#1
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͹ءÃÁ
1) ¼ÅºÇ¡ $n$ ¾¨¹ìáá¢Í§ $1 + \frac{1^2 + 2^2}{1 + 2} + \frac{1^2 + 2^2 + 3^2}{1 + 2 + 3} +...$
2) $\sum_{k = 1}^{\infty} \frac{(2k-1)}{k(k+1)(k+2)}$ ¨§ËÒ $S_n$ áÅÐ $S_\infty$ 3) ¡Ó˹´ $a_n$ à»ç¹¾¨¹ì·ÑèÇ仢ͧÍѹ´ÑºàâҤ³Ôµ â´Â¼ÅºÇ¡Í¹Ñ¹µì¢Í§ $a_n$ ÁÕ¤èÒà»ç¹ $S_1$ áÅмźǡ͹ѹµì¢Í§ $(a_n)^2$ ÁÕ¤èÒà»ç¹ $S_2$ ¨§ËÒ¤èÒÍѵÃÒÊèǹÃèÇÁ¢Í§Íѹ´ÑºàâҤ³Ôµ $a_n$ ¹Ñé¹ã¹ÃÙ»¢Í§ $S_1$ áÅÐ $S_2$ |
#2
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ÍéÒ§ÍÔ§:
1. $\sum_{i = 1}^{n} i = \frac{n(n+1)}{2}$ $\sum_{i = 1}^{n} i(i+1) = \frac{n(n+1)(n+2)}{3}$ $\sum_{i = 1}^{n} i(i+1)(i+2) = \frac{n(n+1)(n+2)(n+3)}{4}$ ... 2. $\frac{1}{ab} = \frac{1}{b-a}(\frac{1}{a} - \frac{1}{b})$ $\frac{1}{abc} = \frac{1}{c-a}(\frac{1}{ab} - \frac{1}{bc})$ 3. $S_1 = \frac{a_1}{1-r}, S_2 = \frac{a_1^2}{1-r^2}$ |
#3
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ÍéÒ§ÍÔ§:
¢éÍ 2 ·Óá¡ÊèǹáÅéǤèÐ áµè¾ÍàÈÉ·ÕèäÁèà·èҡѹ (1, 3, 5, ....) àŵԴÍÂÙè¤èÐ ¢éÍ 3 ä´éẺ¹ÕéáÅéǤèÐ áµè·ÓµèÍãËé r ã¹ÃÙ» $S_1$ áÅÐ $S_2$ äÁèä´é¤èÐ |
#4
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ÍéÒ§ÍÔ§:
¡è͹¤ÃѺ ¨Ò¡¹Ñé¹áµèÅÐÍѹãªé·Õè Hint äÇé 3. $a_1^2 = (1-r)^2S_1^2 = (1-r^2)S_2$ áÅéÇᡵÑÇ»ÃСͺáÅéǵѴ¡Ñ¹ ¨Ò¡¹Ñ鹤ٳ¡ÃШÒÂáÅéÇÂéÒ¢éÒ§¨Ñ´ÃÙ»¡ç¨Ðä´é $r$ ã¹ÃÙ» $S_1, S_2$ ä´é¤ÃѺ. |
#5
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ÍéÒ§ÍÔ§:
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#6
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ÍéÒ§ÍÔ§:
¡è͹Í×è¹ËÒÃÙ»¢Í§¾¨¹ì·ÑèÇ仢ͧ͹ءÃÁ¹Õé ¤×Í $\frac{\sum_{i = 1}^{k}i^2}{\sum_{i = 1}^{k}i}$ $=\frac{\frac{k}{6}(k+1)(2k+1)}{\frac{k}{2}(k+1)}=\frac{2k+1}{3}$ $\therefore $ ¼ÅÃÇÁ¢Í§Í¹Ø¡ÃÁ¹Õé =$\sum_{k = 1}^{n}\frac{2k+1}{3}$ ´Ñ§¹Ñé¹ $=\frac{1}{3}(n(n+1)+n)=\frac{n^2+2n}{3}$ à»ç¹¤ÓµÍº |
#7
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à¾ÔèÁ⨷Âì ¤ÃѺ
