题目

关于数列和函数
已知函数f(x)=x*2+x-1,a、b是方程f(x)=0的两根（a>b）,f'(x)=2x+1.设a1=1,an+1=an-f(an)/f'(an)(n=1,2,3.).记bn=ln(an-b/an-a)(n=1,2,3...),求数列{bn}的前n项和Sn.

提问时间：2021-01-03

答案

A+B=-1,AB=-1
因A＞B,所以B＜0＜A
a(n+1)=an-f(an)/f'(an)
=an-[an^2+an-1]/(2an+1)
=(an^2+1)/(2an+1)
a(n+1)=(an^2+1)/(2an+1)
其特征方程为x^2+x-1=0
其解为A=-1/2+√5/2,B=-1/2-√5/2
a(n+1)-A=(an^2+1)/(2an+1)-A=(an^2-2A*an+1-A)/(2an+1)
a(n+1)-B=(an^2+1)/(2an+1)-B=(an^2-2B*an+1-B)/(2an+1)
两式相除：
[a(n+1)-A]/[a(n+1)-B]=(an^2-2A*an+1-A)/(an^2-2B*an+1-B)
=(an^2-2A*an+A^2)/(an^2-2B*an+B^2)
=(an-A)^2/(an-B)^2
=[(an-A)/(an-B)]^2
设Cn=(an-A)/(an-B),C1=(a1-A)/(a1-B)=(1-A)/(1-B)=A^2/B^2
Cn=[C(n-1)]^2
=[C(n-2)]^4
……
=(C1)^[2^(n-1)]
=[(1-A)/(1-B)]^[2^(n-1)]
(an-A)/(an-B)=Cn=[(1-A)/(1-B)]^[2^(n-1)]^[2^(n-1)]