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  #1  
Old 06 ¾ÄÉÀÒ¤Á 2009, 10:36
kurumi_00 kurumi_00 äÁèÍÂÙèã¹Ãкº
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1+(1+2)+(1+2+3)+.....+(1+2+3+....+n)

ªèÇÂËÒ¾¨¹ì·ÑèÇä»ãËé˹èͤèÒ

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  #2  
Old 06 ¾ÄÉÀÒ¤Á 2009, 10:46
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Scylla_Shadow Scylla_Shadow äÁèÍÂÙèã¹Ãкº
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$\frac{(1)(2)}{2}+\frac{(2)(3)}{2}+\frac{(3)(4)}{2}+...+\frac{(n)(n+1)}{2}$

$\frac{(1)(2)+(2)(3)+(3)(4)+...(n)(n+1)}{2}$

$\frac{1+1^2+2+2^2+3+3^2+..+n+n^2}{2}$

$\frac{\frac{(n)(n+1)}{2}+\frac{(n)(n+1)(2n+1)}{6}}{2}$

$\frac{(n)(n+1)(\frac{n+2}{3})}{2}$

$\frac{(n)(n+1)(n+2)}{6}$
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #3  
Old 06 ¾ÄÉÀÒ¤Á 2009, 20:39
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mercedesbenz mercedesbenz äÁèÍÂÙèã¹Ãкº
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¡Ó˹´ãËé $S_n=a_1+a_2+a_3+\cdots+a_n$
áÅÐ $a_n=1+2+3+\cdots +n=\frac{n(n+1)}{2}$
à¹×èͧ¨Ò¡àÃÒµéͧ¡ÒÃËÒ $S_n=\sum a_n=\sum \frac{n(n+1)}{2}=\frac{1}{2}\sum(n^2+n)=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})$
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