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#1
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¾ÔÊÙ¨¹ìäÁèä´é¤èÐ §§ ÇèÒ«Ô¡ÁèÒ¡¡ÓÅѧ i Áѹ·ÓÂѧä§
$$\sum_{n = 1}^{m} 3\bullet 4^i = 4^{m+1} - 4$$
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#2
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ËÁÒ¶֧Íѹ¹ÕéËÃ×Íà»ÅèÒ¤ÃѺ
$\displaystyle\sum_{n = 1}^{m} 3\bullet 4^n = 4^{m+1} - 4$
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site:mathcenter.net ¤Ó¤é¹ |
#3
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¤Ô´ÍÍ¡áÅéǤèÐ ¢Íº¤Ø³ÁÒ¡
¾ÔÁ¾ì¾Ô´¤èÐ µÃ§ n=1 ¤×Í i = 1 ¹Ð¤Ð ËÃ×Í ¨Ðä»á¡éà»ç¹ 4^n ¡çä´é¤èÐ |
#4
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$\sum_{n = 1}^{m} 3*4^n = 3 \sum_{n=1}^{m} = 3[\frac{4(4^m-1)}{3}] (¼ÅºÇ¡Í¹Ø¡ÃÁàâҤ³Ôµ)
= 4^{m+1} -4 $ 09 ¡Ã¡®Ò¤Á 2012 23:01 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Euler-Fermat |
#5
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¼ÁàʹÍãËéÍÕ¡ÇÔ¸Õ $3=4-1$
$3\cdot 4^i=4^{i+1}-4^i$ «Ö觡ç¤×Í telescopic sum ¹Ñè¹àͧ
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"ªÑèÇâÁ§Ë¹éÒµéͧ´Õ¡ÇèÒà´ÔÁ!" |
#6
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#7
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ÊØ´æ ¤ÒÃÇÐ 10 ¨Í¡¤ÃѺ·èÒ¹
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¤³ÔµÈÒʵÃì ¤×Í ÀÒÉÒÊÒ¡Å ¤³ÔµÈÒʵÃì ¤×Í ¤ÇÒÁÊǧÒÁ ¤³ÔµÈÒʵÃì ¤×Í ¤ÇÒÁ¨ÃÔ§ µÔ´µÒÁªÁ¤ÅÔ»ÇÕ´ÕâÍä´é·Õèhttp://www.youtube.com/user/poperKM |
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