Question

The sum of (n + 1) terms of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+......$ is
[RPET 1999]

• 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 $\frac{2(n+1)}{n+2}$

$T_n=\frac{1}{\left[\frac{n(n+1)}{2}\right]}=2[\frac{1}{n}-\frac{1}{n+1}]$
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 $=\underset{k+1}{\sum ^{n+1}}$
$\Rightarrow S_{n+1}=2[1-\frac{1}{n+2}]\Rightarrow S_{n+1}=\frac{2(n+1)}{(n+2)}.$