Question

在数列{an}中，a1=1，an=1/2（an-1+1/an-1)

Answer 1

an=(1/2)[a(n-1)+1/a(n-1)]
a2=(1/2)[a1+1/a1]
=1
2ana(n-1)-[a(n-1)]^2-1=0
特征方程为2x^2-x^2-1=0
x=±1
2(an+1)[a(n-1)+1]-2an-2a(n-1)-2-[a(n-1)+1]^2+2a(n-1)+1-1=0
2(an-1)[a(n-1)-1]+2an+2a(n-1)-2-[a(n-1)-1]^2-2a(n-1)+1-1=0
2(an+1)a(n-1)=[a(n-1)+1]^2
2(an-1)a(n-1)=[a(n-1)-1]^2
两式相除：
(an-1)/(an+1)={[a(n-1)-1]/[a(n-1)+1]}^2
设bn=(an-1)/(an+1),b1=[(a1-1)/(a1+1)]^2=0,b2=[(a2-1)/(a2+1)]^2=0
bn=b(n-1)^2
bn=b(n-1)^2=b(n-2)^4=b(n-3)^8=……=b3^[2^(n-3)]=b2^[2^(n-2)]=0^[2^(n-2)]=0
bn=(an-1)/(an+1)=0
an=1