Chủ đề này có 1 trả lời

[motdaica]

Chứng minh :$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}=\frac{2^{n+1}-1}{n+1}$

Bài viết đã được chỉnh sửa nội dung bởi hxthanh: 20-10-2013 - 22:27
Tiêu đề

[chanhquocnghiem]

Chứng minh :$\sum_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}=\frac{2^{n+1}-1}{n+1}$

Ta có :

$C_{n+1}^{0}+C_{n+1}^{1}+C_{n+1}^{2}+...+C_{n}^{n}=2^{n+1}$

---> $C_{n+1}^{1}+C_{n+1}^{2}+...+C_{n}^{n}=2^{n+1}-1$

hay $(n+1)+\frac{(n+1).n}{1.2}+\frac{(n+1).n.(n-1)}{1.2.3}+...+\frac{(n+1).n.(n-1)...2.1}{1.2.3...n.(n+1)}=2^{n+1}-1$

---> $1+\frac{n}{1.2}+\frac{n.(n-1)}{1.2.3}+...+\frac{n.(n-1)...2.1}{1.2.3...n.(n+1)}=\frac{2^{n+1}-1}{n+1}$

---> $\frac{C_{n}^{0}}{1}+\frac{C_{n}^{1}}{2}+\frac{C_{n}^{2}}{3}+...+\frac{C_{n}^{n}}{(n+1)}=\frac{2^{n+1}-1}{n+1}$

---> $\sum_{k=0}^{n}C_{n}^{k}.\frac{1}{k+1}=\frac{2^{n+1}-1}{n+1}$ hay $\sum_{k=0}^{n}\binom{n}{k}\frac{1}{k+1}=\frac{2^{n+1}-1}{n+1}$