Question

2!/0!+3!/1!+4!/2!+.... +(n+1)!/(n-1)! = n(n+1)(n+2)/3
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Answer 1

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