计算1+112+122+1+122+132+1+132+142…1+120092+120102=______．

题型：填空题难度：一般来源：不详

计算

1+ 1 12 + 1 22

+

1+ 1 22 + 1 32

+

1+ 1 32 + 1 42

…

1+ 1 20092 + 1 20102

=______．

答案

∵n（n+1）

1+ 1 n2 + 1 (n+1)2

=

n2(n+1)2+2n(n+1)+1

=n（n+1）+1，
∴

1+ 1 n2 + 1 (1+n)2

=

n(n+1)+1

n(n+1)

=1+

1

n

-

1

n+1

，
∴原式=(1+

1

1

-

1

2

)+(1+

1

2

-

1

3

)+(1+

1

3

-

1

4

)+…+(1+

1

2009

-

1

2010

)
=2009+1-

1

2010

=2009

2009

2010

．
故答案为2009

2009

2010

．