设数列 {an} 满足 a0=−1,a1=1,an=−6an−1−9an−2−8,n⩾,则 a_9= _______.
答案 9841.
解析 考虑题中递推式的不动点方程x=-6x-9x-8\iff x=-\dfrac 12,因此设 b_n=a_n+\dfrac12,则b_n=-6b_{n-1}-9b_{n-2},其特征方程为x^2=-6x-9\iff x=-3,于是 b_n=(An+B)\cdot (-3)^n,其中 A,B 为待定系数,满足\begin{cases} b_0=-\dfrac 12,\\ b_1=\dfrac 32,\end{cases}\implies \begin{cases} B=-\dfrac12,\\ (A+B)\cdot (-3)=\dfrac 32,\end{cases} \iff \begin{cases} A=0,\\ B=-\dfrac 12,\end{cases}因此a_n=\dfrac{-1-(-3)^n}2,进而 a_9=\dfrac{-1+3^9}2=\boxed{9841}.