一、形如an+2=pan+1+qan(p,q为常数)的数列,可用特征方程x2=px+q先求出两根(1)若方程有两异根α,β,则可令an=c1αn+c2βn(c1,c2是待定系数)(2)若方程有两重根α=β,则可令an(c1+nc2)an+2二、形如an+1=Aan+B/Can+D的数列,其特征根方程为x=AxB/Cx+D(1)若方程有两异根α,β,则可令an+1-α/an+1-β=c·an-α/an-β(2)若方程有两重根α=β,则可令1/an+1-α-1/-an-α=c.其中,c是待定系数. First, the form as an + 2 = pan + 1 + qan (p, q is a constant) sequence, the characteristic equation can be used x2 = px + q first find two (1) If the equation has two different roots α, β, then (2) If the equation has a double root α = β, then an (c1 + nc2) an + 2 can be formed as an + 1 = Aan + B / Can + D, the eigenvalue equation is x = AxB / Cx + D (1) If the equation has two different roots α and β, then an + 1 -α / an + 1- β = c · an -α / an-β (2) If the equation has a double root α = β, then 1 / an + 1 -α-1 / -an-α = c where c is the coefficient to be determined.