数列{an}\{a_n\}{an}满足a0=−1,a1=1a_0=-1, a_1=1a0=−1,a1=1, an=−6an−1−9an−2−8,n≥2.a_n = -6a_{n-1}-9a_{n-2}-8, n \geq 2.an=−6an−1−9an−2−8,n≥2. 求a9a_9a9.
解:直接计算可知a2=−5,a3=13,a4=−41a_2=-5, a_3=13, a_4=-41a2=−5,a3=13,a4=−41, a5=121,a6=−365,a7=1093a_5=121, a_6=-365, a_7=1093a5=121,a6=−365,a7=1093, a8=−3281,a9=9841a_8=-3281, a_9=9841a8=−3281,a9=9841.
为计算通项公式,令bn=an+1/2b_n = a_n + 1/2bn=an+1/2, 则bn=−6an−1−9an−2.b_n = -6a_{n-1}-9a_{n-2}.bn=−6an−1−9an−2. 用特征方程求出特征值λ1=λ2=−3\lambda_1 = \lambda_2 = -3λ1=λ2=−3. 设通项公式为bn=(c1+c2n)(−3)n.b_n = (c_1 + c_2 n)(-3)^n.bn=(c1+c2n)(−3)n. 代入b0=a0+1/2=−1/2,b1=a1+1/2=3/2b_0=a_0+1/2=-1/2, b_1 = a_1 + 1/2 = 3/2b0=a0+1/2=−1/2,b1=a1+1/2=3/2, 得bn=(−12)(−3)n=an+12b_n = (-\frac{1}{2}) (-3)^n=a_n + \frac{1}{2}bn=(−21)(−3)n=an+21.