(1)Sn=2n2-3n;(2)Sn=3n-2.
解析:(1)当n≥2时,an=Sn-Sn-1=2n2-3n-[2(n-1)2-3(n-1)]=4n-5.
当n=1时,a1=S1=-1满足上式,
∴an=4n-5.
(2)当n≥2时,an=Sn-Sn-1=(3n-2)-(3n-1-2)=2×3n-1,
当n=1时,a1=S1=3-2=1,
∴an=