在数列{an}中,已知a1=3/5,an*a(n-1)+1=2a(n-1)(n>=2,n∈N),数列{bn}满足：bn=

(1)an*a(n-1)+1=2a(n-1)
an=[2a(n-1)-1]/a(n-1)
an -1= [2a(n-1)-1]/a(n-1) -1= [a(n-1)-1]/a(n-1)
1/(an -1) = a(n-1)/[a(n-1)-1] = {[a(n-1)-1]+1} /[a(n-1)-1]
=1+ 1/[a(n-1) -1]
1/(an -1) -1/[a(n-1) -1] =1
即 bn － b(n-1)= 1
所以bn 是等差数列.
(2) bn =b1 +1*(n-1)= 1/(a1 -1) +n-1=1/(-2/5) +n-1= n-7/2,n>=2
an =1/bn +1 =1/(n-7/2) +1= 1+ 2/(2n-7)
当2n-7>0,且2n-7为最小时,an有最大值,
2n-7>0,n>7/2,n＝4时,2n-7>0且最小,此时a4最大,a4＝1+2/(2*4-7)=3
当2n-7