Question

已知数列{an}中，a1=3/5，an=2-1/a（n-1）(n>=2),数列{bn}满足bn=1/an-1，求证bn是等差数列

Answer 1

证：设2bn=b(n+1)+b(n-1) (n>=2)
2bn=2/(an-1)
b(n+1) =1/[a(n+1)-1]
b(n-1) =1/[a(n-1)-1]
即有
2/(an-1) =1/[a(n+1)-1]+ 1/[a(n-1)-1]
=2/(an-1) 显然成立
bn是等差数列 (n>=2)
a1=3/5 b1=-5/2
a2=1/3 b2=-3/2
a3=-1 b3=-1/2
2b2=b3+b1
因此
bn是等差数列 (n>=1)