Given expansion can be re-written as (xx−1)n⋅(1−x)n=(−1)nx−n(1−x)2n Total number of terms will be 2n+1 which is odd ( ∵2n is always even) ∴ Middle term =22n+1+1=(n+1) th Now, Tr+1=nCr(1)rxn−r So, xn⋅(−1)n2nCn⋅x2n−n=2nCn⋅(−1)n Middle term is an odd term. So, n+1 will be odd. So, n will be even. ∴ Required answer is 2nCn.