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#1
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ÁÒªèÇ¡ѹ¤Ô´¤ÃѺ ⨷Âì¢Í§ Putnam
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$a_n = \frac{(2^3 - 1)(3^3 - 1)(4^3 - 1).....(n^3 - 1)}{(2^3 + 1)(3^3 + 1)(4^3 + 1).....(n^3 + 1)}$
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#2
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µÍº $2$ ÃÖà»ÅèÒ äÁèá¹è㨠- -
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#3
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µÍº $\frac{2}{3}$ Åͧ¤Ô´´ÙãËÁè 555+
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ä´é $\frac{2}{3}$ àËÁ×͹¡Ñ¹¤ÃѺ
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¤ÃѺ ¼ÁµÑ´¼Ô´¹Ô´Ë¹èÍ -.-
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#6
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ÍéÒ§ÍÔ§:
¨Ò¡ $x^3 - 1 = (x-1)(x^2 + x + 1)$ àÃÒ¡çáºè§â¨·ÂìÍÍ¡à»ç¹ÊͧÊèǹ¤Ù³¡Ñ¹¤×Í $a_n = (\frac{(2 - 1)(3 - 1)(4 - 1).....(n - 1)}{(2 + 1)(3 + 1)(4 + 1).....(n + 1)}) \bullet (\frac{(2^2+2+1)(3^2+3+1)(4^2+4+1)...(n^2+n+1)}{(2^2-2+1)(3^2-3+1)(4^2-4+1)...(n^2-n+1)}) $ $a_n = (\frac{(1)(2)(3)...(n-1)}{(3)(4)(5)...(n+1)}) \bullet (\frac{(7)(13)(21)...(n^2+n+1)}{(3)(7)(13)...(n^2-n+1)} )$ ; áÅéÇÁѹ¨ÐµÑ´¡Ñ¹ËÁ´ àËÅ×Íá¤è¹Õé $$a_n = \frac{2(n^2+n+1)}{3(n)(n+1)} $$ áÅéÇà·¤ÅÔÁÔµ $n\rightarrow\infty $ ä´é $\frac{2}{3}$ ¹èФÃèÐ |
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ËÑÇ¢éÍ | ¼ÙéµÑé§ËÑÇ¢éÍ | Ëéͧ | ¤ÓµÍº | ¢éͤÇÒÁÅèÒÊØ´ |
¼Å¡ÒÃÊͺá¢è§¢Ñ¹ Putnam »Õ 2009 | SolitudE | §Ò¹ËÃ×Í¢èÒǤÃÒǤ³ÔµÈÒʵÃì·ÑèÇä» | 6 | 10 ¾ÄÉÀÒ¤Á 2010 10:24 |
IMO ¡Ñº Putnam ÃÒ¡ÒÃä˹ÂÒ¡¡Çèҡѹ¤ÃѺ | zzz010307 | ¤³ÔµÈÒʵÃìÍØ´ÁÈÖ¡ÉÒ | 1 | 23 ÊÔ§ËÒ¤Á 2006 16:50 |
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