Question

The sum of $(n+1$) terms of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+......$ is

• A. $\frac{n}{n+1}$
• B. $\frac{2n}{n+1}$
• C. $\frac{2}{n(n+1)}$
• D. $\frac{2(n+1)}{n+2}$

Answer 1

The correct option is D $\frac{2(n+1)}{n+2}$

Given $T_n=\frac{1}{\left[\frac{n(n+1)}{2}\right]}=2[\frac{1}{n}-\frac{1}{n+1}]$
Put $n=1,2,3,....,(n+1)$, we get
$T_1=2[\frac{1}{1}-\frac{1}{2}],T_2=2[\frac{1}{2}-\frac{1}{3}],.....,$
$T_{n+1}=2[\frac{1}{n+1}-\frac{1}{n+2}]$
Hence, sum of $(n+1)$ terms $S_{(n+1)}=\underset{k=1}{\sum ^{n+1}}{T}_k$
$=2[\frac{1}{1}-\frac{1}{2}]+2[\frac{1}{2}-\frac{1}{3}].......+2[\frac{1}{n+1}-\frac{1}{n+2}]$
$=2[\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......\frac{1}{n+1}-\frac{1}{n+2}]$
$\Rightarrow S_{n+1}=2[1-\frac{1}{n+2}]$
$=2\left[\frac{n+2-1}{n+2}\right]$
$\therefore S_{n+1}=\frac{2(n+1)}{(n+2)}$