Question

已知数列{an}的各项均为正数，前n项的和Sn=4分之（an+1）^2

求{an}的通项公式

Answer 1

a1=S1=1/4(a1+1)^2
a1=1
sn=1/4(an+1)^2
s(n-1)=1/4[a(n-1)+1]^2
相减
4an=(an+1)^2-[a(n-1)+1]
(an-1)^2-[a(n-1)+1]=0
[an+a(n-1)][an-a(n-1)-2]=0
∵an＞0
∴an-a(n-1)=2
∴{An}是等差数列
an=1+(n-1)*2
=2n-1

Answer 2

an=sn-s(n-1)=1/4(an+1)^2-1/4[a(n-1)+1]^2
4an=(an+1)^2-[a(n-1)+1]^2
(an-1)^2-[a(n-1)+1]^2=0
[an+a(n-1)][an-a(n-1)-2]=0
又an＞0
∴an-a(n-1)-2=0
故{An}是等差数列
a1=S1=1/4(a1+1)^2
a1=1
an=1+(n-1)*2=2n-1