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

Answer 1

The correct option is D

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

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

put n=1,2,3,.........,(n+1)

$T_1=2[\frac{1}{1}-\frac{1}{2}]$,$T_2=2[\frac{1}{2}-\frac{1}{3}]$,...........,

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

Hence sum of (n+1) terms = $\sum _{k=1}^{n+1}{T}_k$

$\Rightarrow$ $S_{n+1}=2[1-\frac{1}{n+2}]$ $\Rightarrow$ $S_{n+1}=\frac{2(n+1)}{n+2}$