Câu III:
Ta có: $S=\frac{1}{1!2013!}+\frac{1}{2!2012!}+...+\frac{1}{2014!0!}=\frac{1}{2014!}(\frac{2014!}{1!2013!}+\frac{2014!}{2!2012!}+...+\frac{2014!}{2014!0!})$.
Mặt khác ta lại có: $2^{2014}=(1+1)^{2014}=\sum_{k=0}^{2014}C_{2014}^k=\frac{2014!}{1!2013!}+\frac{2014!}{2!2012!}+...+\frac{2014!}{2014!0!}$.
$\implies S=\frac{2^{2014}}{2014!}$
