Question

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

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

Answer 1

Answer: (C)

$\frac{20}{11}$

The sum of series $1+\frac{1}{1+2}+\frac{1}{1+2+3}+----$up tp $10$ term

$\to x=1+\frac{1}{1+2}+\frac{1}{1+2+3}+----$upto $10$ term

So $n=10$

General term $T_n=\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}$

We can simplify it by partial fraction

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

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

Now, we calculate sum of $10$ terms

$x=T_1+T_2+T_3+T_4+\dots +T_{10}$

$2[1-\frac{1}{2}]+[\frac{1}{2}-\frac{1}{3}]+2[\frac{1}{3}-\frac{1}{4}]+------+2[\frac{1}{10}-\frac{1}{11}]$

$=2[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+------+\frac{1}{10}-\frac{1}{11}]$

$=2[1-\frac{1}{11}]$ ($\because$ first & last term remaining )

$=2\left(\frac{11-1}{11}\right)$

$=\frac{2\times 10}{11}$

$x=\frac{20}{11}$