Question

数列{an}中，a1=1,当n>=2时，其前项和Sn满足Sn^2=an(Sn-1/2)

1.求an
2.令bn=Sn/(2n+1)
求bn的前n项Tn

Answer 1

1)Sn^2=an*(Sn-0.5)=(Sn-S(n-1))*(Sn-0.5)=Sn^2-0.5Sn-Sn*S(n-1)+0.5S(n-1)
=>S(n-1)-Sn=2Sn*S(n-1) =>1/Sn-1/S(n-1)=2
可得{1/Sn}为等差数列1/Sn=2n-1 =>Sn=1/2n-1
当n>=2时 an=Sn-S(n-1)=-2/(2n-1)(2n-3)
n=1 a1=1

2)裂项相消
bn=Sn/(2n+1)=0.5*[1/(2n-1)-1/(2n+1)]
Tn=0.5*[1-1/(2n+1)]=n/(2n+1)

Answer 2

题目看的不是很明白，Sn^2=an(Sn-1/2) 这个有歧义，是Sn^2=an{(Sn)-0.5}吗

追问

是的