Question

20．化簡(jiǎn)：${C}_{n}^{1}$+2${C}_{n}^{2}$+3${C}_{n}^{3}$+…+n${C}_{n}^{n}$．

Answer 1

分析 由（1+x）ⁿ=$1+{∁}_{n}^{1}x+$${∁}_{n}^{2}{x}^{2}$+…+${∁}_{n}^{n}{x}^{n}$，兩邊對(duì)x求導(dǎo)可得：n（1+x）^n-1=${C}_{n}^{1}$+2${C}_{n}^{2}$x+3${C}_{n}^{3}$x²+…+n${C}_{n}^{n}$x^n-1．令x=1，即可得出．

解答 解：由（1+x）ⁿ=$1+{∁}_{n}^{1}x+$${∁}_{n}^{2}{x}^{2}$+…+${∁}_{n}^{n}{x}^{n}$，
兩邊對(duì)x求導(dǎo)可得：n（1+x）^n-1=${C}_{n}^{1}$+2${C}_{n}^{2}$x+3${C}_{n}^{3}$x²+…+n${C}_{n}^{n}$x^n-1．
令x=1，可得：n•2^n-1=${C}_{n}^{1}$+2${C}_{n}^{2}$+3${C}_{n}^{3}$+…+n${C}_{n}^{n}$．
即${C}_{n}^{1}$+2${C}_{n}^{2}$+3${C}_{n}^{3}$+…+n${C}_{n}^{n}$=n•2^n-1．

點(diǎn)評(píng) 本題考查了導(dǎo)數(shù)的應(yīng)用、二項(xiàng)式定理的應(yīng)用，考查了推理能力與計(jì)算能力，屬于中檔題．