Correct Answer: √3 n(n+1) / 2
Subject: 2025
Explanation: <p><strong>Key Fact:</strong> Simplify each radical term: • First term: $a_1 = \sqrt{3} = 1\sqrt{3}$ • Second term: $a_2 = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ • Third term: $a_3 = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ • Fourth term: $a_4 = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$</p><p><strong>Step-by-step Calculation:</strong><br>Sum of the first $n$ terms: $$S_n = 1\sqrt{3} + 2\sqrt{3} + 3\sqrt{3} + \dots + n\sqrt{3}$$ $$S_n = \sqrt{3}(1 + 2 + 3 + \dots + n)$$ Since the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$: $$S_n = \sqrt{3} \times \frac{n(n+1)}{2} = \frac{\sqrt{3}\,n(n+1)}{2}$$</p><p><strong>Additional Info:</strong> This is a standard Arithmetic Progression with first term $a = \sqrt{3}$ and common difference $d = \sqrt{3}$.</p><p><strong>Why the other options are wrong:</strong> Option B misses the $\sqrt{3}$ factor; Option C divides by $\sqrt{3}$; Option D multiplies by an extraneous factor of 4.</p><p><strong>Exam Trick:</strong> Factor out $\sqrt{3}$ immediately: $\sqrt{3} \times \Sigma n = \sqrt{3}\frac{n(n+1)}{2}$.</p><p><strong>Final Answer:</strong> √3 n(n+1)/2</p>