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ÊÁѤÃÊÁÒªÔ¡ | ¤ÙèÁ×Í¡ÒÃãªé | ÃÒª×èÍÊÁÒªÔ¡ | »¯Ô·Ô¹ | ¢éͤÇÒÁÇѹ¹Õé | ¤é¹ËÒ |
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à¤Ã×èͧÁ×ͧ͢ËÑÇ¢éÍ | ¤é¹ËÒã¹ËÑÇ¢é͹Õé |
#1
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͹ءÃÁ(ÍÕ¡áÅéÇÍèÒ¤ÃѺ T_T)
1.¨§ËҼźǡ n ¾¨¹ìáá¢Í§ $1\bullet (2+3)+2\bullet(3+4)+3\bullet(4+5)+5\bullet(6+7)+...$
(¼ÁäÁèá¹èã¨ÇèÒ⨷ÂìÁѹ¼Ô´ËÃ×Íà»ÅèÒÍèÒ¤ÃѺ ¾¨¹ì·Õè4 Áѹá»Å¡æ àÅÂäÁèÃÙé¨Ð¤Ô´Âѧä§ÍèÒ¤ÃѺ) 2.¹ÓºÍŷç¡ÅÁÁҨѴàÃÕ§à»ç¹ªÑé¹æà»ç¹ÃÙ»¾ÔÃÐÁÔ´°Ò¹ÊÕèàËÅÕèÂÁ¨ÑµØÃÑÊ â´ÂáµèÅЪÑé¹àÃÕ§à»ç¹ÊÕèàËÅÕèÂÁ¨ÑµØÃÑÊ«é͹¡Ñ¹ ¶éÒªÑé¹°Ò¹ÅèÒ§ÊØ´ ÁÕÅÙ¡ºÍ¡´éÒ¹ÅÐ20ÅÙ¡ áÅЪÑ鹺¹ÊØ´ ÁÕÅÙ¡ºÍÅ 1 ÅÙ¡ ¨§ËҨӹǹÅÙ¡ºÍÅ·Ñé§ËÁ´ã¹¡Í§¹Õé 3.¨§ËҼźǡ n ¾¨¹ìáá¢Í§Í¹Ø¡ÃÁ $ 1^2+(1^2+2^2)+(1^2+2^2+3^2)+...$ 4.¨§ËÒ¤èҢͧ $(\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...)^5 $ 5.ãËé $Sn=1^3+2^3+3^3+...+n^3$ ¨§ËÒ¤èҢͧ $\frac{1}{\sqrt{S1}}+\frac{1}{\sqrt{S2}} +\frac{1}{\sqrt{S3}}+...+\frac{1}{\sqrt{Sn}}$ 6.¨§ËҼźǡ n ¾¨¹ìáá¢Í§Í¹Ø¡ÃÁ $\frac{3}{1\bullet4}+\frac{5}{4\bullet9}+\frac{7}{9\bullet16}+...+\frac{2n+1}{n^2(n+1)^2}+...$ 7.¨§ËҼźǡ n ¾¨¹ìáá¢Í§Í¹Ø¡ÃÁ $\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+...+\frac{2n+1}{2^n}+...$
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ªÕÇÔµ¤×Í¡ÒõèÍÊÙé »ÑËÒ¤×Í¡ÒÃàÃÕ¹ÃÙé ÈѵÃÙ¤×ͤÃ٢ͧàÃÒ 23 ¡Ã¡®Ò¤Á 2011 21:58 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 5 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Kira Yamato à˵ؼÅ: à¾ÔèÁ⨷Âì ⨷Âì¢éÍ7 á¡é⨷Âì¢éÍ5 |
#2
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$1^2 +2 ^2 +3^2 +...+ 20 ^2 = \frac{1}{6} (n)(n+1)(2 n+1) = \frac{1}{6} (20)(20+1)(2 \cdot 20 +1) = 2870$
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) |
#3
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ÍéÒ§ÍÔ§:
$ \frac{1}{1+2+3+...+n} = \frac{2}{n(n+1)} = 2 (\frac{1}{n(n+1)}) = 2 (\frac{1}{n} - \frac{1}{n+1})$ $(\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...\frac{1}{1+2+3+4+...n}) = 2(1- \frac{1}{n+1}) = 2(\frac{n}{n+1})$ $(\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...\frac{1}{1+2+3+4+...n})^5 = 2^5(1- \frac{1}{n+1})^5 = 32(\frac{n}{n+1})^5$
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) |
#4
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ÍéÒ§ÍÔ§:
$\because \frac{5}{4\bullet9} = \frac{1}{2^2} - \frac{1}{3^2}$ . . . $\because \frac{2n+1}{ n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$ ºÇ¡ææææ¡Ñ¹Å§ÁÒ ¨ÐµÑ´æææææ¡Ñ¹ àËÅ×Í $1 - \frac{1}{(n+1)^2} \ $ ËÃ×Í $\frac{n(n+2)}{(n+1)^2}$
