Question

已知数列{a_n}，满足a1=1，

1

an+1

=

1

an

+1，S_n是数列{a_na_n+1}的前n项和，则S₂₀₁₁=______．

Answer 1

∵a1=1，

1

an+1

=

1

an

+1，
∴{

1

an

}是以1为首项以1为公差的等差数列
根据等差数列的通项公式可得，

1

an

=n即an=

1

n

∴anan+1=

1

n(n+1)

=

1

n

-

1

n+1

∴S2011=1-

1

2

+

1

2

-

1

3

+…+

1

2011

-

1

2012

=1-

1

2012

=

2011

2012

故答案为：

2011

2012