Question

16．求和：${C}_{n}^{0}$${C}_{n}^{1}$+${C}_{n}^{1}$${C}_{n}^{2}$+…+${C}_{n}^{n-1}$${C}_{n}^{n}$．

Answer 1

分析 由等式（1+x）ⁿ×（x+1）ⁿ=（1+x）²ⁿ，
取左右兩邊展開(kāi)式中含x^n-1項(xiàng)的系數(shù)，得出${C}_{n}^{0}$•${C}_{n}^{1}$+${C}_{n}^{1}$•${C}_{n}^{2}$+…+${C}_{n}^{n-1}$•${C}_{n}^{n}$的值是多少．

解答 解：∵（1+x）ⁿ×（x+1）ⁿ=（1+x）ⁿ×（1+x）ⁿ=（1+x）²ⁿ，
取左右兩邊展開(kāi)式中含x^n-1項(xiàng)的系數(shù)，則
（${C}_{n}^{0}$•${C}_{n}^{1}$+${C}_{n}^{1}$•${C}_{n}^{2}$+…+${C}_{n}^{n-1}$•${C}_{n}^{n}$）•x^n-1=${C}_{2n}^{n-1}$•x^n-1，
∴${C}_{n}^{0}$•${C}_{n}^{1}$+${C}_{n}^{1}$•${C}_{n}^{2}$+…+${C}_{n}^{n-1}$•${C}_{n}^{n}$=${C}_{2n}^{n-1}$=$\frac{（2n）!}{（n-1）!•（n+1）!}$．

點(diǎn)評(píng) 本題考查了二項(xiàng)式定理的應(yīng)用問(wèn)題，解題的關(guān)鍵是構(gòu)造等式，利用二項(xiàng)式展開(kāi)式定理解答，是基礎(chǔ)題目．