Question

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

Answer 1

Hint: In order to solve this problem we need to sum the $ {{n}^{th}} $ of the series from n = 1 to n = $ \infty $ .Then we need to know the standard formulas for summation of $ {{n}^{2}} $ and for the summation of n. The formulas are given as follows, $ {{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} $ .

Complete step-by-step answer:
We are asked to find the sum of n terms in the series.
We know the $ {{n}^{th}} $ term which is $ 3{{n}^{2}}-n $ .
We can find the sum of n terms by just summing the $ {{n}^{th}} $ term from n = 1 to n = $ \infty $ .
Let the sum of n terms be $ {{S}_{n}} $ .
Therefore the expression for $ {{S}_{n}} $ is as follows,
 $ {{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,
\[\begin{align}
  & {{S}_{n}}=\left( 3-1 \right)+\left( 12-2 \right)+\left( 27-3 \right)+..... \\
 & =2+10+24+....
\end{align}\]
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{align}
  & {{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{align} $
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.