<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 \(n\) can be</p>
Step-by-Step Solution
Key Concept: For p(x) to be a polynomial, the denominator must divide the numerator completely. The numerator is a geometric series sum (1+x²+x⁴+...+x^(2n-2)) and denominator is (1+x+x²+...+x^(n-1)). The degree of numerator is 2n-2 and denominator is n-1, so we need the roots of denominator to be roots of numerator.
<p><strong>Step 1:</strong> Use geometric series formulas:</p><p>Numerator: 1+x²+x⁴+...+x^(2n-2) = (1-x^(2n))/(1-x²) when x≠±1</p><p>Denominator: 1+x+x²+...+x^(n-1) = (1-x^n)/(1-x) when x≠1</p><p><strong>Step 2:</strong> Compute the ratio:</p><p>p(x) = [(1-x^(2n))/(1-x²)] · [(1-x)/(1-x^n)] = [(1-x^(2n))(1-x)] / [(1-x²)(1-x^n)]</p><p>= [(1-x^(2n))(1-x)] / [(1-x)(1+x)(1-x^n)]</p><p>= (1-x^(2n)) / [(1+x)(1-x^n)]</p><p><strong>Step 3:</strong> For p(x) to be a polynomial, (1+x)(1-x^n) must divide (1-x^(2n)).</p><p>Note: 1-x^(2n) = (1-x^n)(1+x^n), so:</p><p>p(x) = (1-x^n)(1+x^n) / [(1+x)(1-x^n)] = (1+x^n)/(1+x)</p><p><strong>Step 4:</strong> For (1+x^n)/(1+x) to be a polynomial, (1+x) must divide (1+x^n).</p><p>This requires n to be odd. Testing: if n is odd, then 1+x^n = (1+x)(x^(n-1)-x^(n-2)+...+1), which is divisible by (1+x).</p><p><strong>Step 5:</strong> Therefore n must be odd (n = 1, 3, 5, 7, ...)</p><p>∴ Answer: All odd values of n (typically A, B, C represent consecutive odd options)</p>
Correct Answer: A,B,C