<p>If \(p(x) = \dfrac{(1 + x^2 + x^4 + \cdots + x^{2n-2})}{(1 + x + x^2 + \cdots + x^{n-1})}\) is a polynomial in \(x\), then find possible values of \(n\).</p>
Step-by-Step Solution
Key Concept: For p(x) to be a polynomial, the denominator must divide the numerator exactly. The numerator is a geometric series with even powers: (1-x^(2n))/(1-x²), and denominator is (1-x^n)/(1-x). The quotient simplifies to (1+x^n)/(1+x) only when n is odd, making it polynomial.
<p><strong>Step 1:</strong> Express both series using geometric series formula.</p><p>Numerator: 1 + x² + x⁴ + ... + x^(2n-2) = (1 - x^(2n))/(1 - x²)</p><p>Denominator: 1 + x + x² + ... + x^(n-1) = (1 - x^n)/(1 - x)</p><p><strong>Step 2:</strong> Simplify the ratio.</p><p>p(x) = [(1 - x^(2n))/(1 - x²)] ÷ [(1 - x^n)/(1 - x)]</p><p>p(x) = [(1 - x^(2n))/(1 - x²)] × [(1 - x)/(1 - x^n)]</p><p>p(x) = [(1 - x^(2n))(1 - x)] / [(1 - x²)(1 - x^n)]</p><p>p(x) = [(1 - x^n)(1 + x^n)(1 - x)] / [(1 - x)(1 + x)(1 - x^n)]</p><p><strong>Step 3:</strong> Cancel common factors (valid when they're non-zero).</p><p>p(x) = (1 + x^n)/(1 + x)</p><p><strong>Step 4:</strong> For p(x) to be a polynomial, (1 + x) must divide (1 + x^n).</p><p>This requires x = -1 to be a root of (1 + x^n), meaning (-1)^n + 1 = 0</p><p>(-1)^n = -1, which holds only when n is odd.</p><p><strong>Verification:</strong> When n is odd: 1 + x^n = (1 + x)(1 - x + x² - x³ + ... + x^(n-1)), which is divisible by (1 + x).</p><p>∴ <strong>n must be odd</strong></p>
Correct Answer: n is odd