Question

The sum of series $\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+....$ upto n terms is

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

Answer 1

The correct option is D $\frac{1}{3}n(n+2)$

$S=\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+...$

$\Rightarrow S=\sum _{r=1}^n\frac{1^2+2^2+3^2+...+r^2}{1+2+3+...+r}$

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

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

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