Question

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

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

Answer 1

Answer: (C)

Let $s_n=c{n}^2$
$S_{n-1}=c(n-1{)}^2=c{n}^2+c-2cn$
$\therefore T_n=2cn-c[\because T_n=S_n-S_{n-1}]$
$T_n^2=(2cn-c{)}^2=4c^2{n}^2+c^2-4c^{2}n$
$\therefore$ Sum $=\sum T_n^2=\frac{4c^2.n(n+1)(2n+1)}{6}+n{c}^2-2c^{2}n(n+1)$
$=\frac{2c^{2}n(n+1)(2n+1)+3n{c}^2-6c^{2}n(n+1)}{3}$
$=\frac{n{c}^2(4n^2+6n+2+3-6n-6)}{3}=\frac{n{c}^2(4n^2-1)}{3}$