题目内容
运用裂项法计算:
+
+
+
+
.
| 1 |
| 3 |
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
分析:因为
=
×(1-
),
=
×(
-
),
=
×(
-
),
=
×(
-
),
=
×(
-
),利用此规律可以把原式进行转化,然后利用分配律提出
,即可解决.
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 15 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 35 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 63 |
| 1 |
| 2 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 99 |
| 1 |
| 2 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 2 |
解答:解:
+
+
+
+
=
×(1-
)+
×(
-
)+
×(
-
)+
×(
-
)+
×(
-
)
=
×(1-
+
-
+
-
+
-
+
-
)
=
×(1-
)
=
×
=
.
| 1 |
| 3 |
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 2 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 2 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 2 |
| 1 |
| 9 |
| 1 |
| 11 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 11 |
=
| 1 |
| 2 |
| 1 |
| 11 |
=
| 1 |
| 2 |
| 10 |
| 11 |
=
| 5 |
| 11 |
点评:此题考查了分数的简便计算,解决的方法是利用裂项法,关键是找到裂项的规律.
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