Question

13．求證：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，比較左邊與右邊的x^n-1的系數(shù)即可得出．

解答 證明：∵（x+1）ⁿ（x+1）ⁿ=（x+1）²ⁿ，
則左邊x^n-1的系數(shù)為：C${\;}_{n}^{0}$${C}_{n}^{1}$+${{C}_{n}^{1}}_{\;}^{\;}$${C}_{n}^{2}$+…+${C}_{n}^{n-1}$${C}_{n}^{n}$，右邊x^n-1的系數(shù)=${∁}_{2n}^{n-1}$=$\frac{（2n）!}{（n-1）!（n+1）!}$．
∴C${\;}_{n}^{0}$${C}_{n}^{1}$+${{C}_{n}^{1}}_{\;}^{\;}$${C}_{n}^{2}$+…+${C}_{n}^{n-1}$${C}_{n}^{n}$=$\frac{（2n）!}{（n-1）!（n+1）!}$．

點(diǎn)評 本題考查了二項(xiàng)式定理的應(yīng)用、組合數(shù)的計(jì)算公式，考查了計(jì)算能力，屬于基礎(chǔ)題．