Question

数列{an}的前n项和为Sn，且满足a1=1，Sn^2=an（Sn-1/2) n>=2

1.证明数列{1/Sn}是等差数列，并求an
2.设bn=sn/2n+1,求{bn}的前n项和Tn，若对任意的n属于正整数都有Tn<log 1/2 m，求m的取值范围

Answer 1

1Sn-S(n-1)=an,带入，化简整理得，Sn=S(n-1)/(1+2S(n-1)两边同时倒过来得1/Sn为等差首项为1，Sn=1/(2n-1)
带入an=1/(2n-1)-1/(2n-3)
2 bn=1/(2n-1)(2n+1)
Tn=1/(1*3)+1/(3*5)+1/(5*7)......
1/(1*3)=(1-1/3)/2 1/(3*5)=(1/3-1/5)/2 相加后+ -抵消得Tn=n/(2n+1)
m<(20n+8)/(2n+1) m最大=28/3