问题
解答题
计算:
|
答案
∵
=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)