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Question

$ S= \frac{C^{8}_{8}}{7.8}+\frac{C^{8}_{9}}{8.9}+\frac{x^{8}_{10}}{9.10}+......+\frac{C^{8}_{2014}}{2013.2014}+\frac{C^{8}_{2015}}{2014.2015} $

Thanh Long · Answer 1 · 6/20/2016 12:40:50 AM

Thực hiện một vài phép biến đổi thì được

$\frac{C^{8}_{8}}{7.8}=\frac{1.2.3.4.5.6}{8!}$;

$\frac{C^{8}_{9}}{8.9}=\frac{2.3.4.5.6.7}{8!}$;

$\frac{C^{8}_{10}}{9.10}=\frac{3.4.5.6.7.8}{8!}$;

$ ...$

$\frac{C^{8}_{2015}}{2014.2015}=\frac{2008.2009.2010.2011.2012.2013}{8!}$.

Suy ra $S=\sum_{k=1}^{2008}\frac{k(k+1)(k+2)(k+3)(k+4)(k+5)}{8!}=\frac{2008.2009.2010.2011.2012.2013.2014}{7.8!}$.

Trả lời 20-06-16 12:40 AM

Thanh Long
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chiemtb98 · Answer 2 · 6/21/2016 8:52:04 PM

$ \frac{C^{8}_{n}}{n(n-1)}=\frac{n!}{8!(n-8)n(n-1)!}=\frac{(n-2)!}{56.6!((n-2)-6)!}=\frac{1}{56}.C^{6}_{n-2} $

$ S=\frac{1}{56}.(C^{6}_{6}+C^{6}_{7}+C^{6}_{8}+...+C^{6}_{2012}+C^{6}_{2013} )$

$ C^{k}_{n}+C^{k+1}_{n}=C^{k+1}_{n+1} $

$ S=\frac{1}{56}.(C^{6}_{6}-C^{7}_{7}+C^{7}_{2014})=\frac{1}{56}.C^{7}_{2014} $

Trả lời 21-06-16 08:52 PM

chiemtb98
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