Question

14．計算C${\;}_{2}^{2}$+C${\;}_{3}^{2}$+C${\;}_{4}^{2}$+C${\;}_{5}^{2}$+…+C${\;}_{99}^{2}$．

Answer 1

分析 根據(jù)組合數(shù)的公式${C}_{n}^{m}$+${C}_{n}^{m-1}$=${C}_{n+1}^{m}$，進行計算即可．

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

點評 本題考查了利用組合數(shù)公式進行化簡、計算的應用問題，是基礎題目．