Question

20．${C}_{3}^{0}$+${C}_{4}^{1}$+${C}_{5}^{2}$+${C}_{6}^{3}$+…+${C}_{20}^{17}$的值为${C}_{21}^{17}$．

Answer 1

分析 根据组合数公式${C}_{n}^{m-1}$+${C}_{n}^{m}$=${C}_{n+1}^{m}$进行计算即可．

解答 解：${C}_{3}^{0}$+${C}_{4}^{1}$+${C}_{5}^{2}$+${C}_{6}^{3}$+…+${C}_{20}^{17}$=（${C}_{4}^{0}$+${C}_{4}^{1}$）+${C}_{5}^{2}$+${C}_{6}^{3}$+…+${C}_{20}^{17}$
=（${C}_{5}^{1}$+${C}_{5}^{2}$）+${C}_{6}^{3}$+…+${C}_{20}^{17}$
=${C}_{6}^{2}$+${C}_{6}^{3}$+…+${C}_{20}^{17}$
=${C}_{7}^{3}$+${C}_{7}^{4}$…+${C}_{20}^{17}$
…
=${C}_{20}^{16}$+${C}_{20}^{17}$
=${C}_{21}^{17}$．
故答案为：${C}_{21}^{17}$．

点评 本题考查了组合数公式的应用问题，也考查了计算能力的应用问题，是基础题目．