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) 27 ÁԶعÒ¹ 2011 16:48 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ banker à˵ؼÅ: á¡é¤Ó¼Ô´ |
#5
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¢Íº¤Ø³ÁÒ¡¤ÃѺ
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ªÕÇÔµ¤×Í¡ÒõèÍÊÙé »ÑËÒ¤×Í¡ÒÃàÃÕ¹ÃÙé ÈѵÃÙ¤×ͤÃ٢ͧàÃÒ |
#6
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ÍéÒ§ÍÔ§:
$\frac{3}{2}+\frac{5}{4}+\frac{7}{6}+...+\frac{2n+1}{2n}+...$ ËÃ×ÍÇèÒ¨Ãԧ澨¹ì·Õè$n$ ã¹$\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+...+\frac{2n+1}{2n}+...$.....¨Ðà»ç¹$\dfrac{2n+1}{2^n}$ ¼ÁÇèÒÅاBanker¡ÓÅѧ·Ó¢éÍ7...ÍÂÙèá¹èàÅ ¼Áá»Å§ $\frac{2n+1}{2n}=1+\frac{1}{2n}$ ¨Ðä´é͹ءÃÁ¨Ò¡$\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+...+\frac{2n+1}{2n}+...$ à»ç¹ $1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+...+1+\frac{1}{2n}+...$ $=n+\left(\,\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}\right) $ ¼ÁËÒ¼ÅÃÇÁ¢Í§$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2n}$...ÂѧäÁèÍÍ¡ ¶éÒà»ç¹$a_n=\dfrac{2n+1}{2^n}$ $S=\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+...+\frac{2n+1}{2^n}$........(1) $\frac{1}{2} S=\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+...+\frac{2n+1}{2^{n+1}}$........(2) (1)-(2);$\frac{1}{2} S=\frac{3}{2}-\frac{2n+1}{2^{n+1}}+\left(\,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^{n-1}}\right) $ $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^{n-1}}=1-\frac{1}{2^{n-1}}$ $\frac{1}{2} S=\frac{3}{2}-\frac{2n+1}{2^{n+1}}+\left(\,1-\frac{1}{2^{n-1}}\right) $ $S=5-\left(\,\frac{2n+1}{2^n}+\frac{1}{2^{n-2}}\right) $ $S=5-\frac{1}{2^n}\left(\,2n+5\right) $
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"¶éÒàÃÒÅéÁºèÍÂæ ã¹·ÕèÊØ´àÃÒ¨ÐÃÙéÇèÒ¶éÒ¨ÐÅéÁ ÅéÁ·èÒä˹¨Ðà¨çº¹éÍ·ÕèÊØ´ áÅÐÃÙéÍÕ¡ÇèÒµèÍä»·ÓÂѧ䧨ÐäÁèãËéÅéÁÍÕ¡ ´Ñ§¹Ñ鹨§ÍÂèÒ¡ÅÑÇ·Õè¨ÐÅéÁ"...ÍÒ¨ÒÃÂìÍӹǠ¢¹Ñ¹ä·Â ¤ÃÑé§áá㹪ÕÇÔµ·ÕèÊͺ¤³ÔµÊÁÒ¤Á¤³ÔµÈÒʵÃìàÁ×èÍ»Õ2533...¼Áä´éá¤è24¤Ðá¹¹(¨Ò¡ÃéͤÐá¹¹) |
#7
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ÍéÒ§ÍÔ§:
⨷Âì¹èÒ¨Ðà»ç¹ $\frac{3}{2}+\frac{5}{4}+\frac{7}{8}+...+\frac{2n+1}{\color{red}{2^n}}+...$ áµè¡çÂѧä»äÁè¶Ù¡
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) |
#8
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àÍèÍ ¢Íâ·É¤ÃѺ ¾¨¹ì·Õèn à»ç¹ $\frac{2n+1}{2^n}$ áËÅФÃѺ ¢Íº¤Ø³ÁÒ¡¤ÃѺ
