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$\sum_{n = 1}^{\infty}$ $\frac{6}{(n)(n+1)(2n+1)}$ ËÒÂÑ§ä§ áÅéÇ¡çÁÕ¤èÒà·èÒäà ¾Í´ÕÍÂÒ¡ÃÙé¹Ð¤ÃѺÇèÒËÒÂÑ§ä§ ÅͧËÒàͧ´ÙáÅéÇ¡çäÁèä´éÊÑ¡·Õ 04 Á¡ÃÒ¤Á 2009 14:15 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 4 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Aphenisol |
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#3
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¼Á¤è͹¢éÒ§ÃÕº æ à¢Õ¹ ÍÒ¨¨ÐÁÕ¼Ô´¾ÅÒ´ºéÒ§ áèµèâ´ÂÃÇÁ¹èҨж١µéͧ
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#4
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#5
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Åͧ¤Ó¹Ç³´Ùã¹ mathematica Áѹä´éà»ç¹ $6(3-2\ln{4})$ ÍèФÃѺ
à·èÒ·Õè´Ùã¹ mathematica ÁѹàÃÔèÁ¼Ô´ä»µÃ§ºÃ÷Ѵ·ÕèÊÅѺàÍÒµÑÇÍÔ¹·Ôà¡Ãµ¡ÑºµÑÇ summation ÍèФÃѺ
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äÁè·ÃÒºÇèҤس beginner01 ãªé¿Ñ§¡ìªÑ¹ÍÐäÃ㹡ÒÃËÒ¤ÃѺ ·ÓäÁ¼Áä´éẺ¹ÕéÍÐ
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#7
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Sum[1/(Sum[k^2, {k, 1, n}]), {n, 1, Infinity}]
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$\dfrac{6}{n(n+1)(2n+1)}=\dfrac{24}{2n(2n+1)}-\dfrac{6}{n(n+1)}$ ´Ñ§¹Ñé¹ $\displaystyle{\sum_{n=1}^{\infty}\dfrac{6}{n(n+1)(2n+1)}=\sum_{n=1}^{\infty}\dfrac{24}{2n(2n+1)}-\sum_{n=1}^{\infty}\dfrac{6}{n(n+1)}}$ µèÍ仾ԨÒóÒ͹ءÃÁÊͧµÑǹÕé $\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)}}=1$ $\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{2n(2n+1)}}=1-\ln{2}$ ͹ءÃÁµÑÇááËÒä´éäÁèÂÒ¡â´Âãªéà·¤¹Ô¤ telescoping sum ¤ÃѺ ÊèǹµÑÇ·ÕèÊͧËÒẺ¹Õé $\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{2n(2n+1)}}=\dfrac{1}{2\cdot 3}+\dfrac{1}{4\cdot 5}+\dfrac{1}{6\cdot 7}+\cdots$ $~~~~~~~~~~~~~~~~~~~=1-\Big(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\cdots\Big)$ $~~~~~~~~~~~~~~~~~~~=1-\ln{2}$ ´Ñ§¹Ñé¹ $\displaystyle{\sum_{n=1}^{\infty}\dfrac{6}{n(n+1)(2n+1)}=24(1-\ln{2})-6(1)}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 18-24\ln{2}$
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site:mathcenter.net ¤Ó¤é¹ |
#9
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¼ÁÅͧµÃǨÊͺ¤ÓµÍº¨Ò¡ÇÔ¸Õ¤Ô´¢Í§Í¹Ø¡ÃÁ·Õè nooonuii áÊ´§äÇé ÃÙéáÅéÇÇèÒÇÔ¸Õ¹Õé¨ÐãªéÍÂèÒ§äö֧¨Ðä´é¼ÅÅѾ¸ì¶Ù¡µéͧ áµèÊÒà˵طÕè¼Ô´¹Ñé¹ ¼ÁÂѧäÁèÃÙé¤ÃѺ
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#10
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à»ç¹ä»ä´éäËÁ¤ÃѺ·ÕèÇèÒÁѹ¢Öé¹ÍÂÙè¡Ñº¡ÒèѴÃÙ»¶Ö§¨Ðä´é¤ÓµÍº·Õè¶Ù¡ÍÍ¡ÁÒ
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#11
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¼Á¤Ô´ÇèÒÇԸբͧ¾Õè Gon ÍÒ¨¨ÐÁÕ»ÑËҵ͹ÊÅѺ·Õè integral ¡Ñº summation ¤ÃѺ
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site:mathcenter.net ¤Ó¤é¹ |
#12
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ÍéÒ§ÍÔ§:
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#13
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¼Áãªé Mathematica 7 ¤ÃѺ
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#14
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