Question

Equação (n+3)! por (n+1)! = 30

Answer 1

Ae mano,

na equação $\dfrac{(n+3)!}{(n+1)!}=30$, com $n\in\mathbb{N}$,

podemos usar o conceito de fatorial..

$\dfrac{(n+3)\cdot(n+2)\cdot(n+1)!}{(n+1)!}=30\\\\ (n+3)(n+2)=30\\ n^2+2n+3n+6=30\\ n^2+5n+6=30\\ n^2+5n-24=0\\\\ \Delta=5^2-4\cdot1\cdot(-24)\\ \Delta=25+96 \\ \Delta=121\\\\ n= \dfrac{-5\pm \sqrt{121} }{2\cdot1}= \dfrac{-5\pm11}{2} \begin{cases}n_1=3\\ n_2=-8~~(\notin\mathbb{N})\end{cases}$

Portanto,


$\huge\boxed{\text{S}=\{3\}}$