I made a mistake in the initial steps.Wait, let me go back.We had:Sum_{n=3}^∞ g(n) = Sum_{n=3}^∞ Sum_{k=3}^∞ 1/k^n = Sum_{k=3}^∞ Sum_{n=3}^∞ 1/k^nWhich is Sum_{k=3}^∞ [1/k^3 + 1/k^4 + 1/k^5 + ...] = Sum_{k=3}^∞ [1/k^3 / (1 - 1/k)] = Sum_{k=3}^∞ 1/(k^2 (k - 1))Then, partial fractions decomposition gives 1/(k - 1) - 1/k - 1/k^2.So, Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2] = [Sum_{k=3}^∞ (1/(k - 1) - 1/k)] - Sum_{k=3}^∞ 1/k^2The first sum telescopes to 1/2, as we saw earlier.The second sum is Sum_{k=3}^∞ 1/k^2 = π^2/6 - 1 - 1/4 = π^2/6 - 5/4.So, the total is 1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4 = 7/4 - π^2/6.So, that's the exact value.But wait, the problem is asking for the sum, and in the initial problem statement, the user provided a solution that ended up with 1/2. But in my calculation, I got 7/4 - π^2/6, which is approximately 0.105, not 1/2.So, perhaps I made a mistake in the partial fractions decomposition or in the telescoping.Wait, let me check the telescoping again.We have:Sum_{k=3}^∞ [1/(k - 1) - 1/k] = lim_{m→∞} [1/2 - 1/m] = 1/2Yes, that's correct.Then, Sum_{k=3}^∞ 1/k^2 = π^2/6 - 1 - 1/4 = π^2/6 - 5/4.So, the total sum is 1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4 = 7/4 - π^2/6.Hmm, so that's the exact value.But wait, in the initial problem, the user had a different approach and got 1/2. Maybe I misapplied the partial fractions.Wait, let me check the partial fractions decomposition again.We had:1 / (k^2 (k - 1)) = A/(k - 1) + B/k + C/k^2Multiplying both sides by k^2 (k - 1):1 = A k^2 + B (k - 1)k + C (k - 1)Expanding:1 = A k^2 + B k^2 - B k + C k - CCombine like terms:(A + B) k^2 + (-B + C) k + (-C) = 1So, equating coefficients:A + B = 0-B + C = 0-C = 1From the last equation, C = -1From the second equation: -B -1 = 0 => B = -1From the first equation: A -1 = 0 => A = 1So, the decomposition is correct: 1/(k - 1) - 1/k - 1/k^2.So, the sum is indeed 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6.Wait, but the user's initial solution got 1/2. Maybe they made a mistake.Alternatively, perhaps I misapplied the partial fractions.Wait, another approach: maybe instead of partial fractions, I can express 1/(k^2 (k - 1)) as the difference of two terms.Let me consider that 1/(k^2 (k - 1)) can be written as [1/(k - 1) - 1/k] / k.Which is the same as before.Alternatively, maybe I can write it as 1/(k - 1) - 1/k - 1/k^2, which is what we did.So, I think the partial fractions are correct.Therefore, the sum is 7/4 - π^2/6.But wait, let me compute this exactly:7/4 is 1.75, π^2 is approximately 9.8696, so π^2/6 ≈ 1.6449.So, 1.75 - 1.6449 ≈ 0.1051.But the user's solution got 1/2, which is 0.5.So, there's a discrepancy here.Wait, perhaps I made a mistake in the initial step of swapping the order of summation.Let me check that.We have Sum_{n=3}^∞ g(n) = Sum_{n=3}^∞ Sum_{k=3}^∞ 1/k^nSwapping the order, it's Sum_{k=3}^∞ Sum_{n=3}^∞ 1/k^nWhich is correct because both are positive terms, so Fubini's theorem allows swapping.So, that step is correct.Then, the inner sum is Sum_{n=3}^∞ 1/k^n = (1/k^3) / (1 - 1/k) = 1/(k^2 (k - 1))That's correct.So, the sum becomes Sum_{k=3}^∞ 1/(k^2 (k - 1)).Then, partial fractions decomposition gives 1/(k - 1) - 1/k - 1/k^2.So, the sum is Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2] = 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6.So, I think my calculation is correct, and the user's initial solution might have been incorrect.Alternatively, perhaps the user's problem was different.Wait, looking back at the problem statement, the user wrote:"For a positive integer n, define g(n) = 1/3^n + 1/4^n + 1/5^n + ... Calculate the value of Sum_{n=3}^∞ g(n)."So, that's exactly the problem I'm solving.But in the initial solution, the user wrote:"We need to find the sum [...] The sum of the numbers in the nth column is an infinite geometric series starting from 1/(n+2)^3 [...] The common ratio of this series is 1/(n+2). Therefore, the sum of these series is 1/((n+1)(n+2)^2). This expression needs further simplification, but by the telescoping nature of the sums, most terms cancel, and the pattern will collapse similarly: Sum_{n=1}^∞ [1/(n+1) - 1/(n+2)] = 1/2."Wait, that seems different from my approach.Wait, in the user's solution, they considered the sum as a double sum where each column is a geometric series starting from 1/(n+2)^3, but I think they might have misapplied the indices.Wait, in their approach, they considered the sum as:Sum_{n=3}^∞ g(n) = Sum_{n=3}^∞ Sum_{k=3}^∞ 1/k^nBut then they reindexed it as columns, where each column corresponds to a fixed k, and each row corresponds to n.Wait, no, in their approach, they seem to have considered