题目内容
探究题:已知:1-1 |
2 |
1 |
1×2 |
1 |
2 |
1 |
3 |
1 |
2×3 |
1 |
3 |
1 |
4 |
1 |
3×4 |
(1)观察上面式子的规律,请你猜测并写出第五项;
(2)上述的规律用一般的式子可以表示为:
1 |
n |
1 |
n+1 |
1 |
n(n+1) |
(3)请直接用上述的结果计算
1 |
2×3 |
1 |
3×4 |
1 |
4×5 |
1 |
x(x+1) |
分析:(1)先总结出通项公式
-
=
(n为正整数);
(2)先通分,再进行分式的加减.
(3)利用上面的规律,将原式分解成分数和的形式,再进行加减即可.
1 |
n |
1 |
n+1 |
1 |
n(n+1) |
(2)先通分,再进行分式的加减.
(3)利用上面的规律,将原式分解成分数和的形式,再进行加减即可.
解答:解:(1)∵1-
=
,
-
=
,
-
=
…
∴第五项:
-
=
;
(2)左边=
-
,
=
-
,
=
,
=
,
∵左边=右边,
∴
-
=
(n为正整数);
(3)原式=
-
+
-
+
-
+…+
-
,
=
-
,
=
,
=
.
1 |
2 |
1 |
1×2 |
1 |
2 |
1 |
3 |
1 |
2×3 |
1 |
3 |
1 |
4 |
1 |
3×4 |
∴第五项:
1 |
5 |
1 |
6 |
1 |
5×6 |
(2)左边=
1 |
n |
1 |
n+1 |
=
n+1 |
n(n+1) |
n |
n(n+1) |
=
n+1-n |
n(n+1) |
=
1 |
n(n+1) |
∵左边=右边,
∴
1 |
n |
1 |
n+1 |
1 |
n(n+1) |
(3)原式=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
4 |
1 |
4 |
1 |
5 |
1 |
x |
1 |
x+1 |
=
1 |
2 |
1 |
x+1 |
=
x+1-2 |
2(x+1) |
=
x-1 |
2(x+1) |
点评:本题是一个找规律的题目,考查了分式的加减,找出通项公式是解题的关键.

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