Question

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

• A. $\frac{n\left(4 n^2-1\right) c^2}{6}$
• B. $\frac{n\left(4 n^2+1\right) c^2}{3}$
• C. $\frac{n\left(4 n^2-1\right) c^2}{3}$
• D. $\frac{n\left(4 n^2+1\right) c^2}{6}$

Answer 1

Answer: (C)

Let $S_n=c n^2$
$S_{n-1}=c(n-1)^2=c n^2+c-2 c n$
$\therefore T_n=2 c n-c \left(\because T_n=S_n-S_{n-1}\right)$
$T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n$
$\therefore$ Required sum,
$\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6}+n c^2-2 c^2 n(n+1)$
$=\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3}$
$=\frac{n c^2\left(4 n^2+6 n+2+3-6 n-6\right)}{3}$
$=\frac{n c^2\left(4 n^2-1\right)}{3}$