the sum as rows being n and columns being k, but their indexing might be off.Wait, in their solution, they wrote:"The sum of the numbers in the nth column is an infinite geometric series starting from 1/(n+2)^3 because our sequence starts from 3^n, and continues with 1/(n+2)^4, 1/(n+2)^5, ... The common ratio of this series is 1/(n+2). Therefore, the sum of these series is 1/((n+1)(n+2)^2)."Wait, that seems incorrect because if we fix n, then the terms are 1/3^n, 1/4^n, 1/5^n, etc., which is a series in k, not in n.So, perhaps the user confused the indices.In other words, in their solution, they seem to have treated n as the column index, but in reality, n is the row index, and k is the column index.Therefore, their approach might have been flawed.In contrast, my approach correctly swapped the order of summation, leading to the correct expression.Therefore, I think my result of 7/4 - π^2/6 is correct.But let me compute this exactly:7/4 - π^2/6 = (21 - 2π^2)/12Because 7/4 is 21/12, and π^2/6 is 2π^2/12.So, 21/12 - 2π^2/12 = (21 - 2π^2)/12.But that's still an exact expression, but it's not a simple fraction like 1/2.Alternatively, maybe I can express it in terms of the Riemann zeta function.We know that Sum_{k=1}^∞ 1/k^2 = ζ(2) = π^2/6.So, Sum_{k=3}^∞ 1/k^2 = ζ(2) - 1 - 1/4 = π^2/6 - 5/4.Therefore, the total sum is 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6.So, that's the exact value.But wait, the problem is asking for the value of the sum, and in the initial problem statement, the user provided a solution that ended up with 1/2, but I think that solution was incorrect because of a misindexing.Therefore, I think the correct answer is 7/4 - π^2/6, which is approximately 0.105.But wait, let me check if there's another way to approach this.Alternatively, maybe I can express the sum as a telescoping series.Wait, from the partial fractions decomposition, we have:Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2] = 1/2 - Sum_{k=3}^∞ 1/k^2Which is 1/2 - (π^2/6 - 1 - 1/4) = 1/2 - π^2/6 + 5/4 = 7/4 - π^2/6.So, that's consistent.Alternatively, maybe I can write this as:7/4 - π^2/6 = (21 - 2π^2)/12But I don't think that simplifies further.Alternatively, perhaps the problem expects an answer in terms of ζ(2), but I think the exact value is 7/4 - π^2/6.Alternatively, maybe I made a mistake in the partial fractions decomposition.Wait, let me try another approach.Let me consider the sum Sum_{k=3}^∞ 1/(k^2 (k - 1)).Let me write this as Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2].Then, the sum becomes:Sum_{k=3}^∞ [1/(k - 1) - 1/k] - Sum_{k=3}^∞ 1/k^2The first sum is telescoping:Sum_{k=3}^∞ [1/(k - 1) - 1/k] = lim_{m→∞} [1/2 - 1/m] = 1/2The second sum is Sum_{k=3}^∞ 1/k^2 = ζ(2) - 1 - 1/4 = π^2/6 - 5/4Therefore, the total sum is 1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4 = 7/4 - π^2/6.So, that's consistent.Therefore, I think the correct answer is 7/4 - π^2/6.But wait, the user's initial solution got 1/2, which is different.So, perhaps the user made a mistake in their approach.Alternatively, maybe the problem was intended to have a telescoping sum that results in 1/2, but due to a misindexing, the user's solution was incorrect.In any case, based on my calculations, the correct answer is 7/4 - π^2/6.But let me check if this can be simplified further or expressed differently.Alternatively, perhaps I can write it as:7/4 - π^2/6 = (21 - 2π^2)/12But that's still not a simple fraction.Alternatively, maybe I can leave it as 7/4 - π^2/6.Alternatively, perhaps the problem expects an answer in terms of ζ(2), but I think 7/4 - π^2/6 is the simplest exact form.Therefore, I think the final answer is 7/4 - π^2/6.But wait, let me compute this exactly:7/4 is 1.75, π^2 is approximately 9.8696, so π^2/6 ≈ 1.6449.So, 1.75 - 1.6449 ≈ 0.1051.But the problem is asking for the exact value, so 7/4 - π^2/6 is the exact value.Therefore, I think that's the answer.But wait, let me check if I can express this as a telescoping series in another way.Wait, another approach: Let me consider that 1/(k^2 (k - 1)) can be written as 1/(k - 1) - 1/k - 1/k^2.So, the sum becomes:Sum_{k=3}^∞ [1/(k - 1) - 1/k - 1/k^2] = Sum_{k=3}^∞ [1/(k - 1) - 1/k] - Sum_{k=3}^∞ 1/k^2As before, the first sum telescopes to 1/2, and the second sum is π^2/6 - 5/4.Therefore, the total is 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6.So, that's consistent.Therefore, I think the correct answer is 7/4 - π^2/6.But wait, the user's initial solution got 1/2, which is different.So, perhaps the user made a mistake in their approach.In conclusion, after carefully considering the problem and verifying each step, I believe the correct value of the sum is 7/4 - π^2/6.