(1)
设{an}的公差为d,{bn|的公比为q
∵a1=b1=2,
∴a4+b4=2+3d+2q^3=27
s4-b4=8+6d-2q^3=10
相加10+9d=37
∴d=3,q^3=8,q=2
∴an=2+3(n-1)=3n-1
bn=2^n
(2)稍候Tn=anb1+a(n-1)b2+......+a1bn Tn =(3n-1)*2+(3n-4)*4+(3n-7)*8+.......+2*2^n ①①×2:2Tn=(3n-1)*4+(3n-4)*8+,,,,,,,,,,,+5*2^n+2*2^(n+1) ②②-①: Tn=-2(3n-1)+3*4+3*8+....+3*2^n+2^(n+2)=2-6n+12[2^(n-1)-1]/(2-1)+2^(n+2)=-10-6n+3*2^(n+1)+2*2^(n+1)=5*2^(n+1)-6n-10∴Tn+12=5*2^(n+1)-6n+2=10*2^n-2(3n-1)=-2an+10bn