已知数列{an}满足a1=3,anan-1=2an-1-1,(1)求证{1/an-1}是等差数列

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(1)
证：
ana(n-1)=2a(n-1)-1
a(n-1)=0时,0=-1,等式恒不成立,因此数列各项均不为0
a(n-1)=1/2时,ana(n-1)=0,an和a(n-1)中至少有1个为0,与各项不为0矛盾,因此数列各项均不等于1/2. (这步判断一定要的,不能直接列分式)
ana(n-1)=2a(n-1)-1
an=[2a(n-1) -1]/a(n-1)
an -1=[2a(n-1)-1-a(n-1)]/a(n-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,为定值.
1/(a1-1)=1/(3-1)=1/2
数列{1/(an -1)}是以1/2为首项,1为公差的等差数列.
(2)

1/(an -1)=1/(a1 -1) +(n-1)=1/2 +n-1= n-1/2=(2n-1)/2
an -1=2/(2n-1)
an=2/(2n-1) +1=(2+2n-1)/(2n-1)=(2n+1)/(2n-1)
bn=an/(2n+1)²=[(2n+1)/(2n-1)]/(2n+1)²=1/[(2n+1)(2n-1)]=(1/2)[1/(2n-1) -1/(2n+1)]
Sn=b1+b2+...+bn
=(1/2)[1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=(1/2)[1-1/(2n+1)]
=n/(2n+1)

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