Question

数列{an}中,a1=1,前n项和Sn满足Sn^2=an(Sn-1/2),(1)求Sn的表达式 (2)设bn=Sn/(2n+1),求bn前n项和Tn.

第一问请详细回答!!

Answer 1

由Sn^2
=an(Sn-1/2)
=[Sn-S(n-1)](Sn-1/2)
=Sn^2-[1/2+S(n-1)]Sn+1/2*S(n-1)
化简为[1/2+S(n-1)]Sn=1/2*S(n-1)
两边同除Sn*S(n-1)/2化为
1/S(n-1)+2=1/Sn
则数列{1/Sn}是首项为1/a1=1,公差为2的等差数列
所以1/Sn=1+(n-1)*2=2n-1
则Sn=1/(2n-1)

bn=Sn/(2n+1)
=1/[(2n-1)(2n+1)]
=1/2*[1/(2n-1)-1/(2n+1)]
则Tn=1/2*[1/1-1/3+1/3-1/5+……+1/(2n-1)-(2n+1)]
=1/2*[1-(2n+1)]
=n/(2n+1)