Question

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

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

Answer 1

Answer: (C)

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