题目内容
由
=
,
-
=
;
=
,
-
=
;
=
,
-
=
;…总结出规律:
= .
并利用这一规律,可知
+
+…+
= .
| 1 |
| 1×2 |
| 1 |
| 2 |
| 1 |
| 1 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 6 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 6 |
| 1 |
| 3×4 |
| 1 |
| 12 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 12 |
| 1 |
| n(n+1) |
并利用这一规律,可知
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| n(n+1) |
考点:有理数的混合运算
专题:规律型
分析:原式利用拆项法总结得到
=
-
,所求式子变形后抵消即可得到结果.
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
解答:解:根据题意得:规律为
=
-
;
则原式=1-
+
-
+…+
-
=1-
=
.
故答案为:
-
;
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
则原式=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n+1 |
| n |
| n+1 |
故答案为:
| 1 |
| n |
| 1 |
| n+1 |
| n |
| n+1 |
点评:此题考查了有理数的混合运算,熟练运用拆项法是解本题的关键.
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