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#2
11 ÁԶعÒ¹ 2010, 20:32
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ÍéÒ§ÍÔ§:
$$ S_n= \sum_{i=1}^n (i+(i+1)+\dots +(3i-2)) $$
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#4
12 ÁԶعÒ¹ 2010, 11:07
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à·ÍÁ·ÕèÍÂÙèã¹ sigma à»ç¹Í¹Ø¡ÃÁàÅ¢¤³Ôµ ·Õèà¾ÔèÁ·ÕÅÐ 1 áÅкǡ¡Ñ¹ 2i-1 à·ÍÁ¤ÃѺ
´Ñ§¹Ñé¹ $$ S_n = \sum_{i=1}^n \,\, \big(\frac{2i-1}{2}\cdot(i+(3i-2)) \big) = \sum_{i=1}^n \,\, (2i-1)^2 $$ ¨Ò¡¹Ñé¹ ¶éÒ¡ÃШÒÂáÅéÇ take sigma ·éÒ·ÕèÊØ´ ¨ÐµÍº $ \frac{n(4n^2-1)}{3}$ ¤ÃѺ
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à¡ÉÕ³µÑÇàͧ »ÅÒÂÁԶعÒ¹ 2557 áµè¨Ð¡ÅѺÁÒà»ç¹¤ÃÑ駤ÃÒÇ 12 ÁԶعÒ¹ 2010 11:21 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ passer-by
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#5
12 ÁԶعÒ¹ 2010, 22:48
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$a_1=1$
$a_2=2+3+4$ $a_3=3+4+5+6+7$ $a_4=4+5+6+7+8+9+10$ .... $a_n=n+(n+1)+(n+2)+...+(3n-2)=(2n-1)^2$ ËÒ $a_n$ ¨Ò¡Êٵà $\frac{n}{2}(a_1+a_n)$ ¼ÅºÇ¡ n ¾¨¹ìáá¢Í§Í¹Ø¡ÃÁàÅ¢¤³Ôµ ´Ñ§¹Ñé¹ $S_n=\sum_{i=1}^{n}a_n=\sum_{i=1}^{n}(2n-1)^2$
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