Question

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

• A. $\frac{2n}{n+1}$
• B. $\frac{n}{2n+1}$
• C. $\frac{n}{n+1}$
• D. None of these

Answer 1

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

Given:

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

The $n^{th}$ term, $T_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

$n=1T_1=\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}n=2T_2=\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}n=3T_3=\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}.....n=n{T}_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

Adding all the terms we see that all the terms gets cancelled accept for:

$S_n={\sum }_{k=1}^n{T}_k=1-\frac{1}{n+1}=\frac{n}{n+1}$