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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 n terms of the series whose nth{{n}^{th}} term is: 3n2−n3{{n}^{2}}-n .

Explanation

Solution

In the question we are asked to find the sum of first n terms of the series whose Tn=3n2−n{{T}_{n}}=3{{n}^{2}}-n . So, take the summation of the general term with limits from n=1 to n. Use the formula ∑n=1nn=n(n+1)2\sum\limits_{n=1}^{n}{n}=\dfrac{n\left( n+1 \right)}{2} and ∑n=1nn2=n(n+1)(2n+1)6\sum\limits_{n=1}^{n}{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} to find the value of the summation and the answer to the above question.

Complete step-by-step answer :
It is given that the nth{{n}^{th}} term of the series is 3n2−n3{{n}^{2}}-n , so we can say that the general term of the series is given by Tn=3n2−n{{T}_{n}}=3{{n}^{2}}-n .
Now we are asked to find the sum of the first n terms of the series, so we will take the summation of the general terms with limits from n=1 to n.
∑n=1nTn=∑n=1n(3n2−n)\sum\limits_{n=1}^{n}{{{T}_{n}}}=\sum\limits_{n=1}^{n}{\left( 3{{n}^{2}}-n \right)}
Let us denote the sum of first n terms of the series by Sn{{S}_{n}} . Also, we know that ∑(a±b)=∑a±∑b\sum{\left( a\pm b \right)}=\sum{a}\pm \sum{b} . SO, if we use this in our equation, we get
Sn=∑n=1n3n2−∑n=1nn{{S}_{n}}=\sum\limits_{n=1}^{n}{3{{n}^{2}}}-\sum\limits_{n=1}^{n}{n}
We know that a constant can be taken out of the summation.
Sn=3∑n=1nn2−∑n=1nn{{S}_{n}}=3\sum\limits_{n=1}^{n}{{{n}^{2}}}-\sum\limits_{n=1}^{n}{n}
Now, we know that ∑n=1nn=n(n+1)2\sum\limits_{n=1}^{n}{n}=\dfrac{n\left( n+1 \right)}{2} and ∑n=1nn2=n(n+1)(2n+1)6\sum\limits_{n=1}^{n}{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} . If we use this in our equation, we get
Sn=3×n(n+1)(2n+1)6−n(n+1)2{{S}_{n}}=3\times \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{n\left( n+1 \right)}{2}
If we take n(n+1) common from both the terms in the right-hand side of the equation, we get
Sn=(3×(2n+1)6−12)n(n+1){{S}_{n}}=\left( 3\times \dfrac{\left( 2n+1 \right)}{6}-\dfrac{1}{2} \right)n\left( n+1 \right)
⇒Sn=((2n+1)2−12)n(n+1)\Rightarrow {{S}_{n}}=\left( \dfrac{\left( 2n+1 \right)}{2}-\dfrac{1}{2} \right)n\left( n+1 \right)
⇒Sn=(2n+1−12)n(n+1)=(2n2)n(n+1)\Rightarrow {{S}_{n}}=\left( \dfrac{2n+1-1}{2} \right)n\left( n+1 \right)=\left( \dfrac{2n}{2} \right)n\left( n+1 \right)
⇒Sn=n2(n+1)\Rightarrow {{S}_{n}}={{n}^{2}}\left( n+1 \right)
Therefore, the answer to the above question is n2(n+1){{n}^{2}}\left( n+1 \right) .

Note :Whenever you are given a series and asked the sum of the n terms, you should use the method of summation as we did above. However, if the given series is an AP or GP, avoid using the summation and go for the formula of sum of n terms of AP and GP because that would be the easier option. Also, it is important that you should learn the formulas of sum of first n natural numbers, sum of square and cubes of first n natural numbers.