If C0,C1,C2,…,Cn are the coefficients of the expansion  of (1 + x) n , then the value of ∑0nCkk+1 is

Here, tr+1= nCrr+1=1r+1⋅nCr             =1n+1⋅n+1Cr+1Putting r = 0, 1, 2, ... n and adding, we get ∑0nCkk+1 =1n+1(n+1C1+n+1C2+n+1C3+…+n+1Cn+1) =1n+1(2n+1−n+1C0)=2n+1−1n+1