# Find the sum of n terms of the series whose ${n^{th}}$ term is $4n\left( {{n^2} + 1} \right) - \left( {6n{}^2 + 1} \right)$.

Answer

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Hint: Simply the expression & apply summation to get sum of n terms of series using standard formulas.

As, given in the question that ${n^{th}}$ term of the series is given by,

\[ \Rightarrow \]${n^{th}}$ term is \[ = {\text{ }}4n\left( {{n^2} + 1} \right) - \left( {6{n^2} + 1} \right) = 4{n^3} - 6{n^2} + 4n - 1\]

Now, we had to find the summation of n terms of the series whose ${n^{th}}$ term is given by

\[ \Rightarrow 4{n^3} - 6{n^2} + 4n - 1\]

So, applying summation to get sum of n terms of the series,

\[ \Rightarrow \]Sum of n terms of the series \[ = {\text{ }}4\Sigma {n^3} - 6\Sigma {n^2} + 4\Sigma n - \Sigma 1{\text{ }}\] (1)

Now, we had to break the above equation to find the sum of series.

And as we know that,

\[ \Rightarrow \]If ${n^{th}}$ term is \[{n^3}\], then sum of n terms will be \[{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\] (statement 1)

\[ \Rightarrow \]If ${n^{th}}$ term is \[{n^2}\], then sum of n terms will be \[\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right)\] (statement 2)

\[ \Rightarrow \]If ${n^{th}}$ term is $n$, then sum of n terms will be \[\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)\] (statement 3)

\[ \Rightarrow \]If all the terms are 1 then the sum of n terms of the series will be $n$. (statement 4)

So, putting values of statement 1, 2, 3 and 4 in equation 1, we get

\[ \Rightarrow \]Sum of n terms of the series \[ = 4{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - 6\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) + 4\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) - n\]

Solving above equation, we get

\[ \Rightarrow \]Sum of n terms of the series \[ = {n^4} + 2{n^3} + {n^2} - 2{n^3} - 3{n^2} - n + 2{n^2} + 2n - n = {n^4}\]

Hence, the sum of n terms of the given series will be \[{n^4}\].

Note:- Whenever we come up with this type of question then to find the solution of the problem efficiently, we must first, expand the given series and then apply the summation to the series. Then put values of \[\sum {{n^3}} \], \[\sum {{n^2}} \], \[\sum n \] and \[\sum 1 \] accordingly.

As, given in the question that ${n^{th}}$ term of the series is given by,

\[ \Rightarrow \]${n^{th}}$ term is \[ = {\text{ }}4n\left( {{n^2} + 1} \right) - \left( {6{n^2} + 1} \right) = 4{n^3} - 6{n^2} + 4n - 1\]

Now, we had to find the summation of n terms of the series whose ${n^{th}}$ term is given by

\[ \Rightarrow 4{n^3} - 6{n^2} + 4n - 1\]

So, applying summation to get sum of n terms of the series,

\[ \Rightarrow \]Sum of n terms of the series \[ = {\text{ }}4\Sigma {n^3} - 6\Sigma {n^2} + 4\Sigma n - \Sigma 1{\text{ }}\] (1)

Now, we had to break the above equation to find the sum of series.

And as we know that,

\[ \Rightarrow \]If ${n^{th}}$ term is \[{n^3}\], then sum of n terms will be \[{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\] (statement 1)

\[ \Rightarrow \]If ${n^{th}}$ term is \[{n^2}\], then sum of n terms will be \[\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right)\] (statement 2)

\[ \Rightarrow \]If ${n^{th}}$ term is $n$, then sum of n terms will be \[\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)\] (statement 3)

\[ \Rightarrow \]If all the terms are 1 then the sum of n terms of the series will be $n$. (statement 4)

So, putting values of statement 1, 2, 3 and 4 in equation 1, we get

\[ \Rightarrow \]Sum of n terms of the series \[ = 4{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} - 6\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) + 4\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right) - n\]

Solving above equation, we get

\[ \Rightarrow \]Sum of n terms of the series \[ = {n^4} + 2{n^3} + {n^2} - 2{n^3} - 3{n^2} - n + 2{n^2} + 2n - n = {n^4}\]

Hence, the sum of n terms of the given series will be \[{n^4}\].

Note:- Whenever we come up with this type of question then to find the solution of the problem efficiently, we must first, expand the given series and then apply the summation to the series. Then put values of \[\sum {{n^3}} \], \[\sum {{n^2}} \], \[\sum n \] and \[\sum 1 \] accordingly.

Last updated date: 27th Sep 2023

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