题目内容
4.计算:${A}_{3}^{2}{+A}_{4}^{2}$+…+${A}_{100}^{2}$.分析 利用排列数公式${A}_{n}^{m}$=${A}_{m}^{m}$•${C}_{n}^{m}$与组合数公式${C}_{n}^{m+1}$+${C}_{n}^{m}$=${C}_{n+1}^{m+1}$,进行计算即可.
解答 解:${A}_{3}^{2}$+${A}_{4}^{2}$+…+${A}_{100}^{2}$=${A}_{2}^{2}$(${C}_{3}^{2}$+${C}_{4}^{2}$+…+${C}_{100}^{2}$)
=2(${C}_{3}^{3}$+${C}_{3}^{2}$+${C}_{4}^{2}$+…+${C}_{100}^{2}$-1)
=2(${C}_{4}^{3}$+${C}_{4}^{2}$+…+${C}_{100}^{2}$-1)
=2(${C}_{100}^{3}$+${C}_{100}^{2}$-1)
=2(${C}_{101}^{3}$-1)
=2${C}_{101}^{3}$-2.
点评 本题考查了排列数与组合数公式的应用问题,是基础题目.
练习册系列答案
相关题目
14.在正方体ABCD-A1B1C1D1中,B1C与对角面DD1B1B所成角的大小是( )
A. | 15° | B. | 30° | C. | 45° | D. | 60° |
14.已知sin55°=m,则cos2015°=( )
A. | $\sqrt{1-{m^2}}$ | B. | -$\sqrt{1-{m^2}}$ | C. | m | D. | -m |