#1
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͹ءÃÁ
$\sum_{k = 1}^{5}\frac{k-1}{k+1} $= ?
$\sum_{k = 2}^{5}\frac{k+1}{k-1} $= ? $\sum_{k = 1}^{10}[1+ (-1)^k] $= ? ªèÇÂÊ͹˹è͹ФР22 Á¡ÃÒ¤Á 2012 19:51 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ Twin_chock |
#2
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á·¹¤èÒ $k=1,2,3,...$ ä»àŤÃѺ
$\sum_{k = 1}^{5}\frac{k-1}{k+1} = 0+\frac{1}{3} +\frac{2}{4}+\frac{3}{5}+\frac{4}{6}$ |
#3
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áÅéǶéÒÍÂÒ¡ãªéÊٵà µéͧ·ÓÂѧ䧤РªèÇÂáÊ´§ãËé´Ù˹èÍÂä´éÃÖà»ÅèÒ â´Â੾ÒÐÊͧ¢éÍÊØ´·éÒÂ
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#4
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â´Â·ÑèÇä» ¡ÒÃËÒ¤èÒ $\Sigma$ ¶éÒá·¹¤èÒ $n$ à¾Õ§äÁè¡Õè¤ÃÑé§ ¡çá·¹¤èÒàÍÒàÅ §èÒ¡ÇèÒ¤ÃѺ
$\sum_{k = 2}^{5}\frac{k+1}{k-1} \rightarrow \Sigma $ ᨡ¼ÅËÒÃäÁèä´é ¡çµé᷹ͧ¤èÒàÍÒ ËÃ×ͨдѴá»Å§¹Ô´Ë¹èÍ àªè¹ $\sum_{k = 2}^{5}\frac{(k-1)+2}{k-1}=\sum_{k = 2}^{5}(1+\frac{2}{k-1})=\sum_{k = 2}^{5}(1)+\sum_{k = 2}^{5}(\frac{2}{k-1})=(1)(5)+\sum_{k = 2}^{5}(\frac{2}{k-1})$ ¶Ö§µÃ§¹Õé¡çµé᷹ͧ¤èÒµèͤÃѺ $\sum_{k = 1}^{10}[1+ (-1)^k] $ ¢é͹Õéá·¹¤èÒàÍÒàÅ¡ç§èÒ áµè¶éÒÍÂÒ¡ãªéÊٵáçä´é àªè¹ $=\sum_{k = 1}^{10}(1)+ \sum_{k = 1}^{10}(-1)^k$ $=(1)(10)+\dfrac{(-1)((-1)^{10}-1)}{(-1)-1} $ ËÁÒÂà赯 $\sum_{n = 1}^{k} c^n=\dfrac{c(c^k-1)}{c-1} $ |
#5
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#4 äÁèà¢éÒã¨Êٵ÷ÕèËÁÒÂà˵ØäÇé
ÍéÍ à¢éÒã¨áÅéǤÃѺ 24 Á¡ÃÒ¤Á 2012 16:02 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ artty60 |
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