<p>The number of real negative terms in the binomial expansion of \((1 + ix)^{4n-2}\), \(n \in \mathbb{N}\), \(x > 0\) is</p>
Step-by-Step Solution
Key Concept: Terms in the binomial expansion of (1 + ix)^(4n-2) are real when the power of i is even (i.e., i^(2k) = (-1)^k). These terms are negative when (-1)^k < 0, which occurs when k is odd. Count the odd values of k from 0 to 4n-2.
<p><strong>Step 1:</strong> General term in expansion of (1 + ix)^(4n-2) is:</p><p>T_(r+1) = C(4n-2, r)(ix)^r = C(4n-2, r)i^r x^r</p><p><strong>Step 2:</strong> For terms to be real, i^r must be real. This happens when r ≡ 0 (mod 4) or r ≡ 2 (mod 4).</p><p>When r ≡ 0 (mod 4): i^r = 1 (positive)</p><p>When r ≡ 2 (mod 4): i^r = i^2 = -1 (negative coefficient)</p><p><strong>Step 3:</strong> Count values of r with r ≡ 2 (mod 4) where 0 ≤ r ≤ 4n-2:</p><p>These are: r = 2, 6, 10, 14, ..., 4n-2</p><p>This is an arithmetic sequence with first term a = 2, common difference d = 4, last term l = 4n-2</p><p><strong>Step 4:</strong> Number of terms = (4n-2-2)/4 + 1 = (4n-4)/4 + 1 = n-1+1 = n</p><p>∴ Answer: A (The answer is <strong>n</strong>)</p>
Correct Answer: A