Question

The sum of n terms of $1^2+(1^2+2^2)+(1^2+2^2+3^2)+....$.

• A. $\frac{n(n+a)(2n+1)}{6}$
• B. $\frac{n(n+1)(2n-1)}{6}$
• C. $\frac{1}{12}n(n+1{)}^2(n+2)$
• D. $\frac{1}{12}{n}^2(n+1{)}^2$

Answer 1

Answer: (C)

$\frac{1}{12}n(n+1{)}^2(n+2)$

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

$=\sum _{r=1}^n(1^2+2^2+3^2+...+r^2)$

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

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

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

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

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

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