Question

The value of $\frac{1^2}{1.3}+\frac{2^2}{3.5}+\cdots +\frac{n^2}{(2n-1)(2n+1)}$ is

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

Answer 1

Answer: (A)

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

$\frac{n^2}{(2n-1)(2n+1)}$
$=\frac{4n^2}{4(2n-1)(2n+1)}$
$=\frac{1}{4}\left[\frac{4n^2-1+1}{4n^2-1}\right]$
$=\frac{1}{4}[1+\frac{1}{4n^2-1}]$
$=\frac{1}{4}[1+\frac{1}{2}(\frac{2n+1-(2n-1)}{4n^2-1}\left)\right]$
$=\frac{1}{4}[1+\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1}\left)\right]$
Consider
$\frac{1}{2n-1}-\frac{1}{2n+1}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}...\frac{1}{2n-1}-\frac{1}{2n+1}$
$=1-\frac{1}{2n+1}$
$=\frac{2n}{2n+1}$

Hence applying summation gives us

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