Question

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

Answer 1

$\text{The given series is},3\times 1^2+5\times 2^2+7\times 3^2+...\text{nth term},a_n=(2n+1)n^2={2n}^3+n^2\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n=({2k}^3+k^2)=2{\sum }_{k=1}^n{k}^3+{\sum }_{k=1}^n{k}^2=2{\left[\frac{n(n+1)}{2}\right]}^2+\frac{n(n+1)(2n+1)}{6}={\frac{n^2(n+1)}{2}}^2+\frac{n(n+1)(2n+1)}{6}=\frac{n(n+1)}{2}\left[n\right(n+1)+\frac{2n+1}{3}]=\frac{n(n+1)}{2}\left[\frac{3n^2+5n+1}{3}\right]=\frac{n(n+1)(3n^2+5n+1)}{6}$