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Question

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


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