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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,
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 :


\[={}^{n}{{C}_{r}}{{\left( {{x}^{2}} \right)}^{n-r}}{{\left( \dfrac{1}{{{x}^{2}}} \right)}^{r}}\]
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,
So simplifying it again, we get,
So now it is given that we have to find the coefficient of ${{x}^{r}}$.
So equating the two powers, we get,
Simplifying further, we get,
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.