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

The correct option is C $\frac{n(4n^2-1)c^2}{3}$

$S_n=c{n}^2$
$S_{n-1}=c(n-1{)}^2=c{n}^2+c-2cn$
$T_n=2cn-c$
$T_n^2=(2cn-c{)}^2=4c^2{n}^2+c^2-4c^{2}n$
Required sum
$=\sum T_n^2=4c^2\sum n^2+n{c}^2-4c^2\sum n$
=$\frac{4c^{2}n(n+1)(2n+1)}{6}+n{c}^2-2c^{2}n(n+1)$
=$\frac{2c^{2}n(n+1)(2n+1)+3n{c}^2-6c^{2}n(n+1)}{3}$
=$\frac{n{c}^2[4n^2+6n+2+3-6n-6]}{3}$
=$\frac{n{c}^2(4n^2-1)}{3}$