Question

Se (n+4)!+(n+3)!=15(n+2)! qual valor de n ?

Answer 1

(n+4)! + (n+3)! = 15(n+2)!

(n+4)! = (n+4).(n+3)(n+2)!

(n+3)! = (n+3).(n+2)!

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

Logo:

(n+4)! + (n+3)! = 15(n+2)!

(n+4).(n+3)(n+2)! + (n+3).(n+2)! = 15(n+2)!, simplificando (n+2)!

(n+4).(n+3) + (n+3) = 15 => n^2 + 3n + 4n + 12 +n + 3 = 15

n^2 + 3n + 4n + 12 +n + 3 = 15

n^2 + 8n + 12 + 3 = 15

n^2 + 8n = 0 => n(n +8) = 0

n1 = 0 e, n2 = -8

n pode assunir o valor de 0 ou -8.