Question

Sum of the series
$S=1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+\frac{1}{4}(1+2+3+4)+...$ upto $20$ terms is

• A. $110.5$
• B. $111$
• C. $115$
• D. $116.5$

Answer 1

The correct option is C $115$

The general term for the above series is $T_n=\frac{\sum n}{n}$$=\frac{n(n+1)}{2n}$$=\frac{n+1}{2}$

Hence, $S_{20}=\sum _{n=1}^{n=20}{T}_n$

$\sum _{n=1}^{n=20}\left(\frac{n+1}{2}\right)$

$\Rightarrow [1+\frac{3}{2}+\frac{4}{2}+.....+\frac{21}{2}]$

$\Rightarrow [\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{21}{2}]$

It is an A.P. with $a=\frac{2}{2},d=\frac{1}{2},andn=20$

We know that the sum of $n$ terms of A.P. is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\therefore S_{20}=\frac{20}{2}[2\times \frac{2}{2}+(20-1\left)\frac{1}{2}\right]$

$\Rightarrow S_{20}=10[2+\frac{19}{2}]=\frac{230}{2}=115$

Hence, the answer is $115$