Question

Determinați numerele naturale nenule n cu proprietatea că $2^{n-1}$+$3^{n-1}$ divide $2^{n+1}$+$3^{n+1}$. Am acces la rezolvare, iar dacă k ar fi raportul dintre cele două sume, k ar putea fi 5, 6, 7, 8. Varianta corectă este 7, dar nu înțeleg cum se ajunge acolo. Dau 40 pct.

Answer 1

$\dfrac{2^{n+1}+3^{n+1}}{2^{n-1}+3^{n-1}}=k \in \mathbb{Z} \\ \\\\ \Rightarrow 2^{n+1}+3^{n+1}= k(2^{n-1}+3^{n-1}) \\ \\ \Rightarrow 2^{n+1}-k\cdot 2^{n-1}+3^{n+1}-k\cdot 3^{n+1} = 0 \\ \\ \Rightarrow 2^{n-1}\cdot(2^2-k)+3^{n-1}\cdot (3^2-k) = 0\\ \\ \Rightarrow 2^{n-1}\cdot(4-k)+3^{n-1}\cdot (9-k) = 0\Big|:2^{n-1}\\ \\ \Rightarrow 4-k+\Big(\dfrac{3}{2}\Big)^{n-1}\cdot (9-k) = 0 \\ \\ \Rightarrow \Big(\dfrac{3}{2}\Big)^{n-1}\cdot (9-k) =k-4$

$\\ \\ \Rightarrow \Big(\dfrac{3}{2}\Big)^{n-1} = \dfrac{k-4}{9-k},\quad k\in \{5,6,7,8\}\quad \text{(din motive existentiale)}\\ \\\\ \boxed{1}\quad k = 5\Rightarrow \Big(\dfrac{3}{2}\Big)^{n-1} = \dfrac{5-4}{9-5}=\dfrac{1}{4}\quad (F)\\ \\\boxed{2}\quad k = 6\Rightarrow\Big(\dfrac{3}{2}\Big)^{n-1} =\dfrac{6-4}{9-6} =\dfrac{2}{3}\quad (F),\quad n = 0\notin \mathbb{N}^*\\ \\\boxed{3}\quad k = 7 \Rightarrow \Big(\dfrac{3}{2}\Big)^{n-1}=\dfrac{7-4}{9-7}=\dfrac{3}{2}\quad (A),\quad n = 2\\ \\ \boxed{4}\quad k = 8\Rightarrow\Big(\dfrac{3}{2}\Big)^{n-1} =\dfrac{8-4}{9-8}=\dfrac{4}{1}\quad (F) \\ \\\\ \Rightarrow \boxed{n = 2}$

Answer 2

Răspuns:

n=2

Explicație pas cu pas:

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