Question

在数列｛an｝中，已知a1=1,当n≥2时，an,Sn,Sn-1/2成等比数列，求｛an｝的通项公式

Answer 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);