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  #1  
Old 13 ÁԶعÒ¹ 2010, 13:07
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1.$¡Ó˹´ãËé S_n = 1\cdot2+ 2\cdot4 + 3\cdot8 + 4\cdot16 + ... + n\cdot2^{n} $

ÍÂÒ¡·ÃÒºÇèÒ $S_{10}$ à·èҡѺà·èÒäÃ

2. ¨§ËҼźǡ¢Í§Í¹Ø¡ÃÁ $2 + 3\cdot2 + 4\cdot2^2 +5\cdot2^3 + ... + n\cdot2^n$

µÍº $(n-1)\cdot2^{n}$

¢Íº¤Ø³ÁÒ¡¤ÃѺ
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #2  
Old 13 ÁԶعÒ¹ 2010, 14:14
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¢éÍ 1 ¼ÅºÇ¡ n ¾¨¹ìááà·èҡѺ $(n-1)2^{n+1}$+2áÅéÇá·¹ n à·èҡѺ 10

13 ÁԶعÒ¹ 2010 14:16 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ NAKHON
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #3  
Old 13 ÁԶعÒ¹ 2010, 16:10
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#2 ¢ÍÇÔ¸Õ·Ó´éǤÃѺ
µÍº¾ÃéÍÁÍéÒ§ÍÔ§¢éͤÇÒÁ¹Õé
  #4  
Old 14 ÁԶعÒ¹ 2010, 10:30
¡ÃкÕèà´ÕÂÇ´ÒÂáÊǧ¾èÒÂ's Avatar
¡ÃкÕèà´ÕÂÇ´ÒÂáÊǧ¾èÒ ¡ÃкÕèà´ÕÂÇ´ÒÂáÊǧ¾èÒ äÁèÍÂÙèã¹Ãкº
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$s_n=1(2)+2(2^2)+3(2^3)+4(2^4)+...+n(2^n)$...(1)
àÍÒ 2 ¤Ù³µÅÍ´
$2s_n=1(2^2)+2(2^3)+3(2^4)+...+n(2^{n+1})$...(2)
(2)-(1)
$s_n=-2-2^2-2^3-...-2^n+n(2^{n+1})$
$s_n=-2(2^n-1)+n(2^{n+1})= (n-1)(2^{n+1})+2$

14 ÁԶعÒ¹ 2010 10:31 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 3 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¡ÃкÕèà´ÕÂÇ´ÒÂáÊǧ¾èÒÂ
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