Sn^2=an×(Sn -1/2)=(Sn-Sn-1)×(Sn -1/2)
整理,得
Sn-1-Sn=2SnSn-1
等式两边同除以SnSn-1
1/Sn-1/Sn-1=2,为定值.
1/S1=1/a1=1/1=1
数列{1/Sn}是以1为首项,2为公差的等差数列.
1/Sn=1+2(n-1)=2n-1
Sn=1/(2n-1)
Sn-1=1/[2(n-1)-1]=1/(2n-3)
an=Sn-Sn-1=1/(2n-1)-1/(2n-3)
n=1时,a1=1/(2-1)-1/(2-3)=1+1=2,与已知不符.
综上,得数列{an}的通项公式为
an= 1 (n=1)
1/(2n-1)-1/(2n-3) (n≥2)