The sum of the series $3\times 1^2+5\times 2^2+7\times 3^2+\dots$ is

Solution:

 Let given series is 5 = 3 x 1² + 5 x 2² + 7 x 3² + ...
First, we will split the given series into two parts which are 3, 5,7, ... and 1²,2²,3², , . . and find the nth term of each part
separately to find the nth term of the given series. 

T_n = (rth term of 3,5, 7,. . ) x (nth term of 1,2,3, . . )²

$\begin{array}{l}=[3+(\mathrm{n}-1\left)2\right][1+(\mathrm{n}-1){]}^2\\ =(3+2\mathrm{n}-2)(\mathrm{n}{)}^2=(2\mathrm{n}+1){\mathrm{n}}^2=2{\mathrm{n}}^3+{\mathrm{n}}^2\\ \text{ Now, }\mathrm{S}={\mathrm{\Sigma T}}_{\mathrm{n}}=2{\mathrm{\Sigma n}}^3+{\mathrm{\Sigma n}}^2\\ =2{\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]}^2+\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}\left[\because {\mathrm{\Sigma n}}^3={\left\{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right\}}^2\right]\\ =\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[2\times \frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{2\mathrm{n}+1}{3}\right]\\ =\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{3\mathrm{n}(\mathrm{n}+1)+2\mathrm{n}+1}{3}\right]\\ =\frac{\mathrm{n}(\mathrm{n}+1)}{6}\times \left(3{\mathrm{n}}^2+3\mathrm{n}+2\mathrm{n}+1\right)\\ =\frac{\mathrm{n}(\mathrm{n}+1)\left(3{\mathrm{n}}^2+5\mathrm{n}+1\right)}{6}\end{array}$