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If $x$ occurs is the expansion of $\left(x+\frac{1}{x}\right)^{n}$, then its coefficient is
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Verified Answer
The correct answer is:
$\frac{n !}{\left(\frac{n+r}{2}\right) !\left(\frac{n-r}{2}\right) !}$
Let $x^{r}$ occurs in the expansion of $\left(x+\frac{1}{x}\right)^{n}$
$T_{r+1}={ }^{n} C_{p} x^{n-p}\left(\frac{1}{x}\right)^{p}$
${ }^{n} C_{p} x^{n-p} x^{-p}={ }^{n} C_{p} x^{n-2 p}$
let $\quad n-2 p=r$
$\Rightarrow \quad n-r=2 p \Rightarrow p=\frac{n-r}{2}$
So, coefficient of $x^{r}={ }^{n} C_{p}=\frac{n !}{p !(n-p) !}$
$=\frac{n !}{\left(\frac{n-r}{2}\right) !\left(n-\frac{n-r}{2}\right) !}$
$=\frac{n !}{\left(\frac{n-r}{2}\right) !\left(\frac{n+r}{2}\right) !}$
$T_{r+1}={ }^{n} C_{p} x^{n-p}\left(\frac{1}{x}\right)^{p}$
${ }^{n} C_{p} x^{n-p} x^{-p}={ }^{n} C_{p} x^{n-2 p}$
let $\quad n-2 p=r$
$\Rightarrow \quad n-r=2 p \Rightarrow p=\frac{n-r}{2}$
So, coefficient of $x^{r}={ }^{n} C_{p}=\frac{n !}{p !(n-p) !}$
$=\frac{n !}{\left(\frac{n-r}{2}\right) !\left(n-\frac{n-r}{2}\right) !}$
$=\frac{n !}{\left(\frac{n-r}{2}\right) !\left(\frac{n+r}{2}\right) !}$
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