#1
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͹ءÃÁ¤ÃѺ
$1\times3\times5\times7+3\times5\times7\times9+5\times7\times9\times11+...+(2n-1)(2n+1)(2n+3)(2n+5) ÁÕ¤èÒà·èÒäËÃè$
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WHAT MAN BELIEVES MAN CAN ACHIEVE |
#2
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$a_n = (2n-1)(2n+1)(2n+3)(2n+5) = 16n^4+64n^3+56n^2-16n-15$
$S_n = \sum_{k=1}^{n}a_k = \sum_{k=1}^{n}16k^4+64k^3+56k^2-16k-15$ áÅéÇ ¡ÃШÒ«ԡÁèÒ á·¹Êٵà à¢éÒä»$ \sum_{k=1}^{n} k^4 = \frac{(n)(n+1)(2n+1)(3n^2+3n-1)}{30}$ |
#3
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ÃÙéä´éä§ËÃͤÃѺ ¾ÔÊÙ¨¹ìãËé´Ù˹èͤÃѺ ¢Íº¤Ø³ÁÒ¡¤ÃѺ
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»Õ¹Õé µéͧäÁè¾ÅÒ´ ÊÙéà¾×èÍ ÁÈÇ »·ØÁÇѹ |
#4
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$(k+1)^4 = k^4+4k^3+6k^2+4k+1$
$(k+1)^4 -k^4 = 4k^3+6k^2+4k+1$ $\sum_{k=1}^{n-1} (k+1)^4 -k^4 = \sum_{k=1}^{n-1} 4k^3+6k^2+4k+1$ ·ÕèàËÅ×ͤ§µèÍàͧä´é¹Ð¤ÃѺ |
#5
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ÍéÒ§ÍÔ§:
http://www.mathcenter.net/sermpra/se...pra18p04.shtml |
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