∵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