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Question: Find the sum of \( n \) terms of the series whose \( {{n}^{th}} \) term is \( 3{{n}^{2}}-n \)...

Find the sum of nn terms of the series whose nth{{n}^{th}} term is 3n2−n3{{n}^{2}}-n

Explanation

Solution

Hint : In order to solve this problem we need to sum the nth{{n}^{th}} of the series from n = 1 to n = ∞\infty .Then we need to know the standard formulas for summation of n2{{n}^{2}} and for the summation of n. The formulas are given as follows, S1=∑n=1nn2=n(n+1)(2n+1)6{{S}_{1}}=\sum\limits_{n=1}^{n}{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} and S2=∑n=1nn=n(n+1)2{{S}_{2}}=\sum\limits_{n=1}^{n}{n}=\dfrac{n\left( n+1 \right)}{2} .

Complete step-by-step answer :
We are asked to find the sum of n terms in the series.
We know the nth{{n}^{th}} term which is 3n2−n3{{n}^{2}}-n .
We can find the sum of n terms by just summing the nth{{n}^{th}} term from n = 1 to n = ∞\infty .
Let the sum of n terms be Sn{{S}_{n}} .
Therefore the expression for Sn{{S}_{n}} is as follows,
Sn=∑n=1n3n2−n................................(i){{S}_{n}}=\sum\limits_{n=1}^{n}{3{{n}^{2}}-n}................................(i)
Solving the equation will look like of we substitute first few terms as follows,

& {{S}_{n}}=\left( 3-1 \right)+\left( 12-2 \right)+\left( 27-3 \right)+..... \\\ & =2+10+24+.... \end{aligned}$$ But we cannot proceed further after that, so instead, we need to use standard formulas, We can write equation (i) as follows, $ {{S}_{n}}=3\sum\limits_{n=1}^{n}{{{n}^{2}}-\sum\limits_{n=1}^{n}{n}} $ Now we must know the standard formulas for summation of $ {{n}^{2}} $ and n. Therefore, $ {{S}_{1}}=\sum\limits_{n=1}^{n}{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} $ and $ {{S}_{2}}=\sum\limits_{n=1}^{n}{n}=\dfrac{n\left( n+1 \right)}{2} $ , By substituting the above relations we get, $ {{S}_{n}}=3{{S}_{1}}-{{S}_{2}} $ , Solving this further we get, $ \begin{aligned} & {{S}_{n}}=3\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)-\left( \dfrac{n\left( n+1 \right)}{2} \right) \\\ & =\dfrac{n\left( n+1 \right)}{2}\left( 2n+1-1 \right) \\\ & ={{n}^{2}}\left( n+1 \right) \end{aligned} $ Therefore, the summation of the series is $ {{n}^{2}}\left( n+1 \right) $ . **Note** : We need to know the standard formula for summation of n and $ {{n}^{2}} $. We cannot use any formula of arithmetic progression and geometric progression because we cannot be sure in which category this series belongs.