题目内容
| 1 |
| 1×2 |
| 1 |
| 1 |
| 1 |
| 2 |
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| n+1 |
根据以上规律计算:
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| n(n+1) |
分析:根据题意,找出规律,把分数拆成两个分数差,然后再进一步解答.
解答:解:
=
-
;
=
-
;
=
-
;
+
+
+
+…+
=1-
+
-
+
-
+…+
-
=1-
=
.
故答案为:
,
,
,
,
,
.
| 1 |
| 2×3 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3×4 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
| 1 |
| 1×2 |
| 1 |
| 2×3 |
| 1 |
| 3×4 |
| 1 |
| 4×5 |
| 1 |
| n(n+1) |
=1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n+1 |
=1-
| 1 |
| n+1 |
=
| n |
| n+1 |
故答案为:
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n |
| 1 |
| n+1 |
点评:考查了有理数的混合运算解答这种题型,要充分利用给出的已知条件找出规律,然后利用这个规律解答后面的问题就简单了.
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