Let A=E[1/(x+1)]=Σ[k=0~n] C(n,k)p^k*q^(n-k)/(k+1). (q=1-p)
Integrate both sides of f(t)=(t+q)^n=Σ[k=0~n] C(n,k)t^k*q^(n-k),
for t=0~x, then
∫[0~x] (t+q)^n dt=Σ[k=0~n] {C(n,k) x^(k+1) *q^(n-k) / (k+1) }
thus,
[(x+q)^(n+1)-q^(n+1) ]/(n+1)=Σ[k=0~n]{C(n,k)x^(k+1)q^(n-k)/(k+1)}
Set x=p, then
[1-(1-p)^(n+1)]/(n+1)=Σ[k=0~n] C(n,k)p^(k+1)q^(n-k)/(k+1)
so,
[1-(1-p)^(n+1)]/[(n+1)p]=Σ[0~n]C(n,k)p^kq^(n-k)/(k+1) =A=E[1/(x+1)]