If S_n is the sum of the first n terms of the | vedclass

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(C) The series is $1^2, 2(2^2), 3^2, 2(4^2), 5^2, 2(6^2), \ldots$ For even $n$,let $n = 2m$. The sum $S_{2m}$ consists of $m$ odd-indexed terms and $m

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$\frac{n(n+1)^2}{2}$

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(C) The series is $1^2, 2(2^2), 3^2, 2(4^2), 5^2, 2(6^2), \ldots$ For even $n$,let $n = 2m$. The sum $S_{2m}$ consists of $m$ odd-indexed terms and $m$ even-indexed terms. $S_{2m} = (1^2 + 3^2 + 5^2 + \ldots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \ldots + (2m)^2)$ $= \sum_{k=1}^{m} (2k-1)^2 + 2 \sum_{k=1}^{m} (2k)^2$ $= \sum_{k=1}^{m} (4k^2 - 4k + 1) + 8 \sum_{k=1}^{m} k^2$ $= 12 \sum_{k=1}^{m} k^2 - 4 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 1$ $= 12 \frac{m(m+1)(2m+1)}{6} - 4 \frac{m(m+1)}{2} + m$ $= 2m(m+1)(2m+1) - 2m(m+1) + m$ $= 2m(m+1)(2m+1-1) + m$ $= 2m(m+1)(2m) + m = 4m^2(m+1) + m = m(4m^2 + 4m + 1) = m(2m+1)^2$ Since $n = 2m$,we have $m = n/2$. Substituting $m = n/2$ into the expression: $S_n = \frac{n}{2}(2(\frac{n}{2}) + 1)^2 = \frac{n}{2}(n+1)^2$.

If $S_n$ is the sum of the first $n$ terms of the series $1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots$,then for even $n$,$S_n$ is equal to:

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