问题
填空题
计算
|
答案
∵n（n+1）
=1+
+1 n2 1 (n+1)2
=n（n+1）+1，n2(n+1)2+2n(n+1)+1
∴
=1+
+1 n2 1 (1+n)2
=1+n(n+1)+1 n(n+1)
-1 n
，1 n+1
∴原式=(1+
-1 1
)+(1+1 2
-1 2
)+(1+1 3
-1 3
)+…+(1+1 4
-1 2009
) 1 2010
=2009+1-
=20091 2010
．2009 2010
故答案为2009
．2009 2010