Given (1+x)(1−x)n =(1−x)n+x(1−x)n General term of (1−x)n=nCr⋅(−1)r⋅xr ∴ Term containing xn in (1−x)n=nCn⋅(−1)n⋅xn So coefficient of xn in (1−x)n=nCn.(−1)n General term of x(1−x)n=nCr.(−1)r⋅xr+1 ∴ Term containing xn in x(1−x)n=nCn−1⋅(−1)n−1⋅xn So coefficient of xn in x(1−x)n=nCn−1⋅(−1)n−1 ∴ coefficient of xn in (1−x)n+x(1−x)n =nCn⋅(−1)n+nCn−1⋅(−1)n−1 =(−1)n−1[nCn−1−nCn] =(−1)n−1[n−1] =(−1)n[1−n]