Question

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

Answer 1

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•${C}_{11}^{r}$•9^r+…-${C}_{11}^{11}$•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．