如题：一直数列｛an｝的前n项和Sn与an满足：an,Sn,Sn-1/2（n大于等于2）成等比数列,且a1=1,

（1）由Sn^2=an(Sn-1/2),an=Sn-Sn-1(n≥2)得Sn^2=(Sn-Sn-1)(Sn- 1/2)即2Sn-1Sn=Sn-1-Sn.由题意知Sn-1Sn≠0,上式两边同除以Sn-1Sn得1/Sn - 1/Sn-1=2∴{1/Sn}是首项为1,公差为2的等差数列,∴1/Sn=1+2(n-1)=2n-1,Sn=1/(...