a1=1,n≥2时  ,sn2=an[sn-(1/2)]求sn

a1=1,n≥2时  ,sn2=an[sn-(1/2)]求sn
a1=1,n≥2时  ,sn2=an[sn-(1/2)]
求sn

a1=1,n≥2时  ,sn2=an[sn-(1/2)]求sn
由题意：（a1+a2)^2=a2*[a1+a2-1/2)]
即(1+a2)^2=a2(a2+1/2)
解得:a2=-2/3
n>=2时,因为(Sn)^2=an*(Sn -1/2)
故 (Sn)^2=(Sn-Sn-1) (Sn -1/2)
化简得到：Sn-1Sn=1/2* Sn-1 -1/2*Sn
即1/Sn -1/Sn-1 =2
故故数列{1/Sn}是以1/S2为首项,2为公差的等差数列
因为S2=a1+a2=1+(-2/3)=1/3
故1/S2 =3,1/Sn =3+(n-2)*2=2n-1
所以Sn=1/(2n-1)(n>=2)
又n=1符合上式.故Sn=1/(2n-1)

解:n>=2时，Sn^2=(Sn-S(n-1))(Sn-1/2)
化简得0=-SnS(n-1)-(1/2)Sn+(1/2)S(n-1)。即1/Sn-1/S(n-1)=2
所以1/Sn是以2为公差的等差数列。首项1/S1=1/A1=1
所以1/Sn=2n-1
Sn=1/(2n-1)