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=

A+B=-1,AB=-1因A＞B,所以B＜0＜Aa(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-√5/2a(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^2Cn=[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)]