Question

15．C${\;}_{4}^{1}$+C${\;}_{5}^{2}$+…+C${\;}_{20}^{17}$等于（　�。�

• A． C${\;}_{21}^{17}$
• B． C${\;}_{21}^{17}$-1
• C． C${\;}_{21}^{18}$-1
• D． C${\;}_{21}^{18}$

Answer 1

分析 利用組合數(shù)的性質(zhì)求解即可．

解答 解：C${\;}_{4}^{1}$+C${\;}_{5}^{2}$+…+C${\;}_{20}^{17}$=1+C${\;}_{4}^{1}$+C${\;}_{5}^{2}$+…+C${\;}_{20}^{17}$-1
=${C}_{4}^{0}$+C${\;}_{4}^{1}$+C${\;}_{5}^{2}$+…+C${\;}_{20}^{17}$-1
=${C}_{5}^{1}$+C${\;}_{5}^{2}$+…+C${\;}_{20}^{17}$-1
=${C}_{6}^{2}$+C${\;}_{6}^{3}$+…+C${\;}_{20}^{17}$-1
=…
=C${\;}_{21}^{17}$-1．
故選：B．

點(diǎn)評(píng) 本題考查組合數(shù)性質(zhì)的應(yīng)用，考查計(jì)算能力．