Question

20．化简：${C}_{n}^{1}$+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}$，两边对x求导可得：n（1+x）^n-1=${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}$，
两边对x求导可得：n（1+x）^n-1=${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．

点评 本题考查了导数的应用、二项式定理的应用，考查了推理能力与计算能力，属于中档题．