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ªÕÇÔµ¤×Í¡ÒõèÍÊÙé »ÑËÒ¤×Í¡ÒÃàÃÕ¹ÃÙé ÈѵÃÙ¤×ͤÃ٢ͧàÃÒ 27 ÁԶعÒ¹ 2011 21:36 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 3 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Kira Yamato |
#9
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ÍéÒ§ÍÔ§:
¤Ø³¡ÔµµÔà´Ò㨶١ ¢éÍ 1 ¡Ñº ¢éÍ 5 à´ÕëÂǤ׹¹ÕéÁÒ·ÓãËé¤ÃѺ ÍéÍ ... Êͧ¢é͹Õé ⨷Âì¼Ô´´éÇ ¨¢¡·. ªèÇÂá¡é´éǤÃѺ
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) 27 ÁԶعÒ¹ 2011 18:34 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ banker à˵ؼÅ: ÁҺ͡ÇèÒ Êͧ¢é͹ÕéàËç¹·Ò§áÅéÇ |
#10
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¢éÍ 1
¶éÒÁÑè¹ã¨ÇèÒ¶Ù¡ ¡çÍÒ¨¨Ðà»ç¹ Fibonacci number ¡çä´é¤ÃѺ |
#11
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ÍéÒ§ÍÔ§:
1.¨§ËҼźǡ n ¾¨¹ìáá¢Í§ $1\bullet (2+3)+2\bullet(3+4)+3\bullet(4+5) $$+4\bullet(5+6)$$+5\bullet(6+7)+...$ ¨Ðä´é $a_n $ $= n[(n+1)+(n+2)]$ $ = n[(n+1)+(n+1)+1]$ $= n[2(n+1)+1]$ $ = 2n(n+1) +n$ $ = 2[\color{blue}{n(n+1)}] +n$ áµè $1\bullet2 + 2\bullet3 + 3\bullet4 + ... + n\bullet(n+1) = \frac{n(n+1)(n+2)}{3} $ $\therefore \ \ S_n =2[\frac{n(n+1)(n+2)}{3}] +n^2 = \frac{2}{3}n(n+1)(n+2) +n^2$ ÁÒ¤Ô´ãËÁè ä´éÍÕ¡Êٵà $S_n = n(n+1)(\frac{4n+11}{6})$ Çѹ¹ÕéªÑ¡ÁÖ¹áÅéÇ ¾ÃØ觹ÕéÁÒ·ÓµèͤÃѺ 9:49 28/6/2554 àªéÒ¹ÕéÁÒµèͤÃѺ ¡àÅÔ¡¢éͤÇÒÁ¢éÒ§µé¹ àÍÒãËÁè¤ÃѺ $1\bullet (2+3)+2\bullet(3+4)+3\bullet(4+5) $$+4\bullet(5+6)$$+5\bullet(6+7)+...$ $ = (1\bullet2 + 1 \bullet3) + (2 \bullet 3 + 2 \bullet 4) + (3 \bullet 4 + 3 \bullet 5) + (4\bullet 5 + 4 \bullet6) + ...$ $ = (1\bullet2 + 2 \bullet 3 + 3 \bullet 4 + 4\bullet 5 + ... + n(n+1)) + (1 \bullet3 + 2 \bullet 4 +3 \bullet 5 +4 \bullet6 + ... ) $ $ = \dfrac{n(n+1)(n+2)}{3} + (1 \bullet3 + 2 \bullet 4 +3 \bullet 5 +4 \bullet6 + ... +n(n+2)) $ $ = \dfrac{n(n+1)(n+2)}{3} + (1 \bullet3 )+ (2 \bullet 4) + (3 \bullet 5) + (4 \bullet6) + ... +n(n+2) $ $ = \dfrac{n(n+1)(n+2)}{3} + (1 \bullet(1+2) )+ (2 \bullet (2+2)) + (3 \bullet (3+2)) + (4 \bullet (4+2)) + ... +n(n+2) $ $ = \dfrac{n(n+1)(n+2)}{3} + (1^2 +1\bullet2) )+ (2^2 + (2\bullet2)) + (3^2 + (3\bullet2)) + (4^2 +(4\bullet2)) + ... +n^2+n\bullet2) $ $ = \dfrac{n(n+1)(n+2)}{3} + (1^2 +1\bullet2) )+ (2^2 + (2\bullet2)) + (3^2 + (3\bullet2) ) + (4^2 +(4\bullet2)) + ... +n^2+n\bullet2) $ $ = \dfrac{n(n+1)(n+2)}{3} + (1^2+2^2+3^2+4^2+...+n^2) + 2(1+2+3+4+...+n)$ $ = \dfrac{n(n+1)(n+2)}{3} + (\frac{n}{6}(n+1)(2n+1)) + 2(\frac{n(n+1)}{2})$ $ = n(n+1) [(\frac{(n+2}{3}) +(\frac{(2n+1}{6}) +1]$ $ S_n = n(n+1)\frac{4n+11}{6}$ Final àÊÕ·Õ
