Question

Cum rezolvi ecuatia (n+2)! +(n+1)! =35n!

Answer 1

(n+2)!=n!*(n+1)*(n+2)
(n+1)!=n!*(n+1)
Ecuatia  devine
n!*(n+1)*(n+2)+n!*(n+1)=35n!
Impartii   ambii   membrii   ai   ecuatiei   prin  n! Obtii
(n+1)*(n+2)+(n+1)=35
Dai  pe  (n+1)  factor   comun
(n+1)*{(n+2)+1]=35
(n+1)(n+3)=5*7=>
n+1=5  si  n+3=7 =>
n=4

Answer 2

Rezolvarea este in atasament.
S final : n = 4.