<p>Since \(a_n + \sqrt{2}\,b_n = \left(2+\sqrt{2}\right)^n\) and \(a_n - \sqrt{2}\,b_n = \left(2-\sqrt{2}\right)^n\), find \(\dfrac{a_n}{b_n}\) as \(n \to \infty\). (Give your answer to 2 decimal places.)</p>
Step-by-Step Solution
Key Concept: Extract $a_n$ and $b_n$ by adding/subtracting the conjugate equations, then recognize that $(2-\sqrt{2})^n \to 0$ as $n \to \infty$ since $|2-\sqrt{2}| < 1$, making the dominant term $(2+\sqrt{2})^n$.
<p><strong>Step 1:</strong> Set up the system from conjugate equations:</p><p>$a_n + \sqrt{2}\,b_n = (2+\sqrt{2})^n$ ... (1)</p><p>$a_n - \sqrt{2}\,b_n = (2-\sqrt{2})^n$ ... (2)</p><p><strong>Step 2:</strong> Add equations (1) and (2):</p><p>$2a_n = (2+\sqrt{2})^n + (2-\sqrt{2})^n$</p><p>$a_n = \frac{(2+\sqrt{2})^n + (2-\sqrt{2})^n}{2}$</p><p><strong>Step 3:</strong> Subtract equation (2) from (1):</p><p>$2\sqrt{2}\,b_n = (2+\sqrt{2})^n - (2-\sqrt{2})^n$</p><p>$b_n = \frac{(2+\sqrt{2})^n - (2-\sqrt{2})^n}{2\sqrt{2}}$</p><p><strong>Step 4:</strong> Note that $2-\sqrt{2} \approx 0.586 < 1$, so $(2-\sqrt{2})^n \to 0$ as $n \to \infty$.</p><p><strong>Step 5:</strong> Find the limit:</p><p>$$\frac{a_n}{b_n} = \frac{(2+\sqrt{2})^n + (2-\sqrt{2})^n}{\frac{(2+\sqrt{2})^n - (2-\sqrt{2})^n}{\sqrt{2}}} \to \frac{(2+\sqrt{2})^n}{\frac{(2+\sqrt{2})^n}{\sqrt{2}}} = \sqrt{2}$$</p><p><strong>Step 6:</strong> Calculate: $\sqrt{2} \approx 1.414213... \approx 1.42$</p><p>∴ Answer: <strong>1.42</strong></p>
Correct Answer: 1.42