\已知等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}为等比数列,b1=2,且b2S2=32,b3S3=120.

b2S2=b2(a1+a2)=b1*q*(2a1+d)=32,b3S3=b3(a1+a2+a3)=b1*q²*(3a1+3d)=120,得 d=2(都是正数）,q=2.
∴an=a1+d(n-1)=2n+1,bn=b1*q^（n-1）=2^n.
Tn=3*2+5*2^2+……+(2n+1)*2^n,
2tn=3*2^2+5*2^3+……+(2n+1)*2^(n+1).
错位相减得：
Tn=（n-1）*2^(n+2)+2^(n+1)+2=(2n-1)*2^(n+1)+2.