数列{ an } 的前n项和记为Sn,a1=2,点(an,Sn)在直线y= 2x-2上 ,n属于N*,

(1) 将(an,Sn)、(a(n-1),s(n-1))代入y= 2x-2,得
Sn=2an-2,S(n-1)=2a(n-1)-2
将上两式相减
Sn-S(n-1)=2an-2a(n-1)
an=2an-2a(n-1)
则得an=2a(n-1)
即{ an } 是等比数列,且q=2,
则an=a1*q^(n-1)=2^n.
(2) bn=log 2 a(n+1),将an=2^n代入得bn=n+1,
则bn/an=(n+1)/(2^n)
数列{bn/an}的前n项和Tn=2/(2^1)+3/(2^2)+4/(2^3)+.+n/[2^(n-1)]+(n+1)/(2^n),(1)
则Tn/2=2/(2^2)+3/(2^3)+.+(n-1)/[2^(n-1)]+n/(2^n)+(n+1)/[2^(n+1)],(2)
(1)-(2)得到 Tn/2=1+1/(2^2)+1/(2^3)+.+1/[2^(n-1)]+1/(2^n)-(n+1)/[2^(n+1)]
=1+(1/2^2)*{[1-(1/2)^(n-1)]/(1-1/2)}-(n+1)/[2^(n+1)]=3-(1/2)^(n-1)-(n+1)/2^n
=3-(n+3)/(2^n).