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[数列] 求项数的最大值

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isee Posted at 2013-9-27 10:20:52 |Read mode
数列$a_1,a_2,\cdots,a_n$中任意连续三项的和是正数,任意的连续五项的和是负数,求$n$的最大值。

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睡神 Posted at 2013-9-27 10:35:06
回复 1# isee
这题非一般的面善,就是它老是把我给遗忘了…
除了不懂,就是装懂

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爪机专用 Posted at 2013-9-27 11:11:22
如果有7项,由a1+a2+2a3+a4+a5>0, a1+a2+a3+a4+a5<0 知 a3>0, 同理 a4, a5>0, 而 a1+a2+a3>0, 故 a1+a2+a3+a4+a5>0, 矛盾。
再举一6项的例子 4, -8, 5, 5, -8, 4.

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地狱的死灵 Posted at 2013-9-27 11:14:37
确实是老题了……
最多只能6项,
比如:4,-6,4,4,-7,4.
假设有第7项,
构造矩阵:
$\begin{array}{*{20}c}
   {a_1 } & {a_2 } & {a_3 } & {a_4 } & {a_5 }  \\
   {a_2 } & {a_3 } & {a_4 } & {a_5 } & {a_6 }  \\
   {a_3 } & {a_4 } & {a_5 } & {a_6 } & {a_7 }  \\
\end{array}$
每行相加都是负数,
每列相加都是正数,
这是不可能的。

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 Author| isee Posted at 2013-9-27 12:33:40
PFPF

完全正确啊,秒了秒了

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kuing Posted at 2013-9-27 14:06:56
嫌老可以考虑推广……

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其妙 Posted at 2013-9-27 14:07:52
回复  isee
这题非一般的面善,就是它老是把我给遗忘了…
睡神 发表于 2013-9-27 10:35

哪里面善?是面目可憎!只可惜被众人揭穿了是一只纸老虎而已

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睡神 Posted at 2013-9-27 16:51:24
回复 7# 其妙
K神解释一下“面善”…
除了不懂,就是装懂

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