Question

The sum of series $\frac{1}{3\times 6}+\frac{1}{6\times 9}+\frac{1}{9\times 12}+$...... is:

Answer 1

$T_n=\frac{1}{3n\times 3(n+1)}=\frac{1}{9\left(n\right)(n+1)}=\frac{1}{9}\left[\frac{n+1-n}{n(n+1)}\right]$

So, $9T_n=\frac{1}{n}-\frac{1}{n+1}$

and $S_n=\sum T_n=T_1+T_2+\cdots +T_n$

So, $9T_1=1-\frac{1}{2}9T_2=\frac{1}{2}-\frac{1}{3}⋮⋮9T_n=\frac{1}{n}-\frac{1}{n+1}$

on adding , we get

$9S_n=1-\frac{1}{n+1}=\frac{n}{n+1}$

So, $S_n=\frac{n}{9(n+1)}$