已知数列{an}的前N项和为Sn与an满足：an,Sn,Sn-1/2（n大于2）成等比数列,且a1=1,求Sn

因为an,Sn,Sn-1/2成等比数列所以an*Sn-1/2=Sn^2因为an=Sn-Sn-1所以（Sn-Sn-1）*Sn-1=2Sn^2等式两边同除以Sn^2得[1-(Sn-1/Sn)](Sn-1/Sn)=2令t=Sn-1/Sn则（1-t）t=2 得t=2或t=-1所以Sn是等比数列 且S1=1 公比为-1或2所...