已知数列{An}的前n项和Sn满足Sn=1/4(An+1)的平方,且An>0.求A1、A2的值

Question

已知数列{An}的前n项和Sn满足Sn=1/4(An+1)的平方,且An>0.求A1、A2的值

Answer 1

a1=S1=1/4(a1+1)^2,
4 a1=(a1+1)^2,
(a1-1)^2=0,
a1=1.
S2=1/4(a2+1)^2,
a1+a2=1/4(a2+1)^2,
1+a2=1/4(a2+1)^2,
4+4a2=(a2+1)^2,
a2^2-2a2-3=0,
(a2+1)(a2-3)=0,
a2=3.
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