Question

12．计算：C${\;}_{3}^{1}$+${C}_{4}^{2}$+${C}_{5}^{3}$+…+${C}_{99}^{97}$．

Answer 1

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

解答 解：C${\;}_{3}^{1}$+${C}_{4}^{2}$+${C}_{5}^{3}$+…+${C}_{99}^{97}$=${C}_{3}^{2}+{C}_{4}^{2}+{C}_{5}^{2}+{C}_{6}^{2}+…+{C}_{99}^{2}$
=（${C}_{3}^{3}$+C₃²）+C₄²+C₅²+…+C₉₉²-${C}_{3}^{3}$
=（${C}_{4}^{3}$+C₄²）+C₅²+…+C₉₉²-1
=（${C}_{5}^{3}$+C₅²）+…+C₉₉²-1
=…=${C}_{99}^{3}$+C₉₉²-1
=${C}_{100}^{3}$-1=161699．

点评 本题考查了利用组合数公式进行化简、计算的应用问题，是基础题目．