已知数列{an}的前n项和为Sn,且Sn=2an-2^(n+1),n∈N*

(1)
n=1时,S1=a1=2a1-2^2
a1=4
n≥2时,
Sn=2an-2^(n+1) S(n-1)=2a(n-1)-2ⁿ
Sn-S(n-1)=an=2an-2^(n+1)-2a(n-1)+2^n=2an-2a(n-1)-2ⁿ
an=2a(n-1)+2ⁿ
等式两边同除以2ⁿ
an/2ⁿ=a(n-1)/2^(n-1) +1
an/2ⁿ-a(n-1)/2^(n-1)=1,为定值.
a1/2^1=4/2=2
数列{an/2ⁿ}是以2为首项,1为公差的等差数列.
an/2^n=2+n-1=n+1
an=(n+1)×2ⁿ
数列{an}的通项公式为an=(n+1)×2ⁿ
(2)
Sn=2an-2^(n+1)=2(an-2ⁿ)=2[(n+1)×2ⁿ-2ⁿ]=n×2^(n+1)
bn=log2(Sn/n)=log2[n×2^(n+1)/n]=log2[2^(n+1)]=n+1
Tn=1/bn+1/b(n+1)+1/(bn+2)+...+1/b(2n-1)
T(n+1)=1/b(n+1)+1/(bn+2)+...+1/b(2n-1)+1/b(2n)+1/b(2n+1)
T(n+1)-Tn=1/b(2n)+1/b(2n+1)-1/bn
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)>0
随n增大,Tn单调递增,因此当n=1时,Tn有最小值(Tn)min=T1=1/b1=1/2,要对任意n,不等式恒成立,则只有
1/2>k/12
k