问题 填空题
数列{an}满足an+an+1=
1
2
(n∈N*),a1=-
1
2
,Sn是{an}的前n项和,则S2011=______.
答案

an+an+1=

1
2
(n∈N*),a1=-
1
2

S2011=a1+(a2+a3)+(a4+a5)+…+(a2010+a2011

=-

1
2
+
1
2
+…+
1
2

=-

1
2
+
1
2
×1005

=502

故答案为:502

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