每日一题[953]上上下下

设数列$a_1,a_2,a_3,\cdots ,a_{21}$满足：$\left|a_{n+1}-a_n\right|=1$($n=1,2,3,\cdots,20$)，$a_1,a_7,a_{21}$成等比数列．若$a_1=1$，$a_{21}=9$，则满足条件的不同数列的个数为_______．

正确答案是$15099$．

分析与解　设$b_n=a_{n+1}-a_n$，$n=1,2,\cdots,20$，则数列$a_1,a_2,a_3,\cdots,a_{21}$与有序数组$$\left(b_1,b_2,\cdots,b_{20}\right)$$一一对应，且$b_i\in\{1,-1\}$($i=1,2,\cdots,20$)．根据题意，有$a_7=3$或$a_7=-3$，于是\[\begin{cases}b_1+b_2+\cdots +b_6=2,\\ b_7+b_8+\cdots+b_{20}=6,\end{cases}\]或\[\begin{cases}b_1+b_2+\cdots+b_6=-4,\\ b_7+b_8+\cdots+b_{20}=12,\end{cases} \]对应的个数为\[{\rm C}_6^4{\rm C}_{14}^{10}+{\rm C}_6^1{\rm C}_{14}^{13}=15099.\]