Question

Arătați că pentru orice număr natural n, numărul:
$a = \dfrac{(-1)^{5n+1} (6n + 9) + 1}{4} \in Z$

Answer 1

[tex]\text{Deci dupa cum ziceam,avem de analizat doua cazuri:}\\ i)n=2k\Rightarrow a=\dfrac{(-1)^{5\cdot2k+1}(6\cdot 2k+9)+1}{4}=\dfrac{(-1)^{10k+1}(12k+9)+1}{4}\\ =\dfrac{-12k-9+1}{4}=\dfrac{-12k-8}{4}=-3k-2\in \mathbb{Z}\\ ii)n=2k+1\Rightarrow a=\dfrac{(-1)^{5(2k+1)+1}(6\cdot(2k+1)+9)+1}{4}=\\ =\dfrac{(-1)^{10k+6}(12k+6+9)+1}{4}=\dfrac{12k+16}{4}=3k+4\in \mathbb{Z}\\ [/tex]