Question

68p ! URGENT!!!!! Dau coroana!!!!! Inducție matematică!!!!

Answer 1

$\it 3^n\ -\ impar,\ \forall\ n\in\mathbb{N} \Rightarrow 3^n-1\ este\ num\breve ar\ par\ \ \ \ (1)\\ \\ 14n\ este\ num\breve ar\ par\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow 3^n+14n-1\ \ e\ num\breve ar\ par \Rightarrow 2\Big|\ 3^n+14n-1,\ \forall n\in\mathbb{N}$

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$\it Fie\ P(n):\ \ 2|(3^n+14n-1)\\ \\ P(1):\ \ 2|(3+14-1) \Rightarrow 2|16\ (A)\\ \\ Presupunem\ P(k)\ adev\breve arat\breve a, adic\breve a:\ \ 2|(3^k+14k-1)\\ \\ Vom\ demonstra\ c\breve a\ P(k+1)\ e\ adev\breve arat\breve a.\\ \\ P(k+1):\ \ 2|3^{k+1}+14(k+1)-1 \Rightarrow 2|3\cdot3^k+14k+14-1 \Rightarrow \\ \\ \Rightarrow 2|(2+1)3^k+14k+14-1 \Rightarrow 2|2\cdot3^k+3^k+14k+14-1 \Rightarrow \\ \\ \Rightarrow 2|(3^k+14k-1)+2(3^k+7)\ \ Adev\breve arat\breve a,\ deoarece:$

$\it 2|(3^k+14k-1)\ \c{s}i\ 2\2(3^k+7).\\ \\ P(k) \Longrightarrow P(k+1),\ atunci\ P(n)\ e\ \ (A),\ \ \forall n\in\mathbb{N}$