Răspuns:
Explicație pas cu pas:
[tex]\texttt{Se demonstreaza prin inductie.}\\P(n): x^n+\dfrac{1}{x^n},n\in\matbb{N}\\\texttt{Etapa 1: Verificarea}\\P(0): x^0+\dfrac{1}{x^0}=1+1=2\in\mathbb{Z}\\\texttt{Etapa 2: Demonstratia propriu-zisa}\\\texttt{Presupunem P(k) si P(k-1) adevarate }\forall~\k\in\mathbb{N}.\texttt{Se demonstreaza ca}\\\texttt{si P(k+1) este adevarat.}\\P(k-1) :x^{k-1}+\dfrac{1}{x^{k-1}}\in\mathbb{Z}\\P(k):x^k+\dfrac{1}{x^k}\in\mathbb{Z}\\P(k+1):x^{k+1}+\dfrac{1}{x^{k+1}}\in \mathbb{Z}[/tex]
[tex]\texttt{Sa observam ca }\left(x+\dfrac{1}{x}\right)\left(x^k+\dfrac{1}{x^k}\right)=x^{k+1}+\dfrac{1}{x^{k-1}}+x^{k-1}+\dfrac{1}{x^{k+1}},\\\texttt{adica } x^{k+1}+\dfrac{1}{x^{k+1}}=\left(x+\dfrac{1}{x}\right)\left(x^k+\dfrac{1}{x^k}\right)-x^{k-1}-\dfrac{1}{x^{k-1}}\in\mathbb{Z},\\\texttt{prin urmare P(k+1) este adevarat.Rezulta deci ca P(n) este}\\\texttt{adevarat }\forall n\in\mathbb{N}[/tex]
