题目内容
已知
+(ab-2)2=0,则
+
+…+
的值为______.
| a-1 |
| 1 |
| ab |
| 1 |
| (a+1)(b+1) |
| 1 |
| (a+2008)(b+2008) |
根据非负数性质可知a-1=0且ab-2=0
解得a=1 b=2
则原式=
+
+…+
裂项得1-
+
-
+
-
+… +
-
=1-
=
;
故答案为
解得a=1 b=2
则原式=
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 2009×2010 |
裂项得1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2009 |
| 1 |
| 2010 |
| 1 |
| 2010 |
| 2009 |
| 2010 |
故答案为
| 2009 |
| 2010 |
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