(1)由数列{an}的前n项和为Sn,数列{Sn}的前n项和为Tn,满足Tn=2Sn-1,T1=S1=a1,由此能求出a1的值.
(2)由Tn=2Sn-1,知Tn-1=2Sn-1-1,n≥2,由此能求出Sn=2an.
(3)由(2)得,n≥2,由此能求出数列{an}的通项公式.
【解析】
(1)∵数列{an}的前n项和为Sn,数列{Sn}的前n项和为Tn,满足Tn=2Sn-1,
∴T1=S1=a1,
∴a1=2a1-1,解得a1=1.
(2)∵Tn=2Sn-1,Tn-1=2Sn-1-1,n≥2,
∵当n≥2时,Sn=Tn-Tn-1,an=Sn-Sn-1,
∴Sn=2an.
(3)由(2)得,n≥2,
故数列{an}是以a2=1为首项,以2为公比的等比数列,
∴.