文[1]用待定系数法求出了由递推式an+1=can+daan+b确定的数列{an}的通项公式(只要方程ax2+(b-c)x-d=0有根(包括复数根),都可用[1]的方法求解;若无根,则a=0,b=c,d≠0,得{an}是等差数列.[1]中对数列{an}的各项取倒数时,应要求an≠0(n∈N*)).受[1]的启发,笔者研究了由递推式an+1=c In [1], we use the undetermined coefficient method to find the general term of the series {an} determined by the recursive an+1=can+daan+b (as long as the equation ax2+(bc)xd=0 has a root (including complex roots ), can be solved by the method of [1]; if there is no root, then a=0, b=c, d≠0, then {an} is an arithmetic series. [1] The entries of {an} In the countdown, an≠0(n∈N*)) should be asked. Inspired by [1], the author studied the recursive an+1=c