已知等差数列[An],Sn=[（An+1）/2]^2,求An的通项公式

∵等差数列{a[n]},S[n]=[(a[n]+1)/2]^2
∴4S[n]=a[n]^2+2a[n]+1
∵4S[n+1]=a[n+1]^2+2a[n+1]+1
∴将上面两式相减,得：
4a[n+1]=a[n+1]^2-a[n]^2+2a[n+1]-2a[n]
2(a[n+1]+a[n])=(a[n+1]+a[n])(a[n+1]-a[n])
如果a[n+1]+a[n]=0,即：a[n+1]=-a[n]
∵a[1]=S[1]=[(a[1]+1)/2]^2
∴a[1]=1
∴{a[n]}是首项为1,公比为-1的等比数列
这与{a[n]}是等差数列的题设条件相矛盾
∴a[n+1]+a[n]≠0
∴a[n+1]-a[n]=2,即公差为2
∵a[1]=1
∴a[n]=1+2(n-1)=2n-1

sn=((an+1)/2)²
s(n-1)=((a(n-1)+1)/2)²
sn-s(n-1)=an
an=((an+1)/2)²-((a(n-1)+1)/2)²
4an=a²n+2an+1-a²(n-1)-2a(n-1)-1
a²n-2an+1-(a²(n-1))+2a(n-1)+1)=0
(an-1)²-(a(n-1)+1)²=0
(an-a(n-1)-2)(an+a(n-1))=0
显然an+a(n-1)不等于零

sn=((an+1)/2)²
s(n-1)=((a(n-1)+1)/2)²
sn-s(n-1)=an
an=((an+1)/2)²-((a(n-1)+1)/2)²
4an=a²n+2an+1-a²(n-1)-2a(n-1)-1
a²n-2an+1-(a²(n-1))+2a(n-1...

等差数列[An]，
设An=pn+q
Sn=[（An+1）/2]^2=[（pn+q+1）/2]^2=[（pn）^2+2p(q+1)n+(q+1)^2]/4
等差数列前n项和没常数项
所以q+1=0 q=-1
An=pn-1 Sn=n(n+1)p/2-n=pn^2/2+n(p-2)/2
Sn=（pn）^2/4
pn^2/2+n(p-2)/2=（pn）^2/4
p=2
An=2n-1