题目：
已知|a-1|+| b-2 |
| 1 |
| ab |
| 1 |
| (a+1)(b+1) |
| 1 |
| (a+2)(b+2) |
| 1 |
| (a+2012)(b+2012) |
| 2013 |
| 2014 |
| 2013 |
| 2014 |
答案
| 2013 |
| 2014 |
解：∵|a-1|+
| b-2 |
∴a=1，b=2，
∴原式=
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 2013×2014 |
∵
| 1 |
| 1×2 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
∴
| 1 |
| n×(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
∴原式=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2013 |
| 1 |
| 2014 |
=1-
| 1 |
| 2014 |
=
| 2013 |
| 2014 |
故答案为：
| 2013 |
| 2014 |
