Question

The sum of the series :
$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots\ldots$ upto $10$ terms, is :

• A. $\frac{18}{11}$
• B. $\frac{22}{13}$
• C. $\frac{20}{11}$
• D. $\frac{16}{9}$

Answer 1

Answer: (C)

$T_{r } = \frac{1}{1 + 2 + 3 + ... + r} = -\frac{2}{r\left(r+1\right)}$
$S_{10} = 2\sum\limits^{10}_{r = 1} \frac{1}{r\left(r+1\right)}= 2\sum\limits^{10}_{r = 1}\left[\frac{r+1}{r\left(r+1\right)}-\frac{r}{r\left(r+1\right)}\right]$
$= 2\sum\limits^{10}_{r = 1}\left(\frac{1}{r}-\frac{1}{r+1}\right)$
$2\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{10}-\frac{1}{11}\right)\right]$
$= 2\left[1-\frac{1}{11}\right]= 2 \times \frac{10}{11} = \frac{20}{11}$