Question

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots \cdots +\frac{1}{n(n+1)}=$

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

Answer 1

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

$S_n=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{n(n+1)}=$

The general term of this series can be written as

$T_n=\sum _{n=1}^n\frac{1}{n(n+1)}$

$S_n=\sum _{n=1}^n\frac{1}{n(n+1)}$

$S_n=\sum _{n=1}^n\frac{(n+1)-\left(n\right)}{n(n+1)}$

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

on putting the values of n

$S_n=\frac{1}{1}-\frac{1}{2}$

$+\frac{1}{2}-\frac{1}{3}$

.

.

.

$+\frac{1}{n-1}-\frac{1}{n}$

$+\frac{1}{n}-\frac{1}{n+1}$

$S_n=1-\frac{1}{n+1}$

$S_n=\frac{n+1-1}{n+1}$

$S_n=\frac{n}{n+1}$.