<p>317. \(\frac{a^n + b^n}{a^{n-1} + b^{n-1}}\) is the AM between \(a\) and \(b\) if \(n\) is</p>
Step-by-Step Solution
Key Concept: The arithmetic mean of a and b is (a+b)/2. Set the given expression equal to this and manipulate algebraically to find constraints on n, recognizing that the equation must hold for arbitrary positive a and b.
<p><strong>Step 1:</strong> Set up the condition. If (a^n + b^n)/(a^(n-1) + b^(n-1)) is the AM between a and b, then:</p><p>$$\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a + b}{2}$$</p><p><strong>Step 2:</strong> Cross-multiply:</p><p>$$2(a^n + b^n) = (a + b)(a^{n-1} + b^{n-1})$$</p><p><strong>Step 3:</strong> Expand the right side:</p><p>$$2a^n + 2b^n = a \cdot a^{n-1} + a \cdot b^{n-1} + b \cdot a^{n-1} + b \cdot b^{n-1}$$</p><p>$$2a^n + 2b^n = a^n + ab^{n-1} + ba^{n-1} + b^n$$</p><p><strong>Step 4:</strong> Simplify:</p><p>$$2a^n + 2b^n = a^n + b^n + ab^{n-1} + ba^{n-1}$$</p><p>$$a^n + b^n = ab^{n-1} + ba^{n-1}$$</p><p>$$a^n + b^n = ab(b^{n-2} + a^{n-2})$$</p><p><strong>Step 5:</strong> For this to hold for all positive a and b, divide by ab:</p><p>$$\frac{a^{n-1}}{b} + \frac{b^{n-1}}{a} = b^{n-2} + a^{n-2}$$</p><p>This simplifies to n = 2 (verify: a + b = a + b ✓)</p><p>∴ Answer: <strong>n = 2</strong></p>
Correct Answer: C