Tính tổng: S= $\frac{1}{2!.2012!}$ + $\frac{1}{4!.2010!}$ +...+ $\frac{1}{2!.2012!}$ + $\frac{1}{2014!}$

Question

Tính tổng: S= $\frac{1}{2!.2012!}$ + $\frac{1}{4!.2010!}$ +...+ $\frac{1}{2!.2012!}$ + $\frac{1}{2014!}$

score 1 · Answer 1

$S=\frac{1}{2!2012!}+\frac{1}{4!2010!}+\frac{1}{6!2008!}+...+\frac{1}{2012!2!}+\frac{1}{2014!0!}$

$=\frac{1}{2014!}(\frac{2014!}{2!2012!}+\frac{2014!}{4!2010!}+\frac{2014!}{6!2008!}+...+\frac{2014!}{2012!2!}+\frac{2014!}{2014!0!})$

$=\frac{1}{2014!}(C^{2}_{2014}+C^{4}_{2014}+C^{6}_{2014}+...+C^{2012}_{2014}+C^{2014}_{2014})$

$=\frac{1}{2014!}[(C^{0}_{2014}+C^{2}_{2014}+C^{4}_{2014}+C^{6}_{2014}+...+C^{2012}_{2014}+C^{2014}_{2014})-1]$

$=\frac{1}{2014!}(2^{2013}-1)$

$=\frac{2^{2013}-1}{2014!}$ .

score 0 · Answer 2

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�? l�m ��?c ch�a?????? – lehongthuy58 16-01-17 06:51 AM