Question:

<p><strong>Step 1:</strong> Set the given expression equal to the geometric mean of a and b:</p><p>$$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab}$$</p><p><strong>Step 2:</strong> Square both sides:</p><p>$$\frac{(a^{n+1}+b^{n+1})^2}{(a^n+b^n)^2} = ab$$</p><p><strong>Step 3:</strong> Expand the numerator and rearrange:</p><p>$$(a^{n+1}+b^{n+1})^2 = ab(a^n+b^n)^2$$</p><p>$$a^{2n+2} + 2a^{n+1}b^{n+1} + b^{2n+2} = ab(a^{2n} + 2a^nb^n + b^{2n})$$</p><p>$$a^{2n+2} + 2a^{n+1}b^{n+1} + b^{2n+2} = a^{2n+1}b + 2a^{n+1}b^{n+1} + ab^{2n+1}$$</p><p><strong>Step 4:</strong> Cancel 2a^{n+1}b^{n+1} from both sides:</p><p>$$a^{2n+2} + b^{2n+2} = a^{2n+1}b + ab^{2n+1}$$</p><p>$$a^{2n+2} + b^{2n+2} = ab(a^{2n} + b^{2n})$$</p><p><strong>Step 5:</strong> Factor strategically. Divide by a^{2n}b^{2n}:</p><p>$$\frac{a^2}{b^{2n}} + \frac{b^2}{a^{2n}} = \frac{a}{b^{2n}} + \frac{b}{a^{2n}}$$</p><p>This is satisfied when <strong>n = -1</strong> (verify by substitution).</p><p>∴ Answer: B (n = -1)</p>