| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
∴f（n+1）=
| 1 |
| n+2 |
| 1 |
| n+3 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
∴f（n+1）-f（n）=（
| 1 |
| n+2 |
| 1 |
| n+3 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
=
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
=
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
故答案为：
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
| 1 |
| n+2 |
| 1 |
| n+3 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+2 |
| 1 |
| n+3 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2n |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| n+1 |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |
| 1 |
| 2n+1 |
| 1 |
| 2n+2 |