Tính tổng chứa công thức tổ hợp

Tính các tổng sau:
a) $A=C_n^1+2C_n^2+3C_n^3+...+nC_n^n$
b) $B=2C_n^0+\dfrac{2^2}{2}C_n^1+\dfrac{2^3}{3}C_n^2+...+\dfrac{2^{n+1}}{n+1}C_n^n$


Trước tiên ta chứng minh : $\mathbf{kC_n^k=nC_{n-1}^{k-1}} \;\;\; (1)$
Ta có: $kC_n^k=k\dfrac{n!}{k!(n-k)!}=\dfrac{k.n.(n-1)!}{k.(k-1)!(n-k)!}=n\dfrac{(n-1)!}{(k-1)!(n-k)!}=nC_{n-1}^{k-1}$
Thay n bằng n+1 và thay k bằng k+1 ta được $(k+1)C_{n+1}^{k+1}=(n+1)C_{n}^{k}$
hay $\mathbf{\dfrac{1}{k+1}C_n^k=\dfrac{1}{n+1}C_{n+1}^{k+1}} \;\;\; (2)$

a) Thay k = 1, 2, ..., n vào (1):
$A=nC_{n-1}^0+nC_{n-1}^1+nC_{n-1}^2+ +nC_{n-1}^{n-1}\\=n \left [ C_{n-1}^0+C_{n-1}^1+C_{n-1}^2+ +C_{n-1}^{n-1}\right ]$
mà $C_{n-1}^0+C_{n-1}^1+C_{n-1}^2+ +C_{n-1}^{n-1}=(1+1)^{n-1}=2^{n-1}$
Vậy $A= n.2^{n-1}$

b) Thay k = 0 , 1, 2, ..., n vào (2):
$B=2\dfrac{1}{n+1}C_{n+1}^1+2^2\dfrac{1}{n+1}C_{n+1}^2+2^3\dfrac{1}{n+1}C_{n+1}^3+...+2^{n+1}\dfrac{1}{n+1}C_{n+1}^{n+1}\\=\dfrac{1}{n+1} \left [ 2C_{n+1}^1+2^2C_{n+1}^2+2^3C_{n+1}^3+...+2^{n+1}C_{n+1}^{n+1} \right ]$
mà $(1+2)^{n+1}=C_{n+1}^0+2C_{n+1}^1+2^2C_{n+1}^2+...+2^{n+1}C_{n+1}^{n+1}$
nên $2C_{n+1}^1+2^2C_{n+1}^2+...+2^{n+1}C_{n+1}^{n+1}=3^{n+1}-1$
Vậy $B=\dfrac{3^{n+1}-1}{n+1}$