(1)∵6Sn=+3an+2,① ∴6a1=+3a1+2,解得a1=1或a1=2. 又6Sn-1=+3an-1+2(n≥2), ② 由①-②,得6an=(-)+3(an-an-1),即(an+an-1)(an-an-1-3)=0. ∵an+an-1>0,∴an-an-1=3(n≥2). 当a1=2时,a2=5,a6=17,此时a1,a2,a6不成等比数列,∴a1≠2; 当a1=1时,a2=4,a6=16,此时a1,a2,a6成等比数列,∴a1=1. ∴{an}是以1为首项3为公差的等差数列,{bn}是以1为首项4为公比的等比数列. ∴an=3n-2,bn=4n-1. (2)由(1)得 Tn=1×4n-1+4×4n-2+…+(3n-5)×41+(3n-2)×40, ③ ∴4Tn=1×4n+4×4n-1+7×4n-2+…+(3n-2)×41. ④ 由④-③,得 3Tn=4n+3×(4n-1+4n-2+…+41)-(3n-2)=4n+12×-(3n-2) =2×4n-(3n+1)-1=2bn+1-an+1-1, ∴3Tn+1=2bn+1-an+1(n∈N*). |