27. Chứng minh rằng:
a) $kC_{n}^{k}=nC_{n-1}^{k-1}$ với 1 ≤ k ≤ n.
b) $\frac{1}{k+1}C_{n}^{k}=\frac{1}{n+1}C_{n+1}^{k+1}$ với 0 ≤ k ≤ n
Bài Làm:
a) Ta có: $kC_{n}^{k}=k\times \frac{n!}{k!(n-k)!}=\frac{kn!}{k(k-1)!(n-k)!}$
$=\frac{n(n-1)!}{(k-1)![(n-1)-(k-1)]!}=nC_{n-1}^{k-1}$
Vậy $kC_{n}^{k}=nC_{n-1}^{k-1}$ với 1 ≤ k ≤ n.
b) Ta có: $\frac{1}{k+1}C_{n}^{k}=\frac{1}{k+1}\times \frac{n!}{k!(n-k)!}=\frac{n!}{(k+1)!(n-k)!}$
$=\frac{1}{n+1}\times \frac{(n+1)n!}{(k+1)![(n+1)-(k+1)]!}$
$=\frac{1}{n+1}\times \frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{1}{n+1}C_{n+1}^{k+1}$
Vậy $\frac{1}{k+1}C_{n}^{k}=\frac{1}{n+1}C_{n+1}^{k+1}$ với 0 ≤ k ≤ n