已知数列{an},Sn为前n项的和,满足关系式2Sn=3an-3,求﹛an﹜的通项公式

2Sn=3an-3, 2S(n-1) = 3a(n-1) -3作差： 2[Sn -S(n-1)]= 3an - 3a(n-1)2an = 3an -3a(n-1)an = 3a(n-1)an/ a(n-1) =3数列{AN}是等比数列,公式是3,当n=1时,a1=3a1-3, a1= 3/2an= (3/2)* 3^(n-1) = 3^n/2