此文将呈现一组与数列有关的姊妹结论,并解决由这组姊妹结论“孕育”出的一类竞赛题,供参考.结论1数列{an}满足:an=a·an-1+b c·an-1+d(c≠0,ad≠bc,n>1,n∈N*),a1≠a·a1+b c·a1+d,若方程x=ax+b cx+d有两相异实根x1,x2,则 This article presents a series of sister-related conclusions and solves a series of competition questions that are “nurtured” by the group of sister conclusions for reference. CONCLUSION 1 The sequence {an} satisfies: an = a · an-1 + bc · an-1 + d (c ≠ 0, ad ≠ bc, n> 1, n∈N *) and a1 ≠ a · a1 + bc · a1 + d. If the equation x = ax + b cx + d Two-phase different root x1, x2, then