<p>If \(n\) is a positive integer, then \((\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n}\) is</p>
<p>an irrational number</p>
<p>an odd positive integer</p>
<p>an even positive integer</p>
<p>a rational number other than positive integers</p>
Step-by-Step Solution
Key Concept: Recognize that (√3+1)^(2n) - (√3-1)^(2n) involves a difference of even powers. Expand using binomial theorem and observe that all rational terms cancel, leaving only irrational terms with √3.
<p><strong>Step 1:</strong> Expand using binomial theorem:</p><p>(√3+1)^(2n) = Σ C(2n,r)(√3)^r(1)^(2n-r)</p><p>(√3-1)^(2n) = Σ C(2n,r)(√3)^r(-1)^(2n-r)</p><p><strong>Step 2:</strong> When subtracting, terms with even r have the same sign in both expansions (since (-1)^(2n-r) = 1 when 2n-r is even), so they cancel.</p><p><strong>Step 3:</strong> Terms with odd r have opposite signs in both expansions (since (-1)^(2n-r) = -1 when 2n-r is odd), so they double when subtracted.</p><p><strong>Step 4:</strong> The remaining sum contains only odd powers of √3, each multiplied by 2·C(2n,r), giving: 2[C(2n,1)√3 + C(2n,3)(√3)^3 + C(2n,5)(√3)^5 + ...]</p><p><strong>Step 5:</strong> Every term is of the form (integer)·√3, making the entire expression a multiple of 2√3 (always irrational and divisible by 2).</p><p>∴ Answer: A (An even multiple of √3 / Always irrational and even)</p>
Correct Answer: A