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#1
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@@@ ͹ءÃÁ 1*2 + 2*3 + 3*4 + ... + 8*9 = ? @@@
ÍÂÒ¡·ÃÒºÇèÒ Í¹Ø¡ÃÁ 1*2 + 2*3 + 3*4 + ... + 8*9 = ? àËÃͤÃѺ ¾ÍÁÕã¤ÃÃÙéºéÒ§äËÁ ¢Í ÇÔ¸Õ·Ó ´éǹФÃѺ
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#2
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¢é͹Õé¨Ó¹Ç¹¾¨¹ìäÁèàÂÍÐ ´Ñ§¹Ñé¹ÇÔ¸Õ·Õè§èÒ·ÕèÊØ´¤×ͨѺ¤Ù³¡Ñ¹áÅéǺǡ·ÕÅо¨¹ì¤ÃѺ
áµè¶éÒ¨Ðà¢Õ¹ãËéÁÕËÅÑ¡¡ÒÃ˹èÍ¡ç¤×ÍÁͧÇèÒ $a_n = n(n+1)$ ´Ñ§¹Ñ鹨Ðä´é $a_i = i(i+1)$ áÅéÇ $$S_n = \sum_{i = 1}^{n}i(i+1) = \frac{n(n+1)(n+2)}{3} $$Note. $$\sum_{i = 1}^{n}i = \frac{n(n+1)}{2} $$ $$\sum_{i = 1}^{n}i(i+1) = \frac{n(n+1)(n+2)}{3} $$ $$\sum_{i = 1}^{n}i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4} $$ ÊÓËÃѺÇÔ¸Õ¾×é¹°Ò¹¤×Í $$S_n = \sum_{i = 1}^{n}i(i+1) = \sum_{i = 1}^{n}i^2 + \sum_{i = 1}^{n}i = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}$$ ã¹·Õè¹Õéµéͧ¡ÒÃËÒ $S_8$ ¡çá·¹ n = 8 ŧã¹ÊÙµÃ
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The Lost Emic <<-- ˹ѧÊ×Íà©Å¢éÍÊͺÃдѺ»ÃжÁ¹Ò¹ÒªÒµÔ EMIC ¤ÃÑ駷Õè 1 - ¤ÃÑ駷Õè 8 ªØ´ÊØ´·éÒ ËŧÁÒ 06 ÁԶعÒ¹ 2010 18:49 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ gon |
#3
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ÅͧãªéÍØ»¹Ñ¾ÔÊÙ¨¹ì¹Ð¤ÃѺ(Íѹ¹Õé¾ÔÁ¾ìµÑé§áµè»Õ¡è͹æ¤ÃѺ)
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