[tex]1)\quad a) \\ \\ \dfrac{3!}{1!}+\dfrac{4!}{2!}+\dfrac{5!}{3!}+...+\dfrac{50!}{48!} = \\ \\ = \sum\limits_{k=1}^{48}\dfrac{(k+2)!}{k!} = \sum\limits_{k=1}^{48}\dfrac{k!(k+1)(k+2)}{k!} = \sum\limits_{k=1}^{48}\Big[(k+1)(k+2)\Big] = \\ \\ =\sum\limits_{k=1}^{48}(k^2+3k+2) = \sum\limits_{k=1}^{48}(k^2) +3\sum\limits_{k=1}^{48}(k)+\sum\limits_{k=1}^{48}(2) = \\ \\ = \dfrac{48(48+1)(2\cdot 48+1)}{6}+3\cdot \dfrac{48(48+1)}{2}+2\cdot 48=...[/tex]
[tex]2)\quad a)\\ \\ (n+2)!-n! = 114 \\ n!(n+1)(n+2) -n! = 114\\ n!\Big((n+1)(n+2)-1\Big) = 114\\ n!(n^2+3n+1) = 3!\cdot 19\\ \\ n=3 \Rightarrow 9+9+1 = 19\quad (A) \\ \\ \text{Nu mai exista alte solutii deoarece pentru n = 2, nu verifica, iar pentru}\\\text{n = 4, 114 nu se imparte exact la 4}!\text{ iar catul oricum va deveni din}\\ \text{ce in ce mai mic, iar }n^2+3n+1 \text{ din ce in ce mai mare, deci e contradictie.}\\ \\ \Rightarrow n = 3.[/tex]
