Sn^2=an×(Sn -1/2)=(Sn-Sn-1)×(Sn -1/2)
整理,得
Sn-1-Sn=2SnSn-1
等式兩邊同除以SnSn-1
1/Sn-1/Sn-1=2,為定值.
1/S1=1/a1=1/1=1
數(shù)列{1/Sn}是以1為首項(xiàng),2為公差的等差數(shù)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時,a1=1/(2-1)-1/(2-3)=1+1=2,與已知不符.
綜上,得數(shù)列{an}的通項(xiàng)公式為
an= 1 (n=1)
1/(2n-1)-1/(2n-3) (n≥2)