Tổng

Câu hỏi: Tổng $C_n^0+\frac{1}{2}{C}_n^1+\frac{1}{3}{C}_n^2+\frac{1}{4}{C}_n^3+...+\frac{1}{n+1}{C}_n^n$  bằng

A. $\frac{2^{n+1}-1}{n+1}$

B. $\frac{2^{n+1}+1}{n+1}$

C. $\frac{2^{n+1}}{n+1}$

D. $\frac{2^{n+1}}{n+2}$