Question

设等差数列{a_n}的前n项和为S_n，已知a₁=-2011，

S2009

2009

-

S2007

2007

=2，则S₂₀₁₁=______．

Answer 1

由

S2009

2009

-

S2007

2007

=2可得  

a1+ a2009

2

-

a1+a2007

2

=2，∴公差d=2．
则S₂₀₁₁ =2011×a₁+

2011(2011-1)

2

×d=-2011，
故答案为：-2011．