题目内容

等差数列{an}的前n项和为Sn,若
S2010
2010
-
S2008
2008
=4
,则
lim
n→∞
Sn
n2
=______.
∵{an}为等差数列,设首项为a1,公差为d,
∴sn=na1+
n(n-1)
2
d,
sn
n
=a1+
n-1
2
d,
S2010
2010
-
S2008
2008
=(a1+
2010-1
2
×d)-(a1+
2008-1
2
×d)=d=4,
∴sn=2n2+(a1-2)n,
lim
n→∞
Sn
n2
=
lim
n→∞
(2+
a1-2
n
)
=2,
故答案为:2.
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