题目内容

计算:
1
2
1+
1
2
+
1
3
(1+
1
2
)(1+
1
3
)
+…+
1
99
(1+
1
2
)(1+
1
3
)…(1+
1
99
)
分析:根据题意,可以根据前几个找出规律,再根据分数的拆项进行巧算.
解答:解:
1
2
1+
1
2
=
1
2
3
2
=
1
2
×
2
3
=
2
2×3

1
3
( 1+
1
2
)(1+
1
3
=
1
3
3
2
×
4
3
=
1
3
×
2×3
3×4
=
2
3×4


1
99
(1+
1
2
)(1+
1
3
)…(1+
1
99
)  
=
1
99
3
2
×
4
3
×…×
100
99
=
1
99
×
2×3×…×99
3×4×…×100
=
2
99×100

所以,
1
2
1+
1
2
+
1
3
(1+
1
2
)(1+
1
3
)
+…+
1
99
(1+
1
2
)(1+
1
3
)…(1+
1
99
)

=
2
2×3
+
2
3×4
+…+
2
99×100

=2×(
1
2×3
+
1
3×4
+…+
1
99×100
),
=2×(
1
2
-
1
3
+
1
3
-
1
4
+…+
1
99
-
1
100
),
=2×(
1
2
-
1
100
),
=1-
1
50

=
49
50
点评:根据题目,找出规律,按照分数的拆项计算即可.
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