In the expansion of the following expression 1+(1+x)+(1+x)2+…+(1+x)n the coefficient of xk  (0≤k≤n) is

The given expression is1+(1+x)+(1+x)2+…+(1+x)n being in GP. LetS=1+(1+x)+(1+x)2+…+(1+x)n=(1+x)n+1−1(1+x)−1=x−1(1+x)n+1−1∴ The coefficient of xk in S. = The coefficient of xk+1 in (1+x)n+1−1=n+Ck+1