Q. The sum to $n$ terms of the infinite series ${1.3}^{2} + {2.5}^{2} + {3.7}^{2} + ...\infty$
A) $\frac{n}{6}(n + 1)(6n^{2} + 14n + 7)$
B) $\frac{n}{6}(n + 1)(2n + 1)(3n + 1)$
C) $4n^{3} + 4n^{2} + n$
D) None of the above
Explanation
(a)
Here, $T_{n} = n(2n + 1)^{2} = 4n^{3} + 4n^{2} + n$
$\therefore S_{n} = \Sigma T_{n} = \Sigma\left( 4n^{3} + 4n^{2} + n \right) = 4\Sigma n^{3} + 4\Sigma n^{2} + \Sigma n$
$= 4\left\{ \frac{n}{2}(n + 1) \right\}^{2} + \frac{4}{6}n(n + 1)(2n + 1) + \frac{n}{2}(n + 1)$
$= n(n + 1)\left\lbrack n^{2} + n + \frac{4}{6}(2n + 1) + \frac{1}{2} \right\rbrack$
$= \frac{n}{6}(n + 1)(6n^{2} + 14n + 7)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.