已知数列an的前n项和为Sn=3n-n²,n∈N*,设bn=2的n次方,求数列anbn的前n项和

an=sn-s(n-1)=-2n+4
anbn=(-2n+4)*2^n=（-n+2)*2^(n+1)
Tn =1*2^2+0*2^3-1*2^4-2*2^5-.+(-n+2)*2^(n+1)
2Tn= 1*2^3+0*2^4-1*2^5-2*2^6-.+(-n+2)*2^(n+2)[上式错位相减得]
-tn=2^2-[2^3+2^4+...+2^(n+1)]-(-n+2)*2^(n+2)=4-[2^(n+2)]-2^3]-(-n+2)*2^(n+2)
Tn=(-n+2)*2^(n+2)+2^(n+2)-12