<p>The sum to the infinity of the series \(1 + \dfrac{2}{