Question

The sum of 'n' terms of series $1^2+(1^2+2^2)+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+.......$ will be

• A. $\frac{1}{4}n(n+1{)}^2(n+2)$
• B. $\frac{1}{3}n(n+1{)}^2(n+2)$
• C. $\frac{1}{12}n(n+1)(n+2{)}^2$
• D. $\frac{1}{12}n(n+1{)}^2(n+2)$

Answer 1

The correct option is D $\frac{1}{12}n(n+1{)}^2(n+2)$

The given series is $1^2+(1^2+2^2)+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+.....$

$a_n=(1^2+2^2+3^2+.......+n^2)$

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

Therefore, we have:

$S_n={\sum }_{k=1}^n{a}_k={\sum }_{k=1}^n(\frac{1}{3}{k}^3+\frac{1}{2}{k}^2+\frac{1}{6}k)=\frac{1}{3}{\sum }_{k=1}^n{k}^3+\frac{1}{2}{\sum }_{k=1}^n{k}^2+\frac{1}{6}{\sum }_{k=1}^{n}k=\frac{1}{3}\times \frac{n^2+(n+1{)}^2}{2^2}+\frac{1}{2}\times \frac{n(n+1)(2n+1)}{6}+\frac{1}{6}\times \frac{n(n+1)}{2}$

$=\frac{n(n+1)}{6}[\frac{n(n+1)}{2}+\frac{(2n+1)}{2}+\frac{1}{2}]=\frac{n(n+1)}{6}\left[\frac{n^2+n+2n+1+1}{2}\right]=\frac{n(n+1)}{6}\left[\frac{n(n+1)+2(n+1)}{2}\right]=\frac{n(n+1)}{6}\left[\frac{(n+1)(n+2)}{2}\right]=\frac{n(n+1{)}^2(n+2)}{12}$

Hence, the sum is $\frac{n(n+1{)}^2(n+2)}{12}$.