Question

9．如果C${\;}_{n}^{0}$+$\frac{1}{2}$C${\;}_{n}^{1}$+$\frac{1}{3}$C${\;}_{n}^{2}$+…+$\frac{1}{n+1}$C${\;}_{n}^{n}$=$\frac{31}{n+1}$，則（1+x）²ⁿ的展開式中系數(shù)最大的項(xiàng)為70x⁴．

Answer 1

分析 由k•${C}_{n+1}^{k}$=（n+1）•${C}_{n}^{k-1}$，得$\frac{1}{k}$•${C}_{n}^{k-1}$=$\frac{1}{n+1}$•${C}_{n+1}^{k}$，化簡C${\;}_{n}^{0}$+$\frac{1}{2}$C${\;}_{n}^{1}$+$\frac{1}{3}$C${\;}_{n}^{2}$+…+$\frac{1}{n+1}$C${\;}_{n}^{n}$，求出n的值，再求二項(xiàng)式展開式中系數(shù)最大的項(xiàng)．

解答 解：由k•${C}_{n+1}^{k}$=（n+1）•${C}_{n}^{k-1}$，得$\frac{1}{k}$•${C}_{n}^{k-1}$=$\frac{1}{n+1}$•${C}_{n+1}^{k}$，
∴C${\;}_{n}^{0}$+$\frac{1}{2}$C${\;}_{n}^{1}$+$\frac{1}{3}$C${\;}_{n}^{2}$+…+$\frac{1}{n+1}$C${\;}_{n}^{n}$
=$\frac{1}{n+1}$•${C}_{n+1}^{1}$+$\frac{1}{n+1}$•${C}_{n+1}^{2}$+$\frac{1}{n+1}$•${C}_{n+1}^{3}$+…+$\frac{1}{n+1}$•${C}_{n+1}^{n+1}$
=$\frac{1}{n+1}$•（${C}_{n+1}^{1}$+${C}_{n+1}^{2}$+${C}_{n+1}^{3}$+…+${C}_{n+1}^{n+1}$）
=$\frac{1}{n+1}$•（2ⁿ⁺¹-1）=$\frac{31}{n+1}$，
解得n=4；
∴（1+x）^2×4的展開式中系數(shù)最大的項(xiàng)為${C}_{8}^{4}$x⁴=70x⁴．
故答案為：70x⁴．

點(diǎn)評(píng) 本題考查了二項(xiàng)式定理的應(yīng)用、組合數(shù)的計(jì)算公式及其性質(zhì)，考查了推理能力與計(jì)算能力，屬于中檔題