The sum of (n+1) terms of 11+11+2+11+2+3+.......... is

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

Options:

A .

\frac{n}{n + 1}

B .

\frac{2 n}{n + 1}

C .

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

D .

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

Answer: Option D
:
D

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

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 = $n + 1 \sum k = 1 T_{k}$

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

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