Question

Sum of the series $S$ upto $20$ terms is

$S=1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+...$

• A. $110$
• B. $111$
• C. $115$
• D. $116$

Answer 1

Answer: (C)

$115$

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

The series can be written as
$S=1+\frac{3}{2}+\frac{6}{3}+\frac{10}{4}+...$ upto 20 terms

Notice that the given series is in A.P. because the common difference of the series is $\frac{1}{2}$.

Since we have $a=1$ and $d=\frac{1}{2}$, so we use the formula for $S_n$ to find sum of n terms which is given by,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Substitute $n=20,a=1,$ and $d=\frac{1}{2}$ in the formula to find the sum of the given series upto 20terms.

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

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

$S_{20}=10\times 11.5$

$S_{20}=115$