问题
填空题
等差数列{an}中，Sn是其前n项和，a1=-2008，
|
答案
∵在等差数列中S2n-1=（2n-1）an，
∴
=a1004，S2007 2007
=a1003，S2005 2005
又∵
-S2007 2007
=2S2005 2005
∴d=2，又由a1=-2008
∴Sn=a1n+
d=n2-n-2008n，n(n-1) 2
∴S2008=-2008
故答案为：-2008
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