解(Ⅰ)设等差数列{an}的公差为d,∵差数列{an}的前n项和为Sn,数列{bn}为等比数列, 且b2+S2=12,q=, ∴,即,解得:. ∴an=a1+(n-1)d=3+(n-1)•3=3n, bn=b1qn-1=1×3n-1=3n-1. (Ⅱ)Tn=anb1+an-1b2+an-2b3+…+a1bn =3n•1+3(n-1)•3+3(n-2)•32+…+3×2×3n-2+3•3n-1 =n•3+(n-1)•32+(n-2)•33+…+2•3n-1+3n. ∴3Tn=n•32+(n-1)•33+…+2•3n+3n+1. ∴3Tn-Tn=-3n+32+33+…+3n+3n+1 =(32+33+…+3n+1)-3n =-3n=-3n-. ∴Tn=-n-. |