数列an的前n项和为Sn,2Sn＋3＝3an,（n属于N＋）.求an通项公式

3a(1)=2s(1)+3=2a(1)+3,a(1)=3.2s(n)+3=3a(n),2s(n+1)+3=3a(n+1),3a(n+1)-3a(n) = [2s(n+1)+3]-[2s(n)+3] = 2[s(n+1)-s(n)] = 2a(n+1),a(n+1) = 3a(n),{a(n)}是首项为a(1)=3,公比为3的等比数列.a(n)=3*3^(n-1) = 3^n...