: Tính tổng $S=C_{n}^{0}+\frac{{{2}^{2}}-1}{2}C_{n}^{1}+…+\frac{{{2}^{n+1}}-1}{n+1}C_{n}^{n}$

: Tính tổng $S=C_{n}^{0}+\frac{{{2}^{2}}-1}{2}C_{n}^{1}+…+\frac{{{2}^{n+1}}-1}{n+1}C_{n}^{n}$

C. $S=\frac{{{3}^{n+1}}-{{2}^{n+1}}}{n+1}$

B. $S=\frac{{{3}^{n}}-{{2}^{n+1}}}{n+1}$

C. $S=\frac{{{3}^{n+1}}-{{2}^{n}}}{n+1}$

D. $S=\frac{{{3}^{n+1}}+{{2}^{n+1}}}{n+1}$

Hướng dẫn

Chọn A

Ta có: $S={{S}_{1}}-{{S}_{2}}$

Trong đó ${{S}_{1}}=\sum\limits_{k=0}^{n}{C_{n}^{k}\frac{{{2}^{k+1}}}{k+1}};\text{ }{{S}_{2}}=\sum\limits_{k=0}^{n}{\frac{C_{n}^{k}}{k+1}}=\frac{{{2}^{n+1}}-1}{n+1}-1$

Mà $\frac{{{2}^{k+1}}}{k+1}C_{n}^{k}=\frac{{{2}^{k+1}}}{n+1}C_{n+1}^{k+1}$$\Rightarrow {{S}_{1}}=\frac{{{3}^{n+1}}-1}{n+1}-1$

Suy ra: $S=\frac{{{3}^{n+1}}-{{2}^{n+1}}}{n+1}$.