Question

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

A
(n+1)
B
4n2(n+1)
C
2n2(n+1)
D
n2(n+1)

Solution

## The correct option is A $$n^2(n+1)$$$$a_{n}=3n^{2}-n$$Sum of n terms = $$\displaystyle\sum_{n=0}^{n}(3n^{2}-n)$$$$=\displaystyle\sum_{n=0}^{n}3n^{2}-\sum_{n=0}^{n}n$$$$=3 \displaystyle\sum_{n=0}^{n}n^{2}-\dfrac{n(n+1)}{2}$$$$=\dfrac{3(n)(n+1)(2n+1)}{62}-\dfrac{n(n+1)}{2}$$$$=\dfrac{n(n+1)}{2}[2n+1-1]$$$$\Rightarrow \dfrac{n(n+1)(2n)}{2}$$$$\Rightarrow n^{2}(n+1)$$$$\therefore \displaystyle\sum_{n=1}^{n}(3n^{2}-n)=n^{2}(n+1)$$Mathematics

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