S-16

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

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 : \(\displaystyle = \frac{100}{101}\)