由(*)得:x=(dx-b)/(a-cx)
[(Fn-x1)/(Fn-x2)]={[aF(n-1)+b]/[cF(n-1)+d]-x1}/{[aF(n-1)+b]/[cF(n-1)+d]-x2}=[(a-cx1)F(n-1)-(dx1-b)]/[(a-cx2)F(n-1)-(dx2-b)]=[(a-cx1)/(a-cx2)]*[F(n-1)-(dx1-b)/(a-cx1)]/[F(n-1)-(dx2-b)/(a-cx2)]=[(a-cx1)/(a-cx2)]*[F(n-1)-x1]/[F(n-1)-x2]
所以:数列[(Fn-x1)/(Fn-x2)]是等比数列,公比是(a-cx1)/(a-cx2)
这已经给了你很大启发,剩下来的你自己想想,不行再call我!