题目内容

若n为正整数,观察下列各式:
1
1×3
=
1
2
(1-
1
3
)
;②
1
3×5
=
1
2
(
1
3
-
1
5
)
;③
1
5×7
=
1
2
(
1
5
-
1
7
)

根据观察计算并填空:
(1)
1
1×3
+
1
3×5
+
1
5×7
=
3
7
3
7

(2)
1
1×3
+
1
3×5
+
1
5×7
+
+
1
(2n-1)(2n+1)
=
n
2n+1
n
2n+1
分析:根据题意可知:
1
(2n-1)(2n+1)
=
1
2
1
2n-1
-
1
2n+1
),据此展开,再正负抵消可求值.
解答:解:(1)
1
1×3
+
1
3×5
+
1
5×7
=
1
2
(1-
1
3
+
1
3
-
1
5
+
1
5
-
1
7
)=
1
2
×
6
7
=
3
7

(2)原式=
1
2
(1-
1
3
+
1
3
-
1
5
+…+
1
2n-1
-
1
2n+1
)=
1
2
×
2n+1-1
2n+1
=
n
2n+1

故答案是
3
7
n
2n+1
点评:本题考查了分式的加减法,解题的关键是找出运算规律,即
1
(2n-1)(2n+1)
=
1
2
1
2n-1
-
1
2n+1
).
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