题目内容
我们道:| 1 |
| 1×2 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n(n+1) |
利用上面的规律计算:
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 2007×2009 |
分析:先找到规律,
=
-
,而
+
+
+…+
=
(1-
+
-
+…+
-
)
再利用这个规律将它展开,计算即可.
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 2007×2009 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2007 |
| 1 |
| 2009 |
再利用这个规律将它展开,计算即可.
解答:解:∵
=
-
,
∴
+
+
+…+
=
(1-
+
-
+…+
-
)
=
(1-
)
=
.
故答案为
-
,
.
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
∴
| 1 |
| 1×3 |
| 1 |
| 3×5 |
| 1 |
| 5×7 |
| 1 |
| 2007×2009 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2007 |
| 1 |
| 2009 |
=
| 1 |
| 2 |
| 1 |
| 2009 |
=
| 1004 |
| 2009 |
故答案为
| 1 |
| n |
| 1 |
| n+1 |
| 1004 |
| 2009 |
点评:本题是一道规律型的题目,考查了有理数的混合运算.
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