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计算:
1
x(x+1)
+
1
(x+1)(x+2)
+
1
(x+2)(x+3)
+…+
1
(x+2008)(x+2009)
人气:242 ℃ 时间:2019-09-06 17:06:35
解答
1
x(x+1)
=
1
x
-
1
x+1
1
(x+1)(x+2)
=
1
x+1
-
1
x+2

1
(x+n)(x+n+1)
=
1
x+n
-
1
x+n+1

∴原式=
1
x
-
1
x+1
+
1
x+1
-
1
x+2
+
1
x+2
-
1
x+3
+…+
1
x+2008
-
1
x+2009

=
1
x
-
1
x+2009

=
2009
x(x+2009)
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