Question

10．求证：${C}_{n}^{0}$+$\frac{1}{2}$${C}_{n}^{1}$+$\frac{1}{3}$${C}_{n}^{2}$+…+$\frac{1}{n+1}$${C}_{n}^{n}$=$\frac{1}{n+1}$（2ⁿ⁺¹-1）

Answer 1

分析 根据组合数公式，$\frac{1}{k+1}$${C}_{n}^{k}$=${C}_{n+1}^{k+1}$•$\frac{1}{n+1}$，再根据二项式定理得到${C}_{n+1}^{1}$+${C}_{n+1}^{2}$+…+${C}_{n+1}^{n+1}$=（1+1）^（n+1）-${C}_{n+1}^{0}$=2ⁿ⁺¹-1，继而得以证明．

解答 证明∵$\frac{1}{k+1}$${C}_{n}^{k}$=$\frac{n！}{k！（n-k）！}$•$\frac{1}{k+1}$=$\frac{n！}{（n-k）！（k+1）！}$=$\frac{（n+1）！}{（n+1-k-1）！（k+1）！}$•$\frac{1}{n+1}$=${C}_{n+1}^{k+1}$•$\frac{1}{n+1}$
∴${C}_{n}^{0}$+$\frac{1}{2}$${C}_{n}^{1}$+$\frac{1}{3}$${C}_{n}^{2}$+…+$\frac{1}{n+1}$${C}_{n}^{n}$=$\frac{1}{n+1}$${C}_{n+1}^{1}$+$\frac{1}{n+1}$${C}_{n+1}^{2}$+…+$\frac{1}{n+1}$${C}_{n+1}^{n+1}$=$\frac{1}{n+1}$（${C}_{n+1}^{1}$+${C}_{n+1}^{2}$+…+${C}_{n+1}^{n+1}$），
∵（1+1）ⁿ=${C}_{n}^{0}$+${C}_{n}^{1}$+${C}_{n}^{2}$+…+${C}_{n}^{n}$，
∴${C}_{n+1}^{1}$+${C}_{n+1}^{2}$+…+${C}_{n+1}^{n+1}$=（1+1）^（n+1）-${C}_{n+1}^{0}$=2ⁿ⁺¹-1，
∴${C}_{n}^{0}$+$\frac{1}{2}$${C}_{n}^{1}$+$\frac{1}{3}$${C}_{n}^{2}$+…+$\frac{1}{n+1}$${C}_{n}^{n}$=$\frac{1}{n+1}$（2ⁿ⁺¹-1），
问题得以证明．

点评 本题考查了组合数公式，以及二项式定理，培养了学生的运算能力，属于中档题．