题目内容
观察下列各式:1 |
1×3 |
1 |
2 |
1 |
3 |
1 |
3×5 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
5×7 |
1 |
2 |
1 |
5 |
1 |
7 |
1 |
1×3 |
1 |
3×5 |
1 |
5×7 |
1 |
(2n-1)×(2n+1) |
分析:根据已知条件,将每一个分数分解成两个负数,寻找抵消规律求解.
解答:解:原式=
(1-
+
-
+
-
+…+
-
)
=
(1-
)
=
.
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
5 |
1 |
7 |
1 |
2n-1 |
1 |
2n+1 |
=
1 |
2 |
1 |
2n+1 |
=
n |
2n+1 |
点评:本题考查的是分式的加减法,根据题意找出规律是解答此题的关键.
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