各项均为正数的数列{an}的前n和为Sn,Sn=1/4an^2+1/2an（n为正整数）,求an

Question

各项均为正数的数列{an}的前n和为Sn,Sn=1/4an^2+1/2an（n为正整数）,求an

Answer 1

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-an-1=2
由Sn=1/4*an^2+1/2*an得a1=1/4a1^2+1/2a1,(a1>0)
即a1=2.
所以数列{an}为首项为2,公差为2的公差数列.从而
an=2+2(n-1)=2n