Question

Questões de fatorial!
1) Calcule:
(n-1)! + n!
________
(n+1)!

2) Resolva a equação:
(n+1)! = 12
________
(n-1)!

Answer 1

1) $\mathtt{\dfrac{(n-1)!+n!}{(n+1)!}}$

 
Expandindo os fatoriais e truncando em $\mathtt{(n-1)!,}$ a expressão fica

$=\mathtt{\dfrac{(n-1)!+n\cdot (n-1)!}{(n+1)\cdot n\cdot (n-1)!}}\\\\\\ =\mathtt{\dfrac{(n-1)!\cdot (1+n)}{(n-1)!\cdot (n+1)\cdot n}}\\\\\\ =\mathtt{\dfrac{(n-1)!\cdot (n+1)}{(n-1)!\cdot (n+1)\cdot n}}\\\\\\ =\mathtt{\dfrac{1}{n}}\quad\quad\texttt{para }\mathtt{n \ge 1,\,n\in\mathbb{N}}$


2) Resolver a equação:

$\mathtt{\dfrac{(n+1)!}{(n-1)!}=12}$


Expandindo os fatoriais e truncando em $\mathtt{(n-1)!,}$ a equação fica

$\mathtt{\dfrac{(n+1)\cdot n\cdot (n-1)!}{(n-1)!}=12}\\\\\\ \mathtt{(n+1)\cdot n=12}\\\\ \mathtt{n^2+n=12}\\\\ \mathtt{n^2+n-12=0}$


Resolvendo via fatoração por agrupamento. Subtraia e some $\mathtt{3n}$ ao lado esquerdo:

$\mathtt{n^2-3n+3n+n-12=0}\\\\ \mathtt{n^2-3n+4n-12=0}\\\\ \mathtt{n(n-3)+4(n-3)=0}\\\\ \mathtt{(n-3)(n+4)=0}\\\\ \begin{array}{rcl} \mathtt{n-3=0}&\texttt{ ou }&\mathtt{n+4=0}\\\\ \mathtt{n=3}&\texttt{ ou }&\mathtt{n=-4}\texttt{ (n\~ao serve)} \end{array}$


Logo, a solução é

$\mathtt{S=\{3\}.}$


Testando:

$\mathtt{\dfrac{(3+1)!}{(3-1)!}}\\\\\\ \mathtt{=\dfrac{4!}{2!}}\\\\\\ \mathtt{=\dfrac{4\cdot 3\cdot \diagup\!\!\!\!\! 2!}{\diagup\!\!\!\!\! 2!}}\\\\\\ \mathtt{=4\cdot 3}\\\\ \mathtt{=12\quad\quad\checkmark}$


Dúvidas? Comente.


Bons estudos! :-)