Binomial Theorem
Divisibility using Mathematical Induction
Grade 11
Question:
<p>For every integer <em>n</em> ≥ 1, which one is correct related to divisibility of \((3^{2^n} - 1)\)?</p>
<p>divisible by \(2^{n+2}\) but not by \(2^{n+3}\)</p>
<p>divisible by \(2^{n+2}\) and \(2^{n+3}\)</p>
<p>divisible by \(2^{n+3}\) but not by \(2^{n+2}\)</p>
<p>not divisible by \(2^{n+2}\) and \(2^{n+3}\)</p>
Step-by-Step Solution
Key Concept: Factor using difference of squares repeatedly: 3^(2^n) - 1 = (3^(2^(n-1)) - 1)(3^(2^(n-1)) + 1). Track which primes divide each factor at each stage to find the pattern of divisibility.
<p><strong>Step 1:</strong> Use difference of squares on 3^(2^n) - 1:</p><p>3^(2^n) - 1 = (3^(2^(n-1)))^2 - 1 = (3^(2^(n-1)) - 1)(3^(2^(n-1)) + 1)</p><p><strong>Step 2:</strong> Check divisibility for small values:</p><p>• n=1: 3^2 - 1 = 8 = 2^3 (divisible by 8)</p><p>• n=2: 3^4 - 1 = 80 = (3^2-1)(3^2+1) = 8 × 10 (divisible by 16)</p><p>• n=3: 3^8 - 1 = (3^4-1)(3^4+1) = 80 × 82 (divisible by 32)</p><p><strong>Step 3:</strong> Observe the pattern: 3^(2^(n-1)) - 1 is always even, and 3^(2^(n-1)) + 1 is always even (since 3^(2^(n-1)) is odd). Each recursive step contributes factors of 2.</p><p><strong>Step 4:</strong> By induction, 3^(2^n) - 1 is divisible by 2^(n+2) for every integer n ≥ 1.</p><p>∴ Answer: A (or verify the option stating divisibility by 2^(n+2))</p>
Correct Answer: A