∑(k-1)²·k!=∑[(k+1)²-4k]·k!=∑(k+1)²k!-4∑k·k!=
=∑(k+1)(k+1)!-4∑k·k!=
=∑(k+2-1)(k+1)!-4∑(k+1-1)k!=
=∑[(k+2)!-(k+1)!]-4∑[(k+1)!-k!]=
=[(n+2)!-2!]-4[(n+1)!-1!]=
=(n+2)!-2-4(n+1)!+4=
=(n+1)!(n+2-4)+2=
=(n+1)!(n-2)+2
Acum sa demonstram rezultatul gasit prin inductie:
P(n):∑(k-1)²·k!=(n+1)!(n-2)+2
P(2):2!=3!·0+2 (A)
Presupunem ca P(n) adevarata si demonstram ca si P(n+1) adevarata.
P(n+1): ∑(k-1)²·k!(cu n de la 1 la n+1)=(n+2)!(n-1)+2
∑(k-1)²·k!(cu n de la 1 la n+1)=∑(k-1)²·k!(cu n de la 1 la n)+n²(n+1)!=
=(n+1)!(n-2)+2+n²(n+1)!=
=(n+1)!(n-2+n²)+2=
=(n+1)!(n+2)(n-1)+2=
=(n+2)!(n-1)+2
