Question

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

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

额 我才初中....

Answer 3

Sn-sn-1=an=1/4*[an^2-an-1^2]+1/2*[an-an-1]
2[an+an-1]=[an+an-1]*[an+an-1]
an+an-1=2 S1=a1=1/4*a1^2+1/2*a1 a1=2
an=a1+(n-1)d=2n