题目内容
等差数列{an}的前n项和为Sn,若
-
=4,则
=______.
| S2010 |
| 2010 |
| S2008 |
| 2008 |
| lim |
| n→∞ |
| Sn |
| n2 |
∵{an}为等差数列,设首项为a1,公差为d,
∴sn=na1+
d,
∴
=a1+
d,
∴
-
=(a1+
×d)-(a1+
×d)=d=4,
∴sn=2n2+(a1-2)n,
∴
=
(2+
)=2,
故答案为:2.
∴sn=na1+
| n(n-1) |
| 2 |
∴
| sn |
| n |
| n-1 |
| 2 |
∴
| S2010 |
| 2010 |
| S2008 |
| 2008 |
| 2010-1 |
| 2 |
| 2008-1 |
| 2 |
∴sn=2n2+(a1-2)n,
∴
| lim |
| n→∞ |
| Sn |
| n2 |
| lim |
| n→∞ |
| a1-2 |
| n |
故答案为:2.
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设等差数列{an}的前n项和为Sn,则a5+a6>0是S8≥S2的( )
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