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¨§à¢Õ¹͹ءÃÁ $\sum_{k = 0}^{n} \frac{(-1)^k \binom{n}{k} }{k^3 +9k^2+26k+24}$ ã¹ÃÙ»¢Í§ $\frac{p(n)}{q(n)}$ â´Â·Õè $p(n)$ ààÅÐ $q(n)$ à»ç¹¾ËعÒÁÊÑÁ»ÃÐÊÔ·¸Ôìà»ç¹¨Ó¹Ç¹àµçÁ
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ÍéÒ§ÍÔ§:
äÁèá¹èã¨ÇèÒ·Óà»ç¹Í¹Ø¡ÃÁâ·Ã·ÃÃȹì¨Ðä´éÁÑé ·Õè¼Á¤Ô´¤×ÍãªéàÍ¡Åѡɳì $\sum_{k=0}^n(-1)^k\binom{n}{k}=0$ ¤ÃѺ ¢Í͸ԺÒ¤ÃèÒèÇæ¹Ð¤ÃѺ(·´ä»·Ñé§ËÁ´ 3 ˹éÒ¡ÃдÒÉ) $\sum_{k = 0}^{n} \frac{(-1)^k \binom{n}{k} }{k^3 +9k^2+26k+24}=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{k=0}^n(-1)^k(k+1)\binom{n+4}{k+4}$ $=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{k=0}^n(-1)^k(k+4)\binom{n+4}{k+4}-3(-1)^k\binom{n+4}{k+4}$ $=\dfrac{1}{(n+1)(n+2)(n+3)(n+4)}\left[-(n+4)\sum_{i=3}^{n+3}(-1)^i\binom{n+3}{i}-3\sum_{i=4}^{n+4}(-1)^i\binom{n+4}{i}\right]$ $=\dfrac{1}{2(n+3)(n+4)}$ |
ʵÑ鹵ç ͹ءÃÁâ·Ã·ÃÃȹì ... 555+:haha:
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§§µÃ§3ºÃ÷ѴÊØ´·éÒÂá¡ k+4 ÁÒà»ç¹ n+4 ä´éä§ÍÐ
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µéͧᵡÊÑÁ»ÃÐÊÔ·¸Ôì·ÇÔ¹ÒÁÍÍ¡ÁÒ¤ÃѺ
$(k+4)\binom{n+4}{k+4}=\dfrac{(k+4)(n+4)(n+3)!}{(k+4)(k+3)!(n-k)!}=(n+4)\binom{n+3}{k+3}$ |
¢Íº¤Ø³ÁÒ¡¤ÃѺ ... ¹Ñè§ÍÖé§ààÅФԴ¹Ò¹ÁÒ¡
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àÇÅÒ·ÕèáÊ´§·Ñé§ËÁ´ à»ç¹àÇÅÒ·Õè»ÃÐà·Èä·Â (GMT +7) ¢³Ð¹Õéà»ç¹àÇÅÒ 00:02 |
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