Question

Arătați ca nr a=2n-9+3 *(-1)^n+1 supra 4 este întreg pt orice nr nat n

Va rog cât mai repede, dau coroana si multe puncte! ​

Answer 1

$a = \dfrac{2n-9+3\cdot(-1)^{n+1}}{4} \\ \\ \\ \boxed{1}\quad n-\text{par}\to n = 2k,\quad k\in \mathbb{N} \\ \\ \Rightarrow a = \dfrac{2\cdot (2k)-9+3\cdot (-1)^{2k+1}}{4} = \dfrac{4k-9-3}{4} = \dfrac{4k-12}{4} \\\\ \Rightarrow a = k-3 \in \mathbb{Z}\\ \\\\\boxed{2}\quad n-\text{impar} \to n = 2k-1,\quad k\in \mathbb{N}\\ \\ \Rightarrow a = \dfrac{2\cdot (2k-1)-9+3\cdot (-1)^{2k-1+1}}{4} = \dfrac{4k-2-9+3}{4} \\ \\ \Rightarrow a = \dfrac{4k-8}{4} = k-2 \in \mathbb{Z}$

$\\\text{Din }\,\boxed{1}\,\text{ si }\,\boxed{2} \Rightarrow a \in \mathbb{Z},\quad \forall n\in \mathbb{N}$