Se rezolva prin inductie
n=1 2!/0!=2/1=2. 1*2*3/3=2
2=2
Presupunem PN adevarata.Verificam daca si Pn+1 adevarata
Pn=2!/0!+3!/1!+4!/2!+(n+1)/(n-1)!=n(n+1)(n+2)/3 A)
Pn+1=[2!/0!+3!/1!+...+(n+1)!/(n-1)!]+(n+2)!/n!=(n+1)(n+2)(n+3) B)
Paranteza dreapta este A)Faci inlocuirea
n(n+1)(n+2)/3+(n+2)/n!=n(n+1)(n+2)/3+(n+1)(n+2)=([n(n+1)(n+2)+3(n+1)(n+2))/3
(n+1)(n+2))(n+3)/3=(B
Pn=>Pn+1
efalitatea este demonstrata
