Ars Mathematica: Solución al problema 68

\begin{matrix} (\binom{m}{n - 1}) = 3003; & (\binom{m}{n}) = 2002; & (\binom{m}{n + 1}) = 1001 \end{matrix}

(\binom{m}{n - 1}) = \frac{m!}{(n - 1)! (m - n + 1)!} = \frac{m!}{(n - 1)! (m - n + 1) (m - n) (m - n - 1)!}

(\binom{m}{n}) = \frac{m!}{(n)! (m - n)!} = \frac{m!}{n (n - 1)! (m - n + 1) (m - n) (m - n - 1)!}

(\binom{m}{n + 1}) = \frac{m!}{(n + 1)! (m - n + 1)!} = \frac{m!}{(n + 1) n (n - 1)! (m - n - 1)!}

\frac{(\binom{m}{n})}{(\binom{m}{n + 1})} = \frac{\frac{m!}{n (n - 1)! (m - n + 1) (m - n) (m - n - 1)!}}{\frac{m!}{(n + 1) n (n - 1)! (m - n - 1)!}} = \frac{m! (n + 1) n (n - 1)! (m - n - 1)!}{m! n (n - 1)! (m - n) (m - n - 1)!} = \frac{n + 1}{m - n} = \frac{2002}{1001} = 2

\frac{(\binom{m}{n - 1})}{(\binom{m}{n})} = \frac{\frac{m!}{(n - 1)! (m - n + 1) (m - n) (m - n - 1)!}}{\frac{m!}{n (n - 1)! (m - n + 1) (m - n) (m - n - 1)!}} = \frac{m! n (n - 1)! (m - n) (m - n - 1)!}{m! (n - 1)! (m - n + 1) (m - n) (m - n - 1)!} =

= \frac{n}{m - n + 1} = \frac{3}{2}

{\begin{matrix} 2 n = 3 (m - n + 1) \\ n + 1 = 2 (m - n) \end{matrix} . \Rightarrow {\begin{matrix} m = 14 \\ n = 9 \end{matrix} .

1, 14, 91, 364, 1001, 2002, 3003, 3003, 2002, 1001, 364, 91, 14, 1

Ars Mathematica