题目内容
| 1 |
| 2 |
| 5 |
| 6 |
| 11 |
| 12 |
| 19 |
| 20 |
| 29 |
| 30 |
| 9701 |
| 9702 |
| 9899 |
| 9900 |
分析:通过分析发现,式中的加数都可表示为1-
r的形式,如
=1-
,所以原式=(1-
)+(1-
)+(1-
)+…+(1-
)=1×99-(
+
+
+…+
),由于括号中的分数都为
的形式,所以可根据分数巧算公式
=
-
进行巧算.
| 1 |
| n |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 9900 |
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 9900 |
| 1 |
| n((n+1) |
| 1 |
| n((n+1) |
| 1 |
| n |
| 1 |
| n+1 |
解答:解:
+
+
+
+
+…+
+
=(1-
)+(1-
)+(1-
)+…+(1-
)
=1×99-(
+
+
+…+
),
=99-(
+
+
…+
),
=99-(1-
+
-
+
-
+…+
-
),
=99-(1-
),
=99-
,
=98
.
| 1 |
| 2 |
| 5 |
| 6 |
| 11 |
| 12 |
| 19 |
| 20 |
| 29 |
| 30 |
| 9701 |
| 9702 |
| 9899 |
| 9900 |
=(1-
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 9900 |
=1×99-(
| 1 |
| 2 |
| 1 |
| 6 |
| 1 |
| 12 |
| 1 |
| 9900 |
=99-(
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 99×100 |
=99-(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 99 |
| 1 |
| 100 |
=99-(1-
| 1 |
| 100 |
=99-
| 99 |
| 100 |
=98
| 1 |
| 100 |
点评:在认真分析式中数据的基础上发现式中数据特点及内在联系是完成本题的关键.
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