Question

The sum of the $n$ terms of the series $\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+..$ is

• A. $\frac{n}{3(2n+3)}$
• B. $\frac{n+1}{2n+3}$
• C. $\frac{n-1}{2n+1}$
• D. $\frac{n}{2n+1}$

Answer 1

The correct option is A $\frac{n}{3(2n+3)}$

$S=\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+..$
Now, $a_n=\frac{1}{(2n+1)(2n+3)}=\frac{1}{2}[\frac{1}{2n+1}-\frac{1}{2n+3}]$
$S={\sum }_{n=1}^n{a}_n=\frac{1}{2}{\sum }_{n=1}^n[\frac{1}{2n+1}-\frac{1}{2n+3}]$
$\Rightarrow S=\frac{1}{2}[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+....+(\frac{1}{2n+1}-\frac{1}{2n+3})]$
$\Rightarrow S=\frac{1}{2}[\frac{1}{3}-\frac{1}{2n+3}]=\frac{n}{3(2n+3)}$
Ans: A