Question

The sum of the $n$terms of the series $\frac{1^2}{1}+\frac{(1^2+2^2)}{(1+2)}+\frac{(1^2+2^2+3^2)}{(1+2+3)}+\dots .$ is

• A. $\frac{(n+1)(n+2)}{3}$
• B. $\frac{n(n+2)}{3}$
• C. $\frac{n(n+1)}{3}$
• D. none of these

Answer 1

The correct option is B

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

Find the sum of the given series

$\frac{1^2}{1}+\frac{(1^2+2^2)}{(1+2)}+\frac{(1^2+2^2+3^2)}{(1+2+3)}+\dots .$

Therefore, $t_n=\frac{\left(1^2+2^2+3^2+......+n^2\right)}{(1+2+3+....n)}$

$$\begin{gathered} =\frac{{\displaystyle \frac{n(n+1)(2n+1)}{6}}}{{\displaystyle \frac{n(n+1)}{2}}} \\ =\frac{2n+1}{3} \end{gathered}$$

Sum of the series

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

Hence, the correct option is $\mathrm{Option}\left(\mathrm{B}\right)$.