(1)由Sn=2an-2得:Sn-1=2an-1-2(n≥2),两式相减可得an=2an-1(n≥2),再求得a1=2,可知数列{an}是以2为首项,2为公比的等比数列,从而可求an=2n;点P(bn,bn+1)在直线x-y+2=0上,可知bn+1-bn=2,又b1=1,从而可求得{bn}的通项公式;
(2))Tn=1×2+3×22+5×23+…+(2n-3)×2n-1+(2n-1)×2n①,2Tn=1×22+3×23+…+(2n-3)×2n+(2n-1)×2n+1②,错位相减即可求得Tn.
【解析】
(1)由Sn=2an-2得:Sn-1=2an-1-2(n≥2),
两式相减得:an=2an-2an-1,即=2(n≥2),
又a1=2a1-2,
∴a1=2,
∴数列{an}是以2为首项,2为公比的等比数列,
∴an=2n.
∵点P(bn,bn+1)在直线x-y+2=0上,
∴bn+1-bn=2,
∴数列{bn}是等差数列,
∵b1=1,
∴bn=2n-1;
(2)Tn=1×2+3×22+5×23+…+(2n-3)×2n-1+(2n-1)×2n①
∴2Tn=1×22+3×23+…+(2n-3)×2n+(2n-1)×2n+1②
①-②得:-Tn=1×2+2(22+23+…+2n)-(2n-1)×2n+1
=2+2×-(2n-1)×2n+1
=2+2×2n+1-8-(2n-1)×2n+1
=(3-2n)2n+1-6,
∴Tn=(2n-3)2n+1+6.