Question

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

• 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 $\frac{n(n+2)}{3}$

General term of the given series is, $T_r=\frac{1^2+2^2+....+r^2}{1+2+.....+r}=\frac{\frac{r(r+1)(2r+1)}{6}}{\frac{r(r+1)}{2}}=\frac{2r+1}{3}$

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

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