Thank you for reporting, we will resolve it shortly
Q.
If $n$ is a positive integer, find the coefficient of $x^{-1}$ in the expansion of $\left(1+x\right)^{n} \left(1+\frac{1}{x}\right)^{n}$.
Binomial Theorem
Solution:
$\because \left(1+x\right)^{n} \left(1+\frac{1}{x}\right)^{n}$
$= \left(1+x\right)^{n}\left(\frac{x+1}{x}\right)^{n} = \frac{\left(1+x\right)^{2n}}{x^{n}}$
Now to find the coefficient of $x^{-1}$ in $\left(1+x\right)^{n} \left(1+\frac{1}{x}\right)^{n}$, it is equivalent to find coefficient of $x^{-1}$ in $\frac{\left(1+x\right)^{2n}}{x^{n}}$ which in turn is equal to the coefficient of $x^{-1}$ in the expansion of $\left(1+x\right)^{2n}$
Since $\left(1 + x\right)^{2} =\,{}^{2n}C_{0}x^{0}+\,{}^{2n}C_{1}x^{1}+\,{}^{2n}C_{2}x^{2}+\ldots$
$+\,{}^{2n}C_{n-1}\,x^{n-1}+\ldots +\,{}^{2n}C_{2n}\,x^{2n}$
Thus, the coefficient of $x^{n-1}$ is $^{2n}C_{n-1} = \frac{\lfloor2n}{\lfloor n-1\,\lfloor n+1}$