已知等差数列{an}的各项均为正数,a1=3,前n项和为Sn,{bn}是等比数列,公比q=2,且a2b2=20,a3b3=56（1）求an与bn (2)求数列{anbn}的前n项和Tn

(1) an=a1+(n-1)d=3+(n-1)d
bn=b1*q^(n-1)=b1*2^(n-1)
a2b2=(3+d)*2b1=20
a3b3=(3+2d)*4b1=56
d=2 b1=2
an=2n+1
bn=2^n
(2) anbn=(2n+1)*2^n
Tn=3*2+5*2^2+7*2^3+...+(2n+1)*2^n
2Tn=3*2^2+5*2^3+...+(2n-1)*2^n+(2n+1)*2^(n+1)
Tn-2Tn=3*2+2(2^2+2^3+...+2^n)-(2n+1)*2^(n+1)
-Tn=6+8[1-2^(n-1)]/(1-2)-(2n+1)*2^(n+1)
Tn=(2n+1)*2^(n+1)-6+8-2*2^(n+1)
=(2n-1)*2^(n+1)+2