Question

6．计算（用数字作答）：${C}_{3}^{2}$+${C}_{4}^{2}$+${C}_{5}^{2}$+…+${C}_{19}^{2}$=1139．

Answer 1

分析 利用${C}_{n}^{m+1}+{C}_{n}^{m}$=${C}_{n+1}^{m+1}$求解．

解答 解：${C}_{3}^{2}$+${C}_{4}^{2}$+${C}_{5}^{2}$+…+${C}_{19}^{2}$
=（${C}_{3}^{3}$+${C}_{3}^{2}$）+${C}_{4}^{2}$+${C}_{5}^{2}$+…+${C}_{19}^{2}$-${C}_{3}^{3}$
=（${C}_{4}^{3}$+${C}_{4}^{2}$）+${C}_{5}^{2}$+…+${C}_{19}^{2}$-1
=（${C}_{5}^{3}$+${C}_{5}^{2}$）+…+${C}_{19}^{2}$-1
=${C}_{6}^{3}$+…+${C}_{19}^{2}$-1
=…=${C}_{19}^{3}$+${C}_{19}^{2}$-1
=${C}_{20}^{3}$-1=1139．
故答案为：1139．

点评 本题考查组合数的求法，是基础题，解题时要认真审题，注意组合数公式的合理运用．