The middle term in the expansion of $\left(1-\frac{1}{x}\right)^n\left(1-x^n\right)$ in powers of $x$ is
The middle term in the expansion of $\left(1-\frac{1}{x}\right)^n\left(1-x^n\right)$ in powers of $x$ is
$-{ }^{2 n} \mathrm{C}_{n-1}$
$-{ }^{2 n} \mathrm{C}_n$
${ }^{2 n} \mathrm{C}_{n-1}$
${ }^{2 n} \mathrm{C}_n$
Solution
Given expansion can be re-written as
$
\left(\frac{x-1}{x}\right)^n \cdot(1-x)^n=(-1)^n x^{-n}(1-x)^{2 n}
$
Total number of terms will be $2 n+1$ which is odd ( $\because 2 n$ is always even)
$\therefore$ Middle term $=\frac{2 n+1+1}{2}=(n+1)$ th
Now, $T_{r+1}={ }^n C_r(1)^r x^{n-r}$
So, $\frac{{ }^{2 n} C_n \cdot x^{2 n-n}}{x^n \cdot(-1)^n}={ }^{2 n} C_n \cdot(-1)^n$
Middle term is an odd term. So, $n+1$ will be odd.
So, $n$ will be even.
$\therefore$ Required answer is ${ }^{2 n} \mathrm{C}_n$