已知数列{an}中a1=1,当n>=2时其前n项和sn满足sn^2=an(sn-1/2).问题1求sn的表达式；2设bn=sn/(2n+1),求{bn}的前n项和Tn.

题目

已知数列{an}中a1=1,当n>=2时其前n项和sn满足sn^2=an(sn-1/2).问题1求sn的表达式；2设bn=sn/(2n+1),求{bn}的前n项和Tn.

答案

1.sn^2=an(sn-1/2)
Sn(Sn-an)+(1/2)an=0
2Sn*S(n-1)+Sn-S(n-1)=0
Sn=S(n-1)/[1+2S(n-1)]
S2=1/3
S3=1/5
S4=1/7
Sn=1/(2n-1)
2.bn=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1)-1/(2n+1)]
2Tn=(1-1/3)+(1/3-1/5)+.+[1/(2n-1)-1/(2n+1)]
=1-1/(2n+1)
=2n/(2n+1)
Tn= n/(2n+1)