Question

$\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....$ upto n terms is

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

Answer 1

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

Let $S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{(2n-1)(2n+1)}$

$=\frac{1}{2}[\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+......+\frac{2n+1-(2n-1)}{(2n-1)(2n+1)}]$

$=\frac{1}{2}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-.........-\frac{1}{(2n-1)}+\frac{1}{(2n-1)}-\frac{1}{(2n+1)}]$

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

Hence option 'B' is the correct choice.