Question

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

• A. $n^2$
• B. ${\left[\frac{n\left(n+1\right)}{2}\right]}^2$
• C. $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$
• D. None of these

Answer 1

Answer: (C)

$\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

$1+(1+3)+(1+3+5)+---$

$n^{th}$ term $=(1+3+5+7+9+---(2n-1\left)\right)$

AP with $a=1$ $d=2$

$s_n=\frac{n}{2}[2a+(n-1\left)d\right]$

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

$=n[1+n-1]=n^2$

$\therefore n^{th}term=n^2$

Sum of series $\left(1\right)+(1+3)+(1+5+5)+--$

is ${\sum }_{n=1}^n{n}^2$

$\Rightarrow \frac{n(n+1)(2n+1)}{6}$