If the sum of first $n$ terms of an $\mathrm{AP}$ is $\mathrm{cn}^2$, then the sum of squares of these $n$ terms is

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Aicharya · Accepted Answer

Let $S_n=c^2$ $ \begin{aligned} & S_{n-1}=c(n-1)^2=c n^2+c-2 c n \ & 	herefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}ight) \ & T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \ & 	herefore 	ext { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \ & =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \ & =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6ight]}{3} \ & =\frac{n c^2\left(4 n^2-1ight)}{3} \end{aligned} $

If the sum of first $n$ terms of an $\mathrm{AP}$ is $\mathrm{cn}^2$, then the sum of squares of these $n$…

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