The sum of all positive integers n for which | vedclass

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(A) The given expression is $S = \frac{\sum_{k=1}^{2n} k^3}{\sum_{k=1}^{n} k^2}$.Using the formulas $\sum_{k=1}^{m} k^3 = \left(\frac{m(m+1)}{2}\right

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$8$

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(A) The given expression is $S = \frac{\sum_{k=1}^{2n} k^3}{\sum_{k=1}^{n} k^2}$.Using the formulas $\sum_{k=1}^{m} k^3 = \left(\frac{m(m+1)}{2}\right)^2$ and $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we have:$S = \frac{\left(\frac{2n(2n+1)}{2}\right)^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(2n+1)^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{6n^2(2n+1)^2}{n(n+1)(2n+1)} = \frac{6n(2n+1)}{n+1}$.Now,simplify the expression:$S = \frac{12n^2+6n}{n+1} = \frac{12n(n+1) - 6n}{n+1} = 12n - \frac{6n}{n+1} = 12n - \frac{6(n+1)-6}{n+1} = 12n - 6 + \frac{6}{n+1}$.For $S$ to be an integer,$n+1$ must be a divisor of $6$.The divisors of $6$ are $1, 2, 3, 6$.Thus,$n+1 \in \{1, 2, 3, 6\}$,which gives $n \in \{0, 1, 2, 5\}$.Since $n$ must be a positive integer,$n \in \{1, 2, 5\}$.The sum of these values is $1+2+5 = 8$.

The sum of all positive integers $n$ for which $\frac{1^3+2^3+\ldots+(2n)^3}{1^2+2^2+\ldots+n^2}$ is also an integer is

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