# Find the value of the coefficient of ${{x}^{r}}$ the expansion ${{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{n}}$.

Last updated date: 26th Mar 2023

•

Total views: 308.4k

•

Views today: 5.86k

Answer

Verified

308.4k+ views

Hint: Find the general term of the binomial expansion of ${{\left( {{x}^{2}}+\dfrac{1}{{{x}^{2}}} \right)}^{n}}$ and then put the power of the term containing $x$ equal to $r$, then equate the two. You will get the coefficient of ${{x}^{r}}$ by doing so. Be careful while equating the powers as it leaves scope for silly mistakes.

According to the binomial theorem, the ${{(r+1)}^{th}}$ term in the expansion of ${{(a+b)}^{n}}$is,

\[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]

The above term is a general term or ${{(r+1)}^{th}}$ term. The total number of terms in the binomial expansion ${{(a+b)}^{n}}$ is $(n+1)$, i.e. one more than the exponent $n$.

In the Binomial expression, we have

\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]

So the coefficients ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},............,{}^{n}{{C}_{n}}$ are known as binomial or combinatorial coefficients.

You can see the general formula ${}^{n}{{C}_{r}}$ being used here which is the binomial coefficient. The sum of the binomial coefficients will be ${{2}^{n}}$ because, we know that,

$\sum\nolimits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right)}={{2}^{n}}$

Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even binomial coefficients and each is equal to ${{2}^{n-1}}$.

The middle term and the total number of terms depends upon the value of $n$.

It $n$ is even: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (odd).

It $n$ is odd: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (even).

It $n$ is a positive integer,

\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]

So here $a={{x}^{2}}$,$b=\dfrac{1}{{{x}^{2}}}$ and $n$ is as it is.

So using the binomial theorem, we can the general term as :

\[{{T}_{p+1}}={}^{n}{{C}_{p}}{{a}^{n-p}}{{b}^{p}}\]

\[={}^{n}{{C}_{r}}{{\left( {{x}^{2}} \right)}^{n-r}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{r}}\]

So the above term is the general term.

But it is mentioned in the question that we have to find the coefficient of term ${{x}^{r}}$.

So for that, we have to simplify the general term, so simplifying the general term we get,

\[{}^{n}{{C}_{r}}{{x}^{2n-2p}}\dfrac{1}{{{x}^{2p}}}\]

So simplifying it again, we get,

\[{}^{n}{{C}_{r}}{{x}^{2n-4p}}\]

So now it is given that we have to find the coefficient of ${{x}^{r}}$.

So equating the two powers, we get,

\[2n-4p=r\]

Simplifying further, we get,

\[p=\dfrac{2n-r}{4}\]

Therefore,

\[p=\dfrac{2n-r}{4}\].

So the \[{}^{n}{{C}_{p}}\] where \[p=\dfrac{2n-r}{4}\] can now be made free of the variable $p$ altogether.

So we get the final answer, that is the coefficient of ${{x}^{r}}$ as \[{}^{n}{{C}_{\dfrac{2n-r}{4}}}\].

Note: Read the question and see what is asked. Using \[{{T}_{p+1}}={}^{n}{{C}_{p}}{{a}^{n-p}}{{b}^{p}}\] will give you the value of r so don’t confuse yourself while substitution. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.

According to the binomial theorem, the ${{(r+1)}^{th}}$ term in the expansion of ${{(a+b)}^{n}}$is,

\[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]

The above term is a general term or ${{(r+1)}^{th}}$ term. The total number of terms in the binomial expansion ${{(a+b)}^{n}}$ is $(n+1)$, i.e. one more than the exponent $n$.

In the Binomial expression, we have

\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]

So the coefficients ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},............,{}^{n}{{C}_{n}}$ are known as binomial or combinatorial coefficients.

You can see the general formula ${}^{n}{{C}_{r}}$ being used here which is the binomial coefficient. The sum of the binomial coefficients will be ${{2}^{n}}$ because, we know that,

$\sum\nolimits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right)}={{2}^{n}}$

Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even binomial coefficients and each is equal to ${{2}^{n-1}}$.

The middle term and the total number of terms depends upon the value of $n$.

It $n$ is even: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (odd).

It $n$ is odd: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (even).

It $n$ is a positive integer,

\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]

So here $a={{x}^{2}}$,$b=\dfrac{1}{{{x}^{2}}}$ and $n$ is as it is.

So using the binomial theorem, we can the general term as :

\[{{T}_{p+1}}={}^{n}{{C}_{p}}{{a}^{n-p}}{{b}^{p}}\]

\[={}^{n}{{C}_{r}}{{\left( {{x}^{2}} \right)}^{n-r}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{r}}\]

So the above term is the general term.

But it is mentioned in the question that we have to find the coefficient of term ${{x}^{r}}$.

So for that, we have to simplify the general term, so simplifying the general term we get,

\[{}^{n}{{C}_{r}}{{x}^{2n-2p}}\dfrac{1}{{{x}^{2p}}}\]

So simplifying it again, we get,

\[{}^{n}{{C}_{r}}{{x}^{2n-4p}}\]

So now it is given that we have to find the coefficient of ${{x}^{r}}$.

So equating the two powers, we get,

\[2n-4p=r\]

Simplifying further, we get,

\[p=\dfrac{2n-r}{4}\]

Therefore,

\[p=\dfrac{2n-r}{4}\].

So the \[{}^{n}{{C}_{p}}\] where \[p=\dfrac{2n-r}{4}\] can now be made free of the variable $p$ altogether.

So we get the final answer, that is the coefficient of ${{x}^{r}}$ as \[{}^{n}{{C}_{\dfrac{2n-r}{4}}}\].

Note: Read the question and see what is asked. Using \[{{T}_{p+1}}={}^{n}{{C}_{p}}{{a}^{n-p}}{{b}^{p}}\] will give you the value of r so don’t confuse yourself while substitution. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry 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

Write a letter to the Principal of your school to plead class 10 english CBSE