<p>Let \(\log_2 n = m\) so that \(n = 2^m\). Then \(\displaystyle\prod_{k=1}^{m}(x^{2^{m-k}}+1)\) equals \(\dfrac{x^4 - B}{x - C}\). Find the values of <em>B</em> and <em>C</em>.</p>
Step-by-Step Solution
Key Concept: Recognize the telescoping product pattern: multiplying (x-1)(x+1)(x²+1)(x⁴+1)... equals (x^(2^n)-1)/(x-1). This comes from the algebraic identity (x-1)(x+1) = x²-1, which generalizes through repeated application.
<p><strong>Step 1:</strong> Use the algebraic identity (a-b)(a+b) = a²-b². Notice that:</p><p>(x-1)(x+1)(x²+1)(x⁴+1)...(x^(2^(m-1))+1) telescopes to give (x^(2^m)-1)/(x-1)</p><p><strong>Step 2:</strong> Our product is ∏_{k=1}^{m}(x^(2^(m-k))+1). Rewriting with j = m-k, when k goes from 1 to m, j goes from m-1 down to 0:</p><p>∏_{j=0}^{m-1}(x^(2^j)+1) = (x+1)(x²+1)(x⁴+1)...(x^(2^(m-1))+1)</p><p><strong>Step 3:</strong> Multiply and divide by (x-1):</p><p>∏_{j=0}^{m-1}(x^(2^j)+1) = [(x-1)·∏_{j=0}^{m-1}(x^(2^j)+1)]/(x-1)</p><p><strong>Step 4:</strong> The numerator telescopes: (x-1)(x+1)(x²+1)...(x^(2^(m-1))+1) = x^(2^m)-1</p><p>∴ ∏_{k=1}^{m}(x^(2^(m-k))+1) = (x^(2^m)-1)/(x-1)</p><p><strong>Step 5:</strong> Since n = 2^m, we have 2^m = n, so:</p><p>∏_{k=1}^{m}(x^(2^(m-k))+1) = (x^n-1)/(x-1)</p><p><strong>Step 6:</strong> Comparing with (x⁴-B)/(x-C): we need the problem statement clarified, but if the answer form matches (x^n-1)/(x-1), then <strong>B = 1 and C = 1</strong></p>
Correct Answer: A