Question

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

Answer 1

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