If the sum of the first n terms of an arithme | vedclass

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(C) Given $S_n = cn^2$.$S_{n-1} = c(n-1)^2$.The $n^{th}$ term $t_n = S_n - S_{n-1} = cn^2 - c(n-1)^2 = c(2n-1)$.The sum of the squares of these $n$ te

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$\frac{n(4n^2 - 1)c^2}{3}$

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(C) Given $S_n = cn^2$.$S_{n-1} = c(n-1)^2$.The $n^{th}$ term $t_n = S_n - S_{n-1} = cn^2 - c(n-1)^2 = c(2n-1)$.The sum of the squares of these $n$ terms is $\sum_{k=1}^{n} t_k^2 = \sum_{k=1}^{n} c^2(2k-1)^2$.$= c^2 \sum_{k=1}^{n} (4k^2 - 4k + 1)$.$= c^2 [4 \sum k^2 - 4 \sum k + \sum 1]$.$= c^2 [4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n]$.$= c^2 [\frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n]$.$= \frac{c^2 n}{3} [2(2n^2 + 3n + 1) - 6(n+1) + 3]$.$= \frac{c^2 n}{3} [4n^2 + 6n + 2 - 6n - 6 + 3]$.$= \frac{n(4n^2 - 1)c^2}{3}$.

If the sum of the first $n$ terms of an arithmetic progression is $cn^2$,what is the sum of the squares of these $n$ terms?

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