题目内容
已知
+(ab-2)2=0,求
+
+
+…+
的值.
| a-1 |
| 1 |
| ab |
| 1 |
| (a+1)(b+1) |
| 1 |
| (a+2)(b+2) |
| 1 |
| (a+2005)(b+2005) |
∵
+(ab-2)2=0,
∴a-1=0且ab-2=0,
解得:a=1,b=2,
∴
+
+…+
=1+
+
+…+
=1+
-
+
-
+…+
-
=1+
-
=
.
| a-1 |
∴a-1=0且ab-2=0,
解得:a=1,b=2,
∴
| 1 |
| a |
| 1 |
| (a+1)(b+1) |
| 1 |
| (a+2005)(b+2005) |
=1+
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 2006×2007 |
=1+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2006 |
| 1 |
| 2007 |
=1+
| 1 |
| 2 |
| 1 |
| 2007 |
=
| 6019 |
| 4014 |
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| A、AB=2AM | ||
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