The value of∑n+1r=1(∑nk=1kCr−1) where r, k, n ϵ N is equal to
2n+1
2n+1−1
2n+1−2
None of these
∑n+1r=1(∑nk=1kCr−1)∑n+1r=1(∑nk=1(k+1Cr−kCr))∑n+1r=1(n+1Cr−1Cr)=2n+1−2
∞∑n=11(n+1)(n+2)(n+3)....(n+k) is equal to