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

Answer 2

Sn-Sn-1=1/4(an+an-1)(an-an-1)+1/2(an-an-1)=an n>=2
1/4(an+an-1)(an-an-1)=1/2(an+an-1) an+an-1>0
an-an-1=2 a1=2 a2=4 an=a2+2(n-2)=2n 当n=1也符合
所以 an=2n

Answer 3

a1=S1=1/4(a1)^2+1/2(a1)
∵an>0
∴a1=2
n≥2时
an=Sn-S(n-1)
=1/4(an)^2-1/4(an-1)^2+1/2(an)-1/2(an-1)
即（an+an-1）(an-an-1)=2(an+an-1)
∴an-an-1=2
∴{an}为等差数列
∴an=2n