Question

find $1+\frac{1}{1+2}+\frac{1}{1+2+3}+......tonterms.$

Answer 1

Solution :-

$1+\frac{1}{1+2}+\frac{1}{1+2+3}+...$ to n terms

Here $T_n=\frac{1}{\sum n}=\frac{2}{n.(n+1)}$

Now $\frac{2}{n.(n+1)}=\frac{A}{n}+\frac{B}{n+1}=\frac{A(n+1)+Bn}{n.(n+1)}$

Now by comparing $A+B=0$ and $A=2$

$\Rightarrow B=-2$

$\therefore t_n=\frac{2}{n}-\frac{2}{n+1}$

on putting $n=1,2,3...n.$

$t_1=\frac{2}{1}-\frac{2}{2}$ $t_2=\frac{2}{2}-\frac{2}{3}...t_n=\frac{2}{n}-\frac{2}{n+1}$

On adding we get

$t_1+t_2+...+t_n=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-.............-\frac{2}{n+1}=\frac{2}{1}-\frac{2}{n+1}=\frac{2(n+1)-2}{n+1}$

$=\frac{2n}{n+1}$

$\therefore 1+\frac{1}{1+2}+\frac{1}{1+2+3}+...=\frac{2n}{n+1}$