Question

Solve :
$1+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+....n)}=\frac{2n}{(n+1)}$.

Answer 1

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

Take L.H.S

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

$S=\frac{2}{n(n+1)}$

$\sum S=\sum _{r=1}^{n}2\left[\frac{(n+1)-n}{n(n-1)}\right]$

$=\sum _{r=1}^{n}2[\frac{1}{n}-\frac{1}{n+1}]$

$r=1,$ $1-\frac{1}{2}$

$r=2,$ $\frac{1}{2}-\frac{1}{3}$

$r=3,$ $\frac{1}{3}-\frac{1}{4}$

$\sum S=2[1-\frac{1}{n+1}]$

$=2\left[\frac{n+1-1}{n+1}\right]$

$=\frac{24}{n+1}$ Hence proved