已知有等比数列{an}和等差数列{bn},且an=2的n次方,bn=2n+1,求数列{an*bn}的前n项和Tn

Tn=2*3+5*2^2+7*2^3+...+(2n+1)*2^n
2Tn=3*2^2+5*2^3+7*2^4+...+(2n+1)*2^(n+1)
Tn-2Tn=2*3+2*2^2+2*2^3+...+2*2^n-(2n+1)*2^(n+1)
-Tn=6+2[4*(2^(n-1)-1)/(2-1)]-(2n+1)*2^(n+1)
-Tn=6+4*2^n-8-2(2n+1)*2^n
Tn=(4n-2)*2^n+2