Question

Find the sum of series to n terms:
$3\times 1^2+5\times 2^2+7\times 3^2+...$

Answer 1

$3\times 1^2+5\times 2^2+7\times 3^2+...$

In each term of sequence, first number forms an A.P. with common difference $2$ and second number base is also forms an A.P. with common difference $1$

Let $T_n$ be the $n^{th}$ term of the given series, then

$T_n=\{n^{th}\text{term of}3,5,7,...\}\times \{n^{th}\text{term of}{1}^2,2^2,3^2,...\}$

$=[3+(n-1)\times 2]\times \left[n^2\right]$

$=[3+2n-2]\times n^2$

$=[2n+1]n^2$

$\Rightarrow T_n=2n^3+n^2\cdots \left(1\right)$

Now, let $S_n=\sum T_n$ terms of the given series.

We have:
$S_n=\sum T_n$

$=\sum (2n^3+n^2)$ [using equation (1)]

$\Rightarrow S_n=2\sum n^3+{\sum n}^2$

Upon simplification we get,

$=2{\left[\frac{n(n+1)}{2}\right]}^2+\left[\frac{n(n+1)(2n+1)}{6}\right]$

$=2\left[\frac{n^2{(n+1)}^2}{4}\right]+\left[\frac{n(n+1)(2n+1)}{6}\right]$

$=n(n+1)[\frac{n(n+1)}{2}+\frac{2n+1}{6}]$

$=n(n+1)\left(\frac{3n^2+3n+2n+1}{6}\right)$

$=\frac{1}{6}n(n+1)(3n^2+5n+1)$

$\therefore \text{Sum of series}{S}_n=\frac{n}{6}(n+1)(3n^2+5n+1)$