(1)由 2an=Sn*S(n-1),an=Sn-S(n-1)
则:2[Sn-S(n-1)]=Sn*S(n-1)
2Sn-2S(n-1)=Sn*S(n-1)
两边同时除以Sn*S(n-1)
2/S(n-1)-2/Sn=1
1/Sn-1/S(n-1)=-1/2
则{1/Sn}为等差数列,公差为-1/2
(2)由于{1/Sn}为等差数列
则:1/Sn=1/S1+(n-1)*(-1/2)
=1/a1+(1-n)/2
=(5-2n)/6
则: Sn=6/(5-3n)
则当n>=2时,
an=Sn-S(n-1)
=6/(5-3n)-6/(8-3n)
=18/[(3n-8)(3n-5)]
又a1=18/[(3-8)(3-5)]=9/5
则: an=18/[(3n-8)(3n-5)] (n>=2)
=3 (n=1)