<p>Let \(S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \cdots\). Find the value of \(S\).</p>
Step-by-Step Solution
Key Concept: The numerators form an arithmetic sequence (1, 2, 6, 10, 14, ...) with differences 1, 4, 6, 8,... requiring decomposition into two series: one geometric and one obtained via differentiation of geometric series.
<p><strong>Step 1:</strong> Identify the pattern in numerators: 1, 2, 6, 10, 14,... The differences are 1, 4, 4, 4,... suggesting we rewrite:</p><p><strong>Step 2:</strong> Regroup as: S = 1 + (2/3 + 6/3² + 10/3³ + ...). For the series starting from second term, numerators are 2, 6, 10, 14,... = 4n-2 (n≥1)</p><p><strong>Step 3:</strong> So S = 1 + Σ(4n-2)/3ⁿ for n=1 to ∞ = 1 + 4Σn/3ⁿ - 2Σ1/3ⁿ</p><p><strong>Step 4:</strong> Use Σ1/3ⁿ = 1/2 and find Σn/3ⁿ. Let T = Σn·xⁿ = x/(1-x)² when |x| < 1. At x=1/3: Σn/3ⁿ = (1/3)/(2/3)² = 3/4</p><p><strong>Step 5:</strong> Therefore: S = 1 + 4(3/4) - 2(1/2) = 1 + 3 - 1 = <strong>3</strong></p><p>∴ Answer: B</p>
Correct Answer: B