Q. If the sum of first $n$ terms of an AP is $cn^{2}$, then the sum of squares of these $n$ terms is
A) $\frac{n\left( 4n^{2} - 1 \right)c^{2}}{6}$
B) $\frac{n\left( 4n^{2} + 1 \right)c^{2}}{3}$
C) $\frac{n\left( 4n^{2} - 1 \right)c^{2}}{3}$
D) $\frac{n\left( 4n^{2} + 1 \right)c^{2}}{6}$
Explanation
(c)
Let $S_{n} = cn^{2}$, then
$S_{n - 1} = c(n - 1)^{2} = cn^{2} + c - 2cn$
$\therefore\ \ T_{n} = 2\ cn - c$ $(\because T_{n} = S_{n} - S_{n - 1})$
$T_{n}^{2} = (2cn - c)^{2} = 4c^{2}n^{2} + c^{2} - 4c^{2}n$
$\therefore Sum\ = \ \sum_{}^{}T_{n}^{2} = \frac{4c^{2}.n(n + 1)(2n + 1)}{6} + nc^{2} - 2c^{2}n(n + 1)$
$= \frac{2c^{2}n(n + 1)(2n + 1) + 3nc^{2} - 6c^{2}n(n + 1)}{3}$
$= \frac{nc^{2}(4n^{2} + 6n + 2 + 3 - 6n - 6)}{3}$
$= \frac{nc^{2}(4n^{2} - 1)}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.