(n+3)! = (n+3).(n+2)!
(n+3)!-(n+2)! = (n+3).(n+2)!-(n+2)!
Ortak paranteze al: (n+2)!(n+3-1)
→ (n+2).(n+2)!
(n+1)! = (n+1).n!
(n+1)!+n! = (n+1).n!+n!
Ortak paranteze al: n!(n+1+1)
→ (n+2).n!
[tex] \frac{(n + 3)! - (n + 2)!}{(n + 1)! + n!} = \frac{(n + 2).(n + 2)!}{(n + 2).n!} [/tex]= [tex] \frac{(n + 2)!}{n!} [/tex]
(n+2)! = (n+2).(n+1).n!
→ (n+2).(n+1).n! / n! = (n+2).(n+1)
Cevap: n²+3n+2
