Question

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

Answer 1

分析 由（x+1）ⁿ（x+1）ⁿ=（x+1）²ⁿ，比较左边与右边的x^n-1的系数即可得出．

解答 证明：∵（x+1）ⁿ（x+1）ⁿ=（x+1）²ⁿ，
则左边x^n-1的系数为：C${\;}_{n}^{0}$${C}_{n}^{1}$+${{C}_{n}^{1}}_{\;}^{\;}$${C}_{n}^{2}$+…+${C}_{n}^{n-1}$${C}_{n}^{n}$，右边x^n-1的系数=${∁}_{2n}^{n-1}$=$\frac{（2n）!}{（n-1）!（n+1）!}$．
∴C${\;}_{n}^{0}$${C}_{n}^{1}$+${{C}_{n}^{1}}_{\;}^{\;}$${C}_{n}^{2}$+…+${C}_{n}^{n-1}$${C}_{n}^{n}$=$\frac{（2n）!}{（n-1）!（n+1）!}$．

点评 本题考查了二项式定理的应用、组合数的计算公式，考查了计算能力，属于基础题．