| an+1+an−1 |
| an+1−an+1 |
∴（1-n）an+1+（1+n）an=1+n
∴an+1＝
| n+1 |
| n−1 |
| n+1 |
| n−1 |
| 1 |
| n−1 |
∴
| an+1 |
| n+1 |
| 1 |
| n−1 |
| 1 |
| n−1 |
| n |
| n−1 |
| an |
| n |
| n |
| n−1 |
| 1 |
| n |
∴
| an+1 |
| n+1 |
| n |
| n−1 |
| an |
| n |
∴
| 1 |
| n |
| an+1 |
| n+1 |
| 1 |
| n−1 |
| an |
| n |
∴{
| 1 |
| n−1 |
| an |
| n |
而
| 1 |
| n−1 |
| an |
| n |
| 1 |
| 2−1 |
| a2 |
| 2 |
an=[2（n-1）+1]n=2n2-n
当n=1时，
| 6+a1−1 |
| 6−a1+1 |
故答案为：2n2-n