Question

Resolva a equação fatorial:
20(n-1)! = (n+1)!

Answer 1

Resolver

$20\,(n-1)!=(n+1)!$

com $n$ natural $\ge 1.$


Como fatorial nunca é igual a zero, podemos passar $(n-1)!$ dividindo. E a equação fica

$20=\dfrac{(n+1)!}{(n-1)!}\\\\\\ 20=\dfrac{(n+1)\cdot n\cdot (n-1)!}{(n-1)!}$


Simplificando os fatores comuns no numerador e no denominador, temos

$20=(n+1)\cdot n\\\\ n^{2}+n=20\\\\ n^{2}+n-20=0~~~\Rightarrow~~\left\{ \!\begin{array}{l} a=1\\b=1\\c=-20 \end{array} \right.$

$\Delta=b^{2}-4ac\\\\ \Delta=1^{2}-4\cdot 1\cdot (-20)\\\\ \Delta=1+80\\\\ \Delta=81\\\\\\ n=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ n=\dfrac{-1\pm \sqrt{81}}{2\cdot 1}\\\\\\ n=\dfrac{-1\pm 9}{2}\\\\\\ \begin{array}{rcl} n=\dfrac{-1+9}{2}&~\text{ ou }~&n=\dfrac{-1-9}{2}\\\\ n=\dfrac{8}{2}&~\text{ ou }~&n=\dfrac{-10}{2}\\\\ n=4&~\text{ ou }~&n=-5~~\text{(n\~{a}o serve, pois }n\text{ \'{e} natural)}\\\\ &\boxed{\begin{array}{c}n=4 \end{array}}& \end{array}$