Question

The sum of $20$ terms of the series $1+\left(1+3\right)+\left(1+3+5\right)+\left(1+3+5+7\right)+....$ is

• A. $400$
• B. $2870$
• C. $5740$
• D. $1540$

Answer 1

Answer: (B)

$T_n=1+3+5+....+\left(2n-1\right)=n^2$
$\therefore S=\sum _{k=1}^n{T}_k$
$=\sum _{k=1}^n{k}^2$
$=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$
Put $n=20$, we get
$S_{20}=\frac{20\left(21\right)\left(41\right)}{6}$
$=2870$