数列（an）,a1=1,当n≥2,其前n项和Sn满足Sn^2=an(Sn-1)证（1/Sn）是等差数列.设bn=log以2为底Sn／S（n＋2）,bn的前n项和Tn,求满足Tn≥6的最小正整数n

n>1时
an=Sn-S(n-1)
(Sn)^2=[Sn-S(n-1)](Sn-1)
=(Sn)^2-Sn+S(n-1)-SnS(n-1)
因此S(n-1)-Sn=SnS(n-1)
由题知道Sn不等于0
两边同时除以SnS(n-1)得
1/Sn-1/S(n-1)=1
因此（1/Sn)是等差数列
首项为1/S1=1/a1=1,公差为1
因此1/Sn=1+（n-1)*1=n
因此Sn=1/n
S（n+2）=1/（n+2）
Sn/S(n+2)=(n+2)/n
bn=log(2)[(n+2)/n]=log(2)(n+2)-log(2)n
当n为奇数时
Tn=1og(2)3-log(2)1+log(2)4-log(2)2+…+log(2)(n+1)-log(2)(n-1)+log(2)(n+2)-log(2)n
=log(2)(n+2)-log(2)1+log(2)(n+1)-log(2)2
=log(2)(n+2)+log(2)(n+1)-1
当n为偶数时
Tn=1og(2)3-log(2)1+log(2)4-log(2)2+…+log(2)(n+1)-log(2)(n-1)+log(2)(n+2)-log(2)n
=log(2)(n+2)-log(2)2+log(2)(n+1)-log(2)1
=log(2)(n+2)+log(2)(n+1)-1
综上所述
Tn
=log(2)(n+2)+log(2)(n+1)-1
=log(2)(n+2)+log(2)(n+1)-log(2)2
=log(2)[(n+2)(n+1)/2]
Tn≥6则
log(2)[(n+2)(n+1)/2]≥6
即log(2)[(n+2)(n+1)/2]≥log(2)(2^6)
又f(x)=log(2)x是增函数,2^6=64
因此
（n+2)(n+1)/2≥64
化简得到
n^2+3n-126≥0
由n>0解得
n≥(-3+√516)/2
又22