已知数列an满足a(n+1)=2an-1，且a1=3，bn=（an-1）⼀ana(n+1)，数列bn前n项和为sn。求sn。

a(n+1)=2an-1
a(n+1)-1=2(an-1)
所以{an-1}为等比数列q=2
an-1=(a1-1)*2^(n-1)=2^n
an=2^n+1
bn=2^n/[2^n+1][2^(n+1)+1]=1/(2^n+1)-1/[2^(n+1)+1]
sn=b1+b2+b3+......+bn=1/3-1/5+1/5-1/9+1/9-1/17+......+1/(2^n+1)-1/[2^(n+1)+1]
所以sn=1/3-1/[2^(n+1)+1]

a(n+1)=2 两边都减1 得
[a(n+1)-1]=2（an-1）再比一下
[a(n+1)-1]/（an-1）=2 则
所以（an-1）为等比数列q=2 a1=2
求得
an-1=2^n(即2的n次方）
an=2^n+1
因为bn=（an-1）/ana(n+1)
所以bn=2^n/[2^n+1][2^(n+1)+1]=1/(2^n+1)-1/[2^(n+1)+1]
sn=b1+b2+b3+......+bn=1/3-1/5+1/5-1/9+1/9-1/17+......+1/(2^n+1)-1/[2^(n+1)+1]
所以sn=1/3-1/[2^(n+1)+1]