题目内容
计算：(
+
+…+
+…+
+
)-
(
+
+…+
+…+
)=
．
| 1 |
| 1×2007 |
| 1 |
| 2×2006 |
| 1 |
| n×(2008-n) |
| 1 |
| 2006×2 |
| 1 |
| 2007×1 |
| 2007 |
| 2008 |
| 1 |
| 1×2006 |
| 1 |
| 2×2005 |
| 1 |
| n×(2007-n) |
| 1 |
| 2006×1 |
| 1 |
| 2015028 |
| 1 |
| 2015028 |
分析：此题根据前边括号与后边括号的特点，将它们进行变形，使之变化成乘法分配律的应用形式，即可进行巧算．
解答：解：(
+
+…+
+…+
+
)-
(
+
+…+
+…+
)
=
×（
+
+…+
）-
×
×（
+
+…+
），
=
×【（1+
）+（
+
）+（
+
）+…+（
+1）】-
×【（1+
）+（
+
）+（
+
）+…+（
+1）】，
=
×（1+
+
+
+…+
）-
×（1+
+
+
…+
），
=
×
，
=
．
故答案为：
| 1 |
| 1×2007 |
| 1 |
| 2×2006 |
| 1 |
| n×(2008-n) |
| 1 |
| 2006×2 |
| 1 |
| 2007×1 |
| 2007 |
| 2008 |
| 1 |
| 1×2006 |
| 1 |
| 2×2005 |
| 1 |
| n×(2007-n) |
| 1 |
| 2006×1 |
=
| 1 |
| 2008 |
| 2008 |
| 1×2007 |
| 2008 |
| 2×2006 |
| 2008 |
| 2007×1 |
| 2007 |
| 2008 |
| 1 |
| 2007 |
| 2007 |
| 1×2006 |
| 2007 |
| 2×2005 |
| 2007 |
| 2006×1 |
=
| 1 |
| 2008 |
| 1 |
| 2007 |
| 1 |
| 2 |
| 1 |
| 2006 |
| 1 |
| 3 |
| 1 |
| 2005 |
| 1 |
| 2007 |
| 1 |
| 2008 |
| 1 |
| 2006 |
| 1 |
| 2 |
| 1 |
| 2005 |
| 1 |
| 3 |
| 1 |
| 2004 |
| 1 |
| 2006 |
=
| 2 |
| 2008 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2007 |
| 2 |
| 2008 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 2006 |
=
| 1 |
| 1004 |
| 1 |
| 2007 |
=
| 1 |
| 2015028 |
故答案为：
| 1 |
| 2015028 |
点评：此题考查了分数巧算的方法．
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