（1）
n=1时,a1=S1=2a1-2 a1=2
Sn=2an-2
Sn-1=2a(n-1)-2
an=Sn-Sn-1=2an-2-2a(n-1)+2
an=2a(n-1)
an/a(n-1)=2,为定值,
数列{an}是以2为首项,2为公比的等比数列,an=2^n
bn-b(n+1)+2=0
b(n+1)-bn=2
b1=1
数列{bn}是以1为首项,2为公差的等差数列.bn=2n-1
(2)
cn=anbn=2^n(2n-1)
Tn=2^1+3*2^2+5*2^3+7*2^4+...+(2n-1)2^n
2Tn=2^2+3*2^3+5*2^4+...+(2n-3)2^n+(2n-1)2^(n+1)
Tn=2Tn-Tn=-2^1-2*2^2-2*2^3-...-2*2^n+(2n-1)2^(n+1)
=2-2(2^1+2^2+2^3+...+2^n)+(2n-1)2^(n+1)
=2-4*(2^n-1)+(4n-2)2^n
=(4n-6)2^n+6
Tn