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) 28 ÁԶعÒ¹ 2011 09:53 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ banker |
#12
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ÍéÒ§ÍÔ§:
$\sqrt{Sn} = \frac{n(n+1)}{2} $ $ \frac{1}{\sqrt{Sn} } = \frac{1 }{\frac{n(n+1)}{2}}$ $ \frac{n}{\sqrt{Sn} } = \frac{2n }{n(n+1)} = \frac{2}{n+1} = 2n(\frac{1}{n} - \frac{1}{n+1})$ ¶éÒà»ç¹ $\frac{1}{\sqrt{S1}}+\frac{1}{\sqrt{S2}} +\frac{1}{\sqrt{S3}}+...+\frac{1}{\sqrt{Sn}}$ ¨Ðä´é ¼ÅÃÇÁà·èҡѺ $2(1- \frac{1}{n+1})$ ¶éÒ⨷Âìà»ç¹ ¨§ËÒ¤èҢͧ $\frac{1}{\sqrt{S1}}+\frac{2}{\sqrt{S2}} +\frac{3}{\sqrt{S3}}+...+\frac{n}{\sqrt{Sn}}$ $ \frac{n}{\sqrt{Sn} } = \frac{2n }{n(n+1)} = \frac{2}{n+1}$ $ \frac{1}{\sqrt{S1} } = \frac{2}{1+1} = \frac{1}{1}$ $ \frac{2}{\sqrt{S2} } = \frac{2}{2+1} = \frac{2}{3}$ $ \frac{3}{\sqrt{S3} } = \frac{2}{3+1} = \frac{2}{4}$ $ \frac{4}{\sqrt{S4} } = \frac{2}{4+1} = \frac{2}{5}$ . . $ \frac{n}{\sqrt{Sn} } = \frac{2}{n+1} $ ¼ÅÃÇÁ¢Í§ $\frac{1}{\sqrt{S1}}+\frac{2}{\sqrt{S2}} +\frac{3}{\sqrt{S3}}+...+\frac{n}{\sqrt{Sn}}$ $ = 1+2(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} +...+ \frac{1}{n+1})$ 仵èÍÂѧä§ËÇèÒ ?
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ÁÒËÒ¤ÇÒÁÃÙéäÇéµÔÇËÅÒ¹ áµèËÅÒ¹äÁèàÍÒàÅ¢áÅéÇ à¢éÒÁÒ·ÓàÅ¢àÍÒÁѹÍÂèÒ§à´ÕÂÇ ¤ÇÒÁÃÙéà»ç¹ÊÔè§à´ÕÂÇ·ÕèÂÔè§ãËé ÂÔè§ÁÕÁÒ¡ ÃÙéÍÐäÃäÁèÊÙé ÃÙé¨Ñ¡¾Í (¡àÇ鹤ÇÒÁÃÙé äÁèµéͧ¾Í¡çä´é ËÒäÇéÁÒ¡æáËÅдÕ) (áµè¡çÍÂèÒãËéÁÒ¡¨¹·èÇÁËÑÇ àÍÒµÑÇäÁèÃÍ´) 28 ÁԶعÒ¹ 2011 15:22 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 3 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ banker |
#13
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¢éÍ 5 ⨷Âì¨ÃÔ§à»ç¹áºº¹Õé¤ÃѺ
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#14
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¢éÍ 3 ¾ÔÁ¾ì¼Ô´ãªèäËÁ¤ÃѺ àÅ¢ªÕé¡ÓÅѧ¾¨¹ì $3^3$ µéͧà»ç¹ $3^2$ ¶éÒãªè¼Á¢ÍµÍºáºº¢éÒ§ÅèÒ§¤ÃѺ
$1^2+(1^2+2^2)+(1^2+2^2+3^2)+...+(1^2+2^2+...+n^2)=\frac{n(n+1)^2(2n+1)}{6}-(\frac{n(n+1)}{2})^2$
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"ªÑèÇâÁ§Ë¹éÒµéͧ´Õ¡ÇèÒà´ÔÁ!" |
#15
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¤ÃѺ ¢Íâ·É¤ÃѺ
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ªÕÇÔµ¤×Í¡ÒõèÍÊÙé »ÑËÒ¤×Í¡ÒÃàÃÕ¹ÃÙé ÈѵÃÙ¤×ͤÃ٢ͧàÃÒ |
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