Question

已知数列{an}中,a1=1,且an,sn,sn一1/2成等比数列,求Sn

Answer 1

1）由Sn^2=an(Sn-1/2),an=Sn-Sn-1(n≥2)得
Sn^2=(Sn-Sn-1)(Sn- 1/2)
即2Sn-1Sn=Sn-1-Sn.
由题意知Sn-1Sn≠0，上式两边同除以Sn-1Sn得1/Sn - 1/Sn-1=2
∴{1/Sn}是首项为1,公差为2的等差数列，
∴1/Sn=1+2(n-1)=2n-1,
Sn=1/(2n-1)(n≥2),
∵S1=1适合Sn=1/(2n-1),∴Sn=1/(2n-1)

an,Sn,Sn-1/2成等比数列
an(Sn-1/2)=Sn^2
a2(S2-1/2)=S2^2
a2(a2+1/2)=(a2+1)^2
a2=-2/3
a3(S3-1/2)=S3^2
a3(a3-1/6)=(a3+1/3)^2
a3=-2/33

[Sn-S(n-1)](Sn-1/2)=Sn^2
-(1/2)Sn-S(n-1)Sn+(1/2)S(n-1)=0
-(1/2)Sn+(1/2)S(n-1)=S(n-1)Sn
1/Sn-1/S(n-1)=-2
1/Sn=(1/S2)+(-2)(n-2)
=[1/(1-2/3)]+(-2)(n-2)
=3+(-2)(n-2)
=-2n+7
Sn=1/(-2n+7)
S(n-1)=1/(-2n+5)
an=Sn-S(n-1)=1/(-2n+7)-1/(-2n+5)
an=1/(-2n+7)-1/(-2n+5);

Answer 2

sn一1/2是S(n-1)/2????????