问题 填空题
观察下列各等式:0<a<1,
1
2×3
=
1
2
-
1
3
1
3×5
=
1
2
(
1
3
-
1
5
)
,…根据你发现的规律,计算:
1
1×4
+
1
4×7
+
1
7×10
+…+
1
(3n-2)(3n+1)
=______(n为正整数).
答案

1
2×3
=
1
2
-
1
3
1
3×5
=
1
2
1
3
-
1
5
),

所以

1
1×4
=
1
3
(1-
1
4
),
1
4×7
=
1
3
1
4
-
1
7
),…,
1
(3n-2)(3n+1)
=
1
3
1
3n-2
-
1
3n+1
),

∴原式=

1
3
(1-
1
4
+
1
4
-
1
7
+…+
1
3n-2
-
1
3n+1
)=
1
3
(1-
1
3n+1
)=
n
3n+1

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