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答案　（1）bn＝（2/3）^n
(2)Sn＝(3n/2)(n+1)+{[(3+n)•2^(n+1)]/3^（n-1）}-18

1. ∵ A(n+2)=(5/3)A(n+1)-(2/3)An,
∴ A(n+2)-A(n+1)=(2/3)[A(n+1)-An], B(n+1)=(2/3)Bn
∴ 数列{Bn}是等比数列.B1=A2-A1=(5/3)-1=2/3, Bn=(2/3)^n
An-A1=(2/3)+(2/3)^+(2/3)^3+…+(2/3)^(n-1),
∴ An=3-2(2/3)^(n-1), nAn=3n-2n(2/3)^(n-1)
2. 设S1=3(1+2+…+n)=3n(n+1)/2
S2=2+2·2(2/3)+2·3(2/3)^…+2n·(2/3)^(n-1),则
(2/3)S2=2·(2/3)+2·2·(2/3)^+…+2·(n-1)·(2/3)^n+2n·(2/3)^(n+1),相减,得
S2/3=2[1+(2/3)+(2/3)^+…+(2/3)^n]-2n·(2/3)^(n+1)
=6[1-(2/3)^(n+1)]-2n·(2/3)^(n+1)
∴ S2=18[1-(2/3)^(n+1)]-6n·(2/3)^(n+1)
Sn=S1-S2=3n(n+1)/2-18[1-(2/3)^(n+1)]+6n·(2/3)^(n+1)