题目内容

2.组合数${C}_{34}^{0}$+${C}_{34}^{2}$+${C}_{34}^{4}$…+${C}_{34}^{34}$被9除的余数是8.

分析 根据组合数的特征,得出${C}_{34}^{0}$+${C}_{34}^{2}$+${C}_{34}^{4}$…+${C}_{34}^{34}$=233=(9-1)11
利用二项展开式即可求出该组合数被9除的余数.

解答 解:∵${C}_{34}^{0}$+${C}_{34}^{2}$+${C}_{34}^{4}$…+${C}_{34}^{34}$=${C}_{34}^{1}$+${C}_{34}^{3}$+${C}_{34}^{5}$+…+${C}_{34}^{33}$,
∴${C}_{34}^{0}$+${C}_{34}^{2}$+${C}_{34}^{4}$…+${C}_{34}^{34}$=$\frac{1}{2}$×234
=233
=811
=(9-1)11
=${C}_{11}^{0}$•911-${C}_{11}^{1}$•910+${C}_{11}^{2}$•92+…+(-1)r•${C}_{11}^{r}$•9r+…-${C}_{11}^{11}$•90
=k×9-1
=(k-1)9+8,其中k∈N;
∴该组合数被9除的余数是8.
故答案为:8.

点评 本题考查了组合数公式的应用问题,也考查了二项式定理的应用问题,是基础题目.

练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网