Question

Find the sum of $n$ terms of the series whose $n^{th}$ term is :
$3n^2-n$

• A. $(n+1)$
• B. $4n^2(n+1)$
• C. $2n^2(n+1)$
• D. $n^2(n+1)$

Answer 1

Answer: (D)

$n^2(n+1)$

$a_n=3n^2-n$

Sum of n terms = $\sum _{n=0}^n(3n^2-n)$

$=\sum _{n=0}^{n}3n^2-\sum _{n=0}^{n}n$

$=3\sum _{n=0}^n{n}^2-\frac{n(n+1)}{2}$

$=\frac{3\left(n\right)(n+1)(2n+1)}{62}-\frac{n(n+1)}{2}$

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

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

$\Rightarrow n^2(n+1)$

$\therefore \sum _{n=1}^n(3n^2-n)=n^2(n+1)$