Question

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

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

Answer 1

Answer: (A)

Given series is $1\cdot 3^2+2\cdot 5^2+3\cdot 7^2+\dots \infty$
This is an arithmetico-geometric series whose $n^{th}$ term is equal to
$T_n=n(2n+1{)}^2=4n^3+4n^2+n$
$\therefore S_n=\sum _1^n{T}_n=\sum _1^n\left(4n^3+4n^2+n\right)$
$=4\sum _1^n{n}^3+4\sum _1^n{n}^2+\sum _1^{n}n$
$=4{\left[\frac{n}{2}(n+1)\right]}^2+\frac{4}{6}n(n+1)(2n+1)+\frac{n}{2}(n+1)$
$=n(n+1)\left[n^2+n+\frac{4}{6}(2n+1)+\frac{1}{2}\right]$
$=\frac{n}{6}(n+1)\left(6n^2+14n+7\right)$