Question

4．${C}_{3}^{0}$+${C}_{4}^{1}$+${C}_{5}^{2}$+${C}_{6}^{3}$+…+${C}_{2013}^{2010}$的值為（　　）

• A． ${C}_{2013}^{3}$
• B． ${C}_{2014}^{3}$
• C． ${C}_{2014}^{4}$
• D． ${C}_{2013}^{4}$

Answer 1

分析 直接運用組合數(shù)的兩條性質(zhì)，${C}_{n}^{m}$=${C}_{n}^{n-m}$和${C}_{n}^{m}$+${C}_{n}^{m+1}$=${C}_{n+1}^{m+1}$，運算求解．

解答 解：根據(jù)組合數(shù)的性質(zhì)一：${C}_{n}^{m}$=${C}_{n}^{n-m}$，
所以，原式=${C}_{3}^{3}$+${C}_{4}^{3}$+${C}_{5}^{3}$+${C}_{6}^{3}$+…+${C}_{2013}^{3}$，
再根據(jù)組合數(shù)的性質(zhì)二：${C}_{n}^{m}$+${C}_{n}^{m+1}$=${C}_{n+1}^{m+1}$，且${C}_{3}^{3}$=${C}_{4}^{4}$，
原式=${C}_{4}^{4}$+${C}_{4}^{3}$+${C}_{5}^{3}$+${C}_{6}^{3}$+…+${C}_{2013}^{3}$，
=${C}_{5}^{4}$+${C}_{5}^{3}$+${C}_{6}^{3}$+…+${C}_{2013}^{3}$，
=${C}_{6}^{4}$+${C}_{6}^{3}$+…+${C}_{2013}^{3}$，
=${C}_{2014}^{4}$，
故選：C．

點評 本題主要考查了組合及組合數(shù)公式的運算，尤其是組合的兩點性質(zhì)，屬于中檔題．