∵Sn=3an+1∴a1=S1=3a1+1, a1=-1/3且S(n-1)=3a(n-1)+1∴an=Sn-S(n-1)=3an-3a(n-1)∴2an=3a(n-1)∴an=(3/2)a(n-1)∴an是以a1=-1/3为首项,q=3/2为公比的等比数列∴an=-1/3×(3/2)^(n-1) 很高兴为您解答,祝你学习进...