在数列an中a1=1 a（n+1）=2an+2^n设bn=an/（2^n-1）.1.证明：数列bn是等差数列

a(n+1)=2an+2^n
a(n+1)/2^(n+1)=2an/2^(n+1)+2^n/2^(n+1)
a(n+1)/2^(n+1)=an/2^n+1/2
a(n+1)/2^(n+1)-an/2^n=1/2
所以an/2^n 是以1/2为公差的等差数列
an/2^n=a1/2^1+1/2(n-1)
an/2^n=1/2+1/2(n-1)
an/2^n=n/2
an=2^n*n/2
an=n*2^(n-1)
sn=1*2^0+2*2^1+.+n*2^(n-1)
2sn=1*2^1+2*2^2+.+n*2^n
sn-2sn=2^0+2^1+2^2+.+2^(n-1)-n*2^n
-sn=(1-2^n)/(1-2)-n*2^n
sn=1-2^n+n*2^n
sn=(n-1)*2^n+1
limSn/[n*2^(n+1)]
=lim[(n-1)*2^n+1]/[n*2^(n+1)]
=1/2
bn=an/(2^n-1)
=n*2^(n-1)/2^(n-1)
=n
bn-b(n-1)=n-(n-1)=1
所以bn是以1为公差的等差数列
cn=2bn-1
=2n-1
Tn=1+3+.+2n-1
=(1+2n-1)*n/2
=n^2
dn=Tn/[(4an)^2-Tn]
=n^2/{4*[n*2^(n-1)]^2-n^2}
=n^2/{4*[n^2*2^(2n-2)-n^2}
=n^2/{n^2*2^2n-n^2}
=1/(2^2n-1)
=1/(4^n-1)
d1=1/(4^1-1)=1/3
[d1+d2+d3+.+dn+d(n+1)]-(d1+d2+d3+.+dn)=d(n+1)
=1/[4^(n+1)-1]>0
∴数列{d1+d2+d3+.+dn}单调递增,
即d1+d2+d3+.+dn>=d1=1/3
要使d1+d2+d3+.+dn>=log8 (2m+t)对任意正整数n成立,
必须且只需1/3>=log8 (2m+t)
即0＜2m+t0,4+t