Derive the solution for ana_nan that satisfies the recurrence equation an=−an−1+12an−2a_n = -a_{n-1} + 12a_{n-2}an=−an−1+12an−2 with a0=3a_0 = 3a0=3 and a1=2a_1 = 2a1=2.
參考答案與解析
an=(−4)n+2⋅3na_n=(-4)^n+2\cdot3^nan=(−4)n+2⋅3n。特徵方程r2+r−12=0r^2+r-12=0r2+r−12=0，即(r+4)(r−3)=0(r+4)(r-3)=0(r+4)(r−3)=0，根為−4,3-4,3−4,3。設an=α(−4)n+β⋅3na_n=\alpha(-4)^n+\beta\cdot3^nan=α(−4)n+β⋅3n，代入a0=3,a1=2a_0=3,a_1=2a0=3,a1=2解得α=1,β=2\alpha=1,\beta=2α=1,β=2（已用遞迴數列逐項驗證吻合）。
