已知数列{an}的前n项和为Sn,且满足Sn=2an-1,n为正整数,求数列{an}的通项公式an

n=1时,S1=a1=2a1-1 ,a1=1
n≥2时,an=Sn-S(n-1)=(2an-1)-(2a(n-1) - 1)
an=2a(n-1),
故an=2^(n-1).