Sum of series 9.

$\displaystyle 16). \;\;\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{100.101}$

SOLUTION

$\displaystyle \Rightarrow \sum_{n=1}^{100} \frac{1}{n(n+1)}$

$\displaystyle \Rightarrow  a_{n}=\frac{1}{n(n+1)}$

$\displaystyle \frac{1}{n(n+1)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)$

Taking $\;\;\displaystyle \sum_{n=1}^{100} \;\;$ On both the sides.

$\displaystyle \sum_{n=1}^{100} \frac{1}{n(n+1)}=\sum_{n=1}^{100} \frac{1}{n}-\sum_{n=1}^{100} \frac{1}{n+1}$

Terms will cancel out and we will get

$\displaystyle = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{100}-\frac{1}{2}-\frac{1}{3}\cdots-\frac{1}{101}$

$\displaystyle = 1-\frac{1}{101}$

$\displaystyle = \frac{100}{101}$

ANSWER : $\displaystyle = \frac{100}{101}$