<p>Given that <em>P(n)</em>: 2n < n!, ∀n ∈ N. For which value of n is P(n) first true? The value of the filler is:</p>
Step-by-Step Solution
Key Concept: Test small natural numbers systematically to find the first value where 2^n < n! holds true, recognizing that factorial grows faster than exponential for sufficiently large n.
<p><strong>Step 1:</strong> Test n = 1: P(1) states 2¹ < 1! → 2 < 1, which is FALSE</p><p><strong>Step 2:</strong> Test n = 2: P(2) states 2² < 2! → 4 < 2, which is FALSE</p><p><strong>Step 3:</strong> Test n = 3: P(3) states 2³ < 3! → 8 < 6, which is FALSE</p><p><strong>Step 4:</strong> Test n = 4: P(4) states 2⁴ < 4! → 16 < 24, which is TRUE ✓</p><p><strong>Step 5:</strong> Verify for n ≥ 4, factorial dominates exponential (n! grows faster than 2ⁿ)</p><p>∴ P(n) is first true at <strong>n = 4</strong></p>
Correct Answer: C