Question

The sum to $n$ terms of the infinite series $1 \cdot 3^2+2 \cdot 5^2+3 \cdot 7^2+\ldots . \infty$ is

• A. $\frac{n}{6}(n+1)\left(6 n^2+14 n+7\right)$
• B. $\frac{n}{6}(n+1)(2 n+1)(3 n+1)$
• C. $4 n^3+4 n^2+n$
• D. None of these

Answer 1

Answer: (A)

Given series is $1 \cdot 3^2+2 \cdot 5^2+3 \cdot 7^2+\ldots \infty$
This is an arithmetico-geometric series whose $n^{t h}$ term is equal to
$T_n =n(2 n+1)^2=4 n^3+4 n^2+n$
$\therefore S_n =\displaystyle\sum_1^n T_n=\displaystyle\sum_1^n\left(4 n^3+4 n^2+n\right) $
$=4 \displaystyle\sum_1^n n^3+4 \displaystyle\sum_1^n n^2+\displaystyle\sum_1^n n$
$ =4\left[\frac{n}{2}(n+1)\right]^2+\frac{4}{6} n(n+1)(2 n+1)+\frac{n}{2}(n+1) $
$ =n(n+1)\left[n^2+n+\frac{4}{6}(2 n+1)+\frac{1}{2}\right] $
$ =\frac{n}{6}(n+1)\left(6 n^2+14 n+7\right)$