设数列{an}的前n项和为Sn,已知Sn=2an-2n+1,（n为下标,n+1为上标）,求通项公式?

Sn=2an-2n+1,得,a1=2a1-2^2,得a1=4
Sn=2an-2^(n+1),得Sn+1=2an+1-2^(n+2)
两式相减,得
an+1=2an+1-2an-2^(n+1)
an+1=2an+2^(n+1)
两边队以2^(n+1),得
an+1/2^(n+1)=an/2^n+1
an/2^n=a1/2+(n-1)=n+1
所以,an=(n+1)2^n