Question

Equação fatorial. (questão em anexo)

Answer 1

         $\frac{(n+2)!-(n+1)!}{n(n-1)!} =25 \\ \\ \frac{(n+2)(n+1)!-(n+1)!}{n(n-1)!} =25 \\ \\ \frac{(n+1)!(n+2-1)}{n(n-1)!} =25 \\ \\ \frac{(n+1)(n)(n-1)!(n+1)}{n(n-1)!} =25 \\ \\ (n+1)(n+1)=25 \\ \\ n^2+2n+1=25 \\ \\ n^2+2n-24=0$

   fatorando
          $(n + 6)(n - 4) = 0 \\ \\ n+6=0 \\ n1=-6 \\ \\ n-4=0 \\ n=4$

   Os números negativos não tem fatorial
   Então
                       n = 4

Answer 2

$\frac{(n+2)\cdot(n+1)\cdot\,n\cdot(n-1)!-(n+1)\cdot\,n\cdot(n-1)!}{n(n-1)!}=25\\\\\\\frac{n(n-1)![(n+2)\cdot(n+1)-(n+1)]}{n(n-1)!}=25$

$[(n+2)\cdot(n+1)-(n+1)]=25\\\\(n+1)[n+2-1]=25\\\\(n+1)(n+1)=25\\\\(n+1)^2=25\\\\(n+1)=\sqrt{25}\\\\n+1=5\\\\\boxed{n=4}$