a(n+1)=（an) +2
這是個等差數(shù)列,公差為2
首項為：（S3)÷3-2=1
n為整數(shù)
an=1+2(n-1)=2n-1
sn=(1+2n-1)*n/2=n^2
二次函數(shù)為：
f(x)=n^2*x^2+(2n-1)x-2
與x軸的兩個交點的橫坐標,為方程n^2*x^2+(2n-1)x-2=0的兩根
方程有兩根,則n^2≠0,n≠0
△=(2n-1)^2+8n^2＞0
4n^2-4n+1+8n^2＞0
12n^2-4n＞0
3n^2-n＞0
n(3n-1)＞0
n＞1/3或n＜0
n≥1或n≤-1
n^2＞0,所以函數(shù)圖像開口向上
f(-3)＞0
f(-1)＜0
f(2)＞0
f(-3):
9n^2-3(2n-1)-2＞0
9n^2-6n+1＞0
(3n-1)^2＞0
n≠1/3
f(-1):
n^2-(2n-1)-2＜0
n^2-2n-1＜0
(n-1)^2＜2
-√2＜n-1＜√2
1-√2＜n＜1+√2
0≤n≤2
f(2):
4n^2+2(2n-1)-2＞0
4n^2+4n-4＞0
n^2+n-1＞0
(n+0.5)^2＞1.25
n+0.5＞√5/2或n+0.5＜-√5/2
n＞（√5-1)/2或n＜-（√5+1)/2
n≥1或n≤-2
綜上,可得：
n=1,2