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Q.
The middle term in the expansion of $\left(1-\frac{1}{x}\right)^{n} \left(1-x^{n}\right)$ in powers of x is
AIEEEAIEEE 2012Binomial Theorem
Solution:
Given expansion can be re-written as
$\left(1-\frac{1}{x}\right)^{n} .\left(1-x^{n}\right) = \left(-1\right)^{n}x^{-n} \left(1-x\right)^{2n}$
Total number of terms will be $2n + 1$ which is odd ($\because 2n$ is always even)
$\therefore $ Middle term $= \frac{2n+1+1}{2} = \left(n+1\right) \,th$
Now, $T_{r+1}=\,{}^{n}C_{r}\left(1\right)^{r} x^{n-r}$
So, $\frac{^{2n}C_{n} . x^{2n-n}}{x^{n}.\left(-1\right)^{n}} = ^{2n}C_{n} .\left(-1\right)^{n}$
Middle term is an odd term. So, $n + 1$ will be odd.
So, n will be even.
$\therefore $ Required answer is $^{2n}C_{n}.$