Question

已知各项均为正数的数列an的前n项和,an^2+an=2sn,求证:Sn

Answer 1

证：
n=1时,a1²晌漏+a1=2S1=2a1
a1²-a1=0
a1(a1-1)=0
a1=0(舍去明谨碰)或a1=1
n≥2时,
Sn=(an²+an)/2 S(n-1)=[a(n-1)²+a(n-1)]/2
an=Sn-S(n-1)=(an²+an)/2 -[a(n-1)²+a(n-1)]/2
an²-a(n-1)²-an-a(n-1)=0
[an+a(n-1)][an-a(n-1)]-[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-1]=0
an+a(n-1)恒>0,因此只有an-a(n-1)-1=0
an-a(n-1)=1,为定值.数列{an}是以1为首项,1为公比的等比数列.
an=1+n-1=n
Sn=1+2+...+n=n(n+1)/2
Sn-[an²+a(n+1)²激谈]/4
=n(n+1)/2-[n²+(n+1)²]/4
=-1/4