Question

O valor de n, tal que (n 2)! n! / (n 1)! (n-1)! = 15

Answer 1

$\frac{(n+2)!n!}{(n+1)!(n-1)!} = 15$


Obs:

(n + 2)! = (n + 2) . (n + 1)!
n! = n . (n - 1)!

Desenvolvendo:

$\frac{(n+2).(n+1)!.n.(n-1)!}{(n+1)!(n-1)!} = 15$

$\frac{(n+2).\not{(n+1)!}.n.\not{(n-1)!}}{\not{(n+1)!}\not{(n-1)!}} = 15$

$(n + 2).n = 15$
$n^2 + 2n = 15$
$n^2 + 2n - 15 = 0$


Δ = (2)² - 4.(1).(-15)
Δ = 4 + 60
Δ = 64


$n = \frac{-(2)^+_-\sqrt{64}}{2.(1)}$

$n = \frac{-2^+_-8}{2}$


$n' = \frac{-2+8}{2}$

$n' = \frac{6}{2}$

n' = 3



$n'' = \frac{-2-8}{2}$

$n'' = \frac{-10}{2}$

n'' = -5


S = {-5, 3}