¨§ËҼźǡ n ¾¨¹ìáá¢Í§ $1+\frac{3}{4}+\frac{3\cdot5}{4\cdot8}+\frac{3\cdot5\cdot7}{4\cdot8\cdot12}+...$
__________________
WHAT MAN BELIEVES MAN CAN ACHIEVE |
#8
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ÍéÒ§ÍÔ§:
__________________
You may face some difficulties in your ways But it’s “Good” right ? |
#9
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à¾ÔèÁ⨷Âì¤èÐ
¨§ËÒ $S_n = \frac{1 \cdot 2}{1 \cdot 3 \cdot 5} + \frac{2 \cdot 3}{5 \cdot 7 \cdot 9} +\frac{3 \cdot 4}{9 \cdot 11 \cdot 13} +... $ Åͧ·Ó´Ùä´é $\sum_{n = 1}^{n} \frac{n \cdot (n+1)}{(4n-3) \cdot (4n-1) \cdot (4n+1)} $ áµè仵èÍäÁèä´é¤èÐ |
#10
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#10
diverge ¹Ð¤ÃѺ |
#11
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#11 ⨷ÂìãËéËÒ $S_n$ ú¡Ç¹´éǤèÐ
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#12
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ÍéÒ§ÍÔ§:
à¾ÃÒжéҨѴÃÙ»¨Ðä´éà»ç¹ $$\frac{1}{16}[\frac{1}{4n-1} + \frac{3}{(4n-1)(4n+1)} + \frac{9}{(4n-3)(4n-1)(4n+1)}]$$ «Ö觾¨¹ìáá¤×Í $\frac{1}{4n-1}$ ¶éÒËҼźǡ¨ÐäÁèÊÒÁÒöà¢Õ¹ã¹ÃÙ»ÍÂèÒ§§èÒÂä´é µéͧµÔ´¼ÅºÇ¡µÃ§ æ ËÃ×Íà¢Õ¹ã¹ÃÙ»¿Ñ§¡ìªÑ¹ºÒ§ÍÂèÒ§ áµè¶éÒ¤Ù³·ÕèµÑÇÊèǹ¢Í§â¨·Âì´éÇ $(4n+3)$ ËÃ×Í $(4n-5)$ Ẻ¹Õé¡ç¨ÐËÒà»ç¹ÊٵçèÒ æ ä´é¤ÃѺ. |
#13
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ËÁÒ¤ÇÒÁÇèÒÂѧ䧤Фس gon ·ÕèÇèÒ¶éÒ¤Ù³·ÕèµÑÇÊèǹ¢Í§â¨·Âì´éÇ (4n+3) ËÃ×Í (4n-5) ¨Ðà»ç¹ÊٵçèÒÂæ¹èФèÐ??
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#14
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¼ÁËÁÒ¤ÇÒÁÇèÒ ¶éÒà»ÅÕè¹⨷Âìà»ç¹ àªè¹ $$\sum_{i = 1}^n\frac{i(i+1)}{(4i-3)(4i-1)(4i+1)(4i+3)}$$
Ẻ¹Õé¡ç¨ÐËÒ $S_n$ ä´é¤ÃѺ à¾ÃÒÐÁѹ¨Ð¨Ñ´ÃÙ»ãËéÍÂÙèã¹ÃÙ»¼ÅµèÒ§¢Í§àÈÉÊèǹ áÅéǵѴ¡Ñ¹ä´é |
#15
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µÃǨÊͺáÅéÇ⨷Âìà»ç¹áºº¹Õé¨ÃÔ§æ¤èÐ ËÃ×ͤÇõͺÇèÒ divergence ¤Ð? áµèà¤éÒ¡çäÁèä´é¶ÒÁ $S_\infty$ ¹Ð¤ÐàÅÂäÁèÁÑè¹ã¨ÇèÒ¤ÇõͺÍÂèÒ§äÃ
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