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

Last updated date: 29th Mar 2023

•

Total views: 309.9k

•

Views today: 5.87k

Answer

Verified

309.9k+ views

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE