Question

Find the sum of the series $\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+......$ up to 10 terms.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

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

Here, numbers in denominator are consecutive odd numbers and difference between between them is 2

Multiplying 2 in numerator and denominator

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

$\frac{1}{2}[\frac{3-5}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+......\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}]$
$\frac{1}{2}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....\frac{1}{(2n-1)}-\frac{1}{(2n+1)}]$
$\frac{1}{2}[1-\frac{1}{(2n+1)}]$
Here value of n = 10
$\frac{1}{2}[1-\frac{1}{(20+1)}]$
= 10/21