Question

19．計(jì)算下列各題．
（1）（C${\;}_{100}^{98}$+C${\;}_{100}^{97}$）÷A${\;}_{101}^{3}$；
（2）C${\;}_{2}^{2}$+C${\;}_{3}^{2}$+C${\;}_{4}^{2}$+…+C${\;}_{10}^{2}$．

Answer 1

分析 （1）利用${C}_{n}^{m}={C}_{n}^{n-m}$和${C}_{n}^{m}+{C}_{n}^{m+1}={C}_{n+1}^{m+1}$，能求出結(jié)果．
（2）利用${C}_{n}^{m}+{C}_{n}^{m+1}={C}_{n+1}^{m+1}$，能得到C${\;}_{2}^{2}$+C${\;}_{3}^{2}$+C${\;}_{4}^{2}$+…+C${\;}_{10}^{2}$=${C}_{11}^{3}$，由此能求出結(jié)果．

解答 解：（1）（C${\;}_{100}^{98}$+C${\;}_{100}^{97}$）÷A${\;}_{101}^{3}$
=（${C}_{100}^{2}+{C}_{100}^{3}$）÷${A}_{101}^{3}$
=${C}_{101}^{3}÷{A}_{101}^{3}$
=$\frac{1}{3！}$
=$\frac{1}{6}$．
（2）C${\;}_{2}^{2}$+C${\;}_{3}^{2}$+C${\;}_{4}^{2}$+…+C${\;}_{10}^{2}$
=${C}_{4}^{3}+{C}_{4}^{2}+{C}_{5}^{2}+…+{C}_{10}^{2}$
=${C}_{5}^{3}+{C}_{5}^{2}+{C}_{6}^{2}+…+{C}_{10}^{2}$
=${C}_{6}^{3}+{C}_{6}^{2}+{C}_{7}^{2}+…+{C}_{10}^{2}$
=${C}_{7}^{3}+{C}_{7}^{2}+{C}_{8}^{2}+…+{C}_{10}^{2}$
=${C}_{8}^{3}+{{C}_{8}^{2}+C}_{9}^{2}+{C}_{10}^{2}$
=${C}_{9}^{3}+{C}_{9}^{2}+{C}_{10}^{2}$
=${C}_{10}^{3}+{C}_{10}^{2}$
=${C}_{11}^{3}$
=$\frac{11×10×9}{3×2×1}$
=165．

點(diǎn)評(píng) 本題考查組合數(shù)公式化簡(jiǎn)求值，是基礎(chǔ)題，解題時(shí)要認(rèn)真審題，注意組合數(shù)公式的性質(zhì)的合理運(yùn)用．