Question

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

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

Answer 1

The correct option is D

$\frac{n(4n^2-1)}{3}{C}^2$

In a sequence, we get the sum of n terms by adding $n^{th}$ term to the sum of (n-1) terms. we can write this as
$S_n=S_{n-1}+t_n$. We will first find the nth term $t_n$, of this AP. Once we have the nth term, we can find the nth term of the second AP which is the square of the first one. We will add the terms of the second A.P to get the required sum.

$\Rightarrow t_n=S_n-S_{(n-1)}{T}_n=S_n-S_{n-1}$
$=c{n}^2-(cn-1{)}^2$
$=c(n^2-(n^2-2n+1\left)\right)$
$=(2n-1)c$
We want to find ${\Sigma }_{n=1}^{2}T{n}^2$
$=\Sigma (2n-1{)}^2{c}^2$
$=c^2\Sigma (4n^2-4n+1)$
$=c^2(4\Sigma n^2-4\Sigma n+\Sigma 1)$
$=c^2\frac{2(n^2+n)(2n+1)-6(n^2+n)+3n}{3}$
$=\frac{c^2}{3}(4n^3+2n^2+4n^2+2n-6n^2-6n+3n)$
$=\frac{c^2}{3}(4n^3-n)$
$=\frac{c^2}{3}n(4n^2-1)-D$ is right option