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. $\frac{n(4n^2-1)c^2}{6}$
• 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 $\frac{n(4n^2-1)c^2}{3}$

$t_n=S_n-S_{n-1}=c\{n^2-(n-1{)}^2\}$$=c(2n-1)$

$\Rightarrow t_n^2=c^2(4n^2-4n+1)$
${\sum }_{n=1}^n{t}_n^2=c^2\{\frac{4n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}+n\}$
$=\frac{c^{2}n}{6}\left\{4\right(n+1\left)\right(2n+1)-12(n+1)+6\}$

$=\frac{c^{2}n}{3}\{4n^2+6n+2-6n-6+3\}=\frac{c^2}{3}n(4n^2-1)$