Question

The sum of the n terms of the series $\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+.....+\frac{n^2}{(2n-1)(2n+1)}$ is

• A. $\frac{n(n+1)}{2(2n+1)}$
• B. $\frac{(2n^2+1)}{(2n+1)}$
• C. $\frac{(2n^2+1)}{4(2n+1)}$
• D. None of these

Answer 1

The correct option is A

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

Find the sum of the given series

$\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+.....+\frac{n^2}{(2n-1)(2n+1)}$

Here, $t_n=\frac{n^2}{(2n-1)(2n+1)}$

$\Rightarrow t_n=\frac{1}{4}+\frac{1}{8(2n-1)}-\frac{1}{8(2n+1)}$

Let the sum of the given series be $S_n$

$$\begin{gathered} \Rightarrow S_n=\frac{n}{4}+\frac{1}{8}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+........-\frac{1}{2n-1}+\frac{1}{2n-1}-\frac{1}{2n+1}\right) \\ \Rightarrow S_n=\frac{n}{4}+\frac{1}{8}\left(1-\frac{1}{2n+1}\right) \\ \Rightarrow S_n=\frac{n}{4}+\frac{n}{4(2n+1)} \\ \Rightarrow S_n=\frac{n}{4}\left(\frac{2n+2}{2n+1}\right) \\ \Rightarrow S_n=\frac{n\left(n+1\right)}{2(2n+1)} \end{gathered}$$

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