Question

3．证明：C$\left.\begin{array}{l}{0}\\{n}\end{array}\right.$+$\frac{1}{2}$C$\left.\begin{array}{l}{1}\\{n}\end{array}\right.$+$\frac{1}{3}$C$\left.\begin{array}{l}{2}\\{n}\end{array}\right.$+…+$\frac{1}{k}$C$\left.\begin{array}{l}{k-1}\\{n}\end{array}\right.$+…+$\frac{1}{n+1}$C$\left.\begin{array}{l}{n}\\{n}\end{array}\right.$=$\frac{1}{n+1}$（2ⁿ⁺¹-1）．

Answer 1

分析 利用组合数的性质，化$\frac{1}{k}$${C}_{n}^{k-1}$=$\frac{1}{n+1}$${C}_{n+1}^{k}$，代入题目中等式的左边，即可得出右边．

解答 证明：∵$\frac{1}{k}$${C}_{n}^{k-1}$=$\frac{1}{k}$•$\frac{n!}{（k-1）!•（n-k+1）!}$
=$\frac{n!}{k!•（n+1-k）!}$
=$\frac{1}{n+1}$•$\frac{（n+1）!}{k!•（n+1-k）!}$
=$\frac{1}{n+1}$${C}_{n+1}^{k}$，
∴C$\left.\begin{array}{l}{0}\\{n}\end{array}\right.$+$\frac{1}{2}$C$\left.\begin{array}{l}{1}\\{n}\end{array}\right.$+$\frac{1}{3}$C$\left.\begin{array}{l}{2}\\{n}\end{array}\right.$+…+$\frac{1}{k}$C$\left.\begin{array}{l}{k-1}\\{n}\end{array}\right.$+…+$\frac{1}{n+1}$C$\left.\begin{array}{l}{n}\\{n}\end{array}\right.$
=$\frac{1}{n+1}$•${C}_{n+1}^{1}$+$\frac{1}{n+1}$•${C}_{n+1}^{2}$+$\frac{1}{n+1}$•${C}_{n+1}^{3}$+…+$\frac{1}{n+1}$•${C}_{n+1}^{k}$+…+$\frac{1}{n+1}$•${C}_{n+1}^{n+1}$
=$\frac{1}{n+1}$•（${C}_{n+1}^{1}$+${C}_{n+1}^{2}$+${C}_{n+1}^{3}$+…+${C}_{n+1}^{k}$+…+${C}_{n+1}^{n+1}$）
=$\frac{1}{n+1}$（2ⁿ⁺¹-1）．

点评 本题考查了组合数公式的应用问题，也考查了二项式定理的应用问题，是中档题目．