Question

$1^2+(1^2+2^2)+(1^2+2^2+3^2)+....$

Answer 1

$\text{The given series is},1^2+(1^2+2^2)+(1^2+2^2+3^2)+...a_n=(1^2+2^2+3^2+...+n^2)=\frac{n(n+1)(2n+1)}{6}=\frac{n({2n}^2+3n+1)}{6}=\frac{2^3+{3n}^2+n}{6}=\frac{1}{3}{n}^3+\frac{1}{2}{n}^2+\frac{1}{6}n\therefore S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n\left(\frac{1}{3}{k}^2+\frac{1}{2}{k}^2+\frac{1}{6}k\right)=\frac{1}{3}{\sum }_{k=1}^n{k}^3+\frac{1}{2}{\sum }_{k=1}^n{k}^2+\frac{1}{36}{\sum }_{k=1}^{n}k=\frac{1}{3}\frac{n^2(n+1{)}^2}{(2{)}^2}+\frac{1}{2}\times \frac{n(n+1)(2n+1)}{6}+\frac{1}{6}\times \frac{n(n+1)}{2}=\frac{n(n+1)}{6}[\frac{n(n+1)}{2}+\frac{(2n+1)}{2}+\frac{1}{2}]=\frac{n(n+1)}{6}\left[\frac{n^2+n+2n+1+1}{2}\right]=\frac{n(n+1)}{6}\left[\frac{n^2+n+2n+2}{2}\right]=\frac{n(n+1)}{6}\left[\frac{n(n+1)+2(n+1)}{2}\right]=\frac{n(n+1)}{6}\left[\frac{(n+1)(n+2)}{2}\right]=\frac{n{(n+1)}^2(n+2)}{12}$