题目内容
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:根据数字特点,前一个分数都是后一个分数的2倍,因此可把每个分数拆成两个分数相减的形式,然后通过加减相互抵消,求得结果.
解答:
解:
-
-
-
-
-
-
=
-(
+
+
+
+
+
)
=
-(
-
+
-
+
-
+
-
+
-
+
-
)
=
-(
-
)
=
-
+
=
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
=
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
=
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 128 |
| 1 |
| 256 |
=
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 256 |
=
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 256 |
=
| 1 |
| 256 |
点评:此题采用了裂项消项法,先进行分数裂项,然后通过加减相互抵消,求出结果.
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