If the sum of the first n terms of an A.P. is | vedclass

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(C) Given the sum of the first $n$ terms $S_n = c n^2$.The $n^{th}$ term $T_n = S_n - S_{n-1} = c n^2 - c(n-1)^2 = c(n^2 - (n^2 - 2n + 1)) = 2cn - c$.

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

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(C) Given the sum of the first $n$ terms $S_n = c n^2$.The $n^{th}$ term $T_n = S_n - S_{n-1} = c n^2 - c(n-1)^2 = c(n^2 - (n^2 - 2n + 1)) = 2cn - c$.We need to find the sum of the squares of these $n$ terms,i.e.,$\sum_{k=1}^n T_k^2$.$T_k^2 = (2ck - c)^2 = c^2(2k - 1)^2 = c^2(4k^2 - 4k + 1)$.Sum $= \sum_{k=1}^n c^2(4k^2 - 4k + 1) = c^2 [4 \sum k^2 - 4 \sum k + \sum 1]$.Using standard summation formulas:$= c^2 [4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n]$.$= c^2 [\frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n]$.$= \frac{c^2}{3} [2n(2n^2 + 3n + 1) - 6n^2 - 6n + 3n]$.$= \frac{c^2}{3} [4n^3 + 6n^2 + 2n - 6n^2 - 3n] = \frac{c^2}{3} [4n^3 - n] = \frac{n c^2(4n^2 - 1)}{3}$.

If the sum of the first $n$ terms of an $A.P.$ is $c n^2$,then the sum of the squares of these $n$ terms is

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