Mathematics - Binomial Theorem Question with Solution | TestHub

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