Question

The sum of first $n$ terms of the series
$\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+\cdots$ is equal to

• A. $\frac{n+2}{3}$
• B. $\frac{n(n+2)}{3}$
• C. $\frac{n(n-2)}{3}$
• D. $\frac{n(n-2)}{6}$

Answer 1

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

$\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+\cdots$ upto $n$ terms
$=\sum _{r=1}^n\frac{1^2+2^2+\cdots +r^2}{1+2+\cdots +r}$

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

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

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

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