各项均为正数的数列{an}的前n项和为Sn,Sn=1/4*an^2+1/2*an

an=Sn-Sn-1=1/4an^2+1/2an-(1/4an-1^2-1/2an-1)=1/4[(an^2-an-1^2)+2(an-an-1)] 从而(an^2-an-1^2)+2(an-an-1)=4an (an^2-an-1^2)-2(an+an-1)=0 (an+an-1)(an-an-1-2)=0 由{an}为正数数列得an>0,an-1>0,即an+an-1>0 从而an-an-1-2=0即an-