Basic Mathematics & Logarithm
Logarithmic expressions and products
Grade 11
Question:
<p>Let \(\log_2 n\) is an integer. If \(\displaystyle\prod_{k=1}^{\log_2 n}\left(x^{\frac{n}{2^k}} + 1\right) = \dfrac{x^A - B}{x - C}\), where \(A\), \(B\) and \(C\) are positive integers. Then the value of \((B + C + \log_2 A)\) for \(n = 2^{92}\) is:</p>
<p>(a) 90</p>
<p>(b) 92</p>
<p>(c) 94</p>
<p>(d) 100</p>
Step-by-Step Solution
Key Concept: Use the algebraic identity (a-b)(a+b) = a²-b² repeatedly by telescoping: multiply numerator and denominator by (x^(n/2^k) - 1) to collapse the product into a simple rational form.
<p><strong>Step 1:</strong> Recognize the telescoping pattern. Multiply and divide by (x^(n/2⁰) - 1) = (x^n - 1):</p><p>∏ₖ₌₁^(log₂ n) (x^(n/2ᵏ) + 1) = [(x^n - 1)] / [(x^(n/2⁰) - 1)] × ∏ₖ₌₁^(log₂ n) (x^(n/2ᵏ) + 1) / (x^n - 1)</p><p><strong>Step 2:</strong> Apply the identity (a - b)(a + b) = a² - b² telescopically:</p><p>(x^n - 1) = (x^(n/2) - 1)(x^(n/2) + 1)</p><p>(x^(n/2) - 1) = (x^(n/4) - 1)(x^(n/4) + 1), and so on...</p><p><strong>Step 3:</strong> The product telescopes:</p><p>∏ₖ₌₁^(log₂ n) (x^(n/2ᵏ) + 1) = (x^n - 1)/(x - 1)</p><p><strong>Step 4:</strong> For n = 2⁹²: A = n = 2⁹², B = 1, C = 1</p><p>Therefore: B + C + log₂ A = 1 + 1 + log₂(2⁹²) = 1 + 1 + 92 = <strong>94</strong></p><p>∴ Answer: C</p>
Correct Answer: C