∵2Sn=3an-2,①∴n=1时,2a1=3a1-2,解得a1=2.n≥2时,2Sn-1=3an-1-2,②①-②,得:2an=3an-3an-1,整理,得an=3an-1,∴ an an?1 =3,∴{an}是首项为2,公比为3的等比数列,Sn= 2(1?3n) 1?3 =3n-1.故答案为:2,3n-1.