Question

Am si eu nevoie urgenta de ajutor la 3 si 4!!!

Answer 1

$3)\\ \\ \displaystyle\sum\limits_{k=1}^{n}\dfrac{k+2}{k!+(k+1)!+(k+2)!}= \sum\limits_{k=1}^{n}\dfrac{k+2}{k!+k!(k+1)+(k+1)!(k+2)}=\\ \\ =\sum\limits_{k=1}^{n}\dfrac{k+2}{k!(1+k+1)+(k+1)!(k+2)}= \sum\limits_{k=1}^{n}\dfrac{k+2}{k!(k+2)+(k+1)!(k+2)}= \\ \\ =\sum\limits_{k=1}^{n}\dfrac{k+2}{(k+2)\Big(k!+(k+1)!\Big)}=$

$\displaystyle =\sum\limits_{k=1}^{n}\dfrac{1}{k!+(k+1)!}= \sum\limits_{k=1}^{n}\dfrac{1}{k!+k!(k+1)}=\\ \\=\sum\limits_{k=1}^{n}\dfrac{1}{k!(1+k+1)}=\sum\limits_{k=1}^{n}\dfrac{1}{k!(k+2)}= \sum\limits_{k=1}^{n}\dfrac{k+1}{k!(k+1)(k+2)} =\\ \\ = \sum\limits_{k=1}^{n}\dfrac{k+2-1}{(k+2)!}= \sum\limits_{k=1}^{n}\dfrac{k+2}{(k+2)!}-\sum\limits_{k=1}^{n}\dfrac{1}{(k+2)!} = \\ \\ =\sum\limits_{k=1}^{n}\dfrac{1}{(k+1)!} -\sum\limits_{k=1}^{n}\dfrac{1}{(k+2)!}=$

$= \dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{(n+1)!}-\dfrac{1}{3!}-\dfrac{1}{4!}-...-\dfrac{1}{(n+1)!}-\dfrac{1}{(n+2)!}= \\ \\ = \dfrac{1}{2}-\dfrac{1}{(n+2)!}$

$4)\\\\ \text{card}(A) = n\\ \\ \text{card}(A\times A) = \text{card}\Big(P(A)\Big) \Leftrightarrow \\ \\\Leftrightarrow n\times n = 2^n \Leftrightarrow \\ \\ \Leftrightarrow n^2 = 2^n \Leftrightarrow \\ \\ \Leftrightarrow n \in\{2;4\}\\ \\ \Rightarrow \boxed{\text{card(A)}= \{2;4\}}$