<p>Consider: <span>\((\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n}\)</span>. Which of the following is true?</p>
Step-by-Step Solution
Key Concept: When you expand using binomial theorem, all even-powered terms in both expressions are identical and cancel out, leaving only odd-powered terms with irrational parts. The result is always even since odd-powered terms from conjugate expressions combine to give rational multiples of 2.
<p><strong>Step 1:</strong> Apply binomial theorem to both terms:</p><p>$(\sqrt{3}+1)^{2n} = \sum_{r=0}^{2n} \binom{2n}{r}(\sqrt{3})^{2n-r}(1)^r$</p><p>$(\sqrt{3}-1)^{2n} = \sum_{r=0}^{2n} \binom{2n}{r}(\sqrt{3})^{2n-r}(-1)^r$</p><p><strong>Step 2:</strong> Subtract the expressions. Terms with even $r$ have $(-1)^r = 1$ (same in both), so they cancel. Terms with odd $r$ have $(-1)^r = -1$, so they double:</p><p>$(\sqrt{3}+1)^{2n} - (\sqrt{3}-1)^{2n} = 2\sum_{\text{odd } r} \binom{2n}{r}(\sqrt{3})^{2n-r}$</p><p><strong>Step 3:</strong> Each odd $r$ term has $(\sqrt{3})^{2n-r}$ where $2n-r$ is odd (even minus odd). This makes the irrational part cancel in summation, yielding an even integer multiple of 2.</p><p><strong>Conclusion:</strong> The expression is always an even integer (divisible by 2).</p><p>∴ Answer: A</p>
Correct Answer: A