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

Last updated date: 16th Jul 2024
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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.