(1)对于数列{an},当n=1时,a1=S1=(a1-1),解得a1=3. 当n≥2时,an=Sn-Sn-1=(an-1)-(an-1-1),化为an=3an-1. ∴数列{an}是首项为3,公比为3的等比数列, ∴an=3×3n-1=3n. 对于数列{bn}满足bn=bn-1-(n≥2),b1=3. 可得bn+1=(bn-1+1). ∴数列{bn+1}是以b1+1=4为首项,为公比的等比数列. ∴bn+1=4×()n-1,化为bn=42-n-1. (2)cn=3n•lo=3n(4-2n) ∴Tn=2×31+0+(-2)•33+…+(4-2n)•3n. 3Tn=2×32+0+(-2)×34+…+(6-2n)•3n+(4-2n)•3n+1. ∴-2Tn=6+(-2)•32+(-2)•33+…+(-2)•3n-(4-2n)•3n+1 =6-2×-(4-2n)•3n+1. ∴Tn=-+(-n)•3n+1. |