<p>If the sum of the first <i>n</i> terms of the series \(\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \cdots\) is \(435\sqrt{3}\), then <i>n</i> equals</p>
Step-by-Step Solution
Key Concept: Recognize that each term can be simplified by factoring out perfect squares: √3, √(75)=5√3, √(243)=9√3, √(507)=13√3, revealing an arithmetic sequence with first term a=√3 and common difference d=4√3. Use the arithmetic sum formula Sₙ = n/2[2a + (n-1)d].
<p><strong>Step 1:</strong> Simplify each term by factoring perfect squares:</p><p>• √3 = √3</p><p>• √75 = √(25×3) = 5√3</p><p>• √243 = √(81×3) = 9√3</p><p>• √507 = √(169×3) = 13√3</p><p>The series is: √3 + 5√3 + 9√3 + 13√3 + ...</p><p><strong>Step 2:</strong> Factor out √3: √3(1 + 5 + 9 + 13 + ...)</p><p>The sequence in parentheses is arithmetic with first term a = 1 and common difference d = 4.</p><p><strong>Step 3:</strong> The nth term is 1 + (n-1)×4 = 4n - 3</p><p><strong>Step 4:</strong> Sum of arithmetic series: Sₙ = n/2[2a + (n-1)d] = n/2[2(1) + (n-1)×4] = n/2[2 + 4n - 4] = n/2[4n - 2] = n(2n - 1)</p><p><strong>Step 5:</strong> Therefore: √3 × n(2n - 1) = 435√3</p><p>n(2n - 1) = 435</p><p>2n² - n - 435 = 0</p><p><strong>Step 6:</strong> Using the quadratic formula or factoring: (2n + 29)(n - 15) = 0</p><p>Since n must be positive: n = 15</p><p><strong>Verification:</strong> 15(2×15 - 1) = 15(29) = 435 ✓</p><p>∴ Answer: <strong>D (n = 15)</strong></p>
Correct Answer: D