(Ⅰ)当n=1,a1=2…(1分) 当n≥2时,an=Sn-Sn-1=2an-2an-1…(2分) ∴an=2an-1(n≥2),∴{an}是等比数列,公比为2,首项a1=2 ∴an=2n…(3分) 又点P(bn,bn+1) (n∈N*)在直线y=x+2上,∴bn+1=bn+2, ∴{bn}是等差数列,公差为2,首项b1=1,∴bn=2n-1…(5分) (Ⅱ)∵an•bn=(2n-1)×2n ∴Dn=1×21+3×22+5×23+7×24+…(2n-3)×2n-1+(2n-1)×2n① 2Dn=1×22+3×23+5×24+7×25+…(2n-3)×2n+(2n-1)×2n+1② ①-②得-Dn=1×21+2×22+2×23+2×24+…2×2n-(2n-1)×2n+1…(7分) =2+2×-(2n-1)×2n+1=2n+1(3-2n)-6…(8分) Dn=(2n-3)2n+1+6…(9分) (Ⅲ)cn=…(11分) T2n=(a1+a3+…+a2n-1)-(b2+b4+…b2n) =2+23+…+22n-1-[3+7+…+(4n-1)]=-2n2-n…(13分) |