Answer

Verified

410.4k+ 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

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

What happens when entropy reaches maximum class 11 chemistry JEE_Main

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Write the difference between soap and detergent class 10 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

Differentiate between calcination and roasting class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE