1)由题得an=Sn−Sn−1=
3
2
(an−an−1)(n≥2)所以an=3an-1故有
an
an−1
=3(n≥2)
又S1=
3
2
(a1−1)=a1,解得a1=3,所以数列{an}成等比数列
由(1)得an=3n,则bn=log3an=log33n=n故有 cn=anbn=n3n
设Tn=1•31+2•32+3•33+…+(n-1)3n-1+n•3n
3Tn=1•32+2•33+3•34+…+(n-1)3n+n•3n+1
则 −2Tn=(31+32+33+…+3n)−n•3n+1=
3(1−3n)
1−3
−n•3n+1
所以Tn=
(2n−1)3n+1+3
4