题目内容
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 63 |
| 64 |
| 63 |
| 64 |
分析:根据题意,令m=
+
+
+
+
+
,两边同时乘2,得:2m=1+
+
+
+
+
;然后两式想减即可求出答案.
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
解答:解:令m=
+
+
+
+
+
;两边同时乘2,得:2m=1+
+
+
+
+
;
两式相减,得:
2m-m,
=(1+
+
+
+
+
)-(
+
+
+
+
+
),
=1+
+
+
+
+
-
-
-
-
-
-
,
=1-
,
=
.
故答案为:
.
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
两式相减,得:
2m-m,
=(1+
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
=1+
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
=1-
| 1 |
| 64 |
=
| 63 |
| 64 |
故答案为:
| 63 |
| 64 |
点评:根据题意,前一个分数是后一个分数的2倍,这是解决本题的关键,根据这一特点,令原式乘上2,然后再进一步解答即可.
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