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. $\raisebox{1ex}{$n(4n^2-1)c^2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$
• B. $\raisebox{1ex}{$n(4n^2+1)c^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• C. $\raisebox{1ex}{$n(4n^2-1)c^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• D. None of these

Answer 1

The correct option is C

$\raisebox{1ex}{$n(4n^2-1)c^2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Explanation for the correct option.

Sum of first $n$ terms of an AP is ${\mathit{S}}_{\mathbf{n}}\mathbf{=}\mathit{c}{\mathit{n}}^{\mathbf{2}}$

$\begin{array}{rcl}& \Rightarrow & S_{n-1}=c{\left(n-1\right)}^2\\ & =& c{n}^2-2cn+c\end{array}$

$\begin{array}{rcl}{T}_n& =& S_n-S_{n-1}\\ & =& 2cn-c\\ & =& c(2n-1)\end{array}$

$\begin{array}{rcl}{T}_{n^2}& =& {\left[c(2n-1)\right]}^2\\ & =& c^2\left(4n^2-4n+1\right)\end{array}$

The sum of squares $=\Sigma T_{n^2}$

$\begin{array}{rcl}\Sigma T_{n^2}& =& c^2\left[4\Sigma n^2-4\Sigma n+\left(1\times n\right)\right]\\ & =& c^2\left[4\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)-4\left(\frac{n\left(n+1\right)}{2}\right)+\left(n\right)\right]\\ & =& n{c}^2\left[2\left(\frac{2n^2+3n+1}{3}\right)-2\left(n+1\right)+1\right]\\ & =& n{c}^2\left[\frac{4n^2+6n+2-6n-6+3}{3}\right]\\ & =& n{c}^2\left[\frac{4n^2-1}{3}\right]\end{array}$

Hence, option C is correct.