Q13/108) Evaluate the given sum k from 0 to n ∑ (nCk) /(k+1)

(nCk)/(k+1) = n!/(k! * (n-k)!) * (k+1) = 1/(n+1) ( n+ 1 C k+1)

So ∑ (nCk)/(k+1) = 1/(n+1) (∑ (n+1 C k) - (n+1 C 0))
= 1/(n+1) (∑(n+1Ck) - 1)
= 1/(n+1) ( 2^n+1) – 1)