Question

$\frac{1^2}{1.3}+\frac{2^2}{3.5}+\dots +\frac{n^2}{(2n-1)(2n+1)}=$

• A. $\frac{n(n+1)}{2n+1}$
• B. $\frac{n(n+1)}{2(2n+1)}$
• C. $\frac{n}{4(2n+1)}$
• D. $\frac{n+1}{2n+1}$

Answer 1

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

$\frac{1^2}{1.3}+\frac{2^2}{3.5}+\dots +\frac{n^2}{(2n-1)(2n+1)}$

this can be written as

$\sum _{k=1}^n\frac{k^2}{(2k-1)(2k+1)}$

$\Rightarrow \frac{k^2}{(2k-1)(2k+1)}=\frac{4k^2+1-1}{4(2k-1)(2k+1)}$

$\Rightarrow \frac{(2k+1)(2k-1)+1}{4(2k+1)(2k-1)}$

Splitting the fraction, we get

$\frac{1}{4}+\frac{1}{4(2k+1)(2k-1)}$

$\Rightarrow \frac{1}{4}+\frac{1}{8}[\frac{1}{2k-1}-\frac{1}{2k+1}]$

$\therefore \sum _{k=1}^n\frac{k^2}{(2k-1)(2k+1)}=\sum _{k=1}^n\frac{1}{4}+\frac{1}{8}[\frac{1}{2k-1}-\frac{1}{2k+1}]$

$\Rightarrow \frac{n}{4}+\frac{1}{8}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{2n-1}-\frac{1}{2n+1}]$

$\Rightarrow \frac{n}{4}+\frac{1}{8}[1-\frac{1}{2n+1}]$

$\Rightarrow \frac{n}{4}+\left[\frac{2n}{8(2n+1)}\right]=\frac{n}{4}[1+\frac{1}{2n+1}]$

$\Rightarrow \frac{n}{4}\left[\frac{2n+2}{2n+1}\right]=\frac{n(n+1)}{2(2n+1)}$

Hence, the correct answer is $\frac{n(n+1)}{2(2n+1)}$