问题 解答题
观察下列成立的式子:
1
1×2
=1-
1
2
1
2×3
=
1
2
-
1
3
1
3×4
=
1
3
-
1
4
1
4×5
=
1
4
-
1
5

(1)则第n个算式为______=______.
(2)如果将上列式子左右相加得:
1
1×2
+
1
2×3
+
1
3×4
+
1
4×5
=1-
1
2
+
1
2
-
1
3
+
1
3
-
1
4
+
1
5
=1-
1
5
=
4
5
根据这个结果,则请你直接写出下列式子的结果:①
1
1×2
+
1
2×3
+
1
3×4
+…+
1
2008×2009
=______;
1
1×2
+
1
2×3
+
1
3×4
+…+
1
n×(n+1)
=______;
(3)探究并计算
1
2
+
1
6
+
1
12
+
1
20
+
1
30
+
1
42
+
1
56
+
1
72
+
1
90
答案

(1)观察给出的式子可知:第n个算式为

1
n(n+1)
=
1
n
-
1
n+1
(n为正整数);

(2)①

1
1×2
+
1
2×3
+
1
3×4
+…+
1
2008×2009

=1-

1
2
+
1
2
-
1
3
+
1
3
-
1
4
+…+
1
2008
-
1
2009

=1-

1
2009

=

2008
2009

1
1×2
+
1
2×3
+
1
3×4
+…+
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

(3)

1
2
+
1
6
+
1
12
+
1
20
+
1
30
+
1
42
+
1
56
+
1
72
+
1
90

=

1
1×2
+
1
2×3
+
1
3×4
+
1
4×5
+
1
5×6
+
1
6×7
+
1
7×8
+
1
8×9
+
1
9×10

=1-

1
2
+
1
2
-
1
3
+
1
3
-
1
4
+…+
1
9
-
1
10

=1-

1
10

=

9
10

故答案为:

2008
2009
n
n+1

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