一、形如an+2 =pan+1 +qan(p ,q是常数 )的数列形如a1 =m1 ,a2 =m2 ,an+2 =pan+1 +qan(p、q是常数 )的二阶递归数列都可用特征根法求得通项an.其特征方程为x2 =px +q . ①若①有二异根α ,β ,则可令an =c1 an +c2 βn(c1 ,c2 是待定常数 ) ;若①有两重根α =β ,则可令an =(c1 First, the form of an + 2 = pan + 1 + qan (p, q is a constant) the number of forms as a1 = m1, a2 = m2, an +2 = pan +1 + qan (p, q is a constant) Order recursive sequence can be obtained by eigenvalue general term an. The characteristic equation is x2 = px + q. ① If there are two different roots α, β, then an = c1 an + c2 βn (c1, c2 is To be determined constant); If there is a double root α = β, you can make an = (c1