Question

Sabendo que 8N !     (N + 2) ! + (N + 1)!,  o valor de N é:

                                      N + 1

Answer 1

Sendo que:  (n+2)! = (n+2)*(n+1)*n!    e (n+1)= (n+1)*n!

$\frac{8n! = ((n+2)*(n+1)*n! + (n+1)*n!}{(n+1)} \\ \\ \frac{8n! = (n+2)*(n+1)+ (n+1)}{(n+1)} \\ \\ 8 = (n+2) +1 \\ \\ 8 = n + 3 \\ \\ -n = -8 + 3 *(-1) \\ \\ n = 8 - 3 \\ \\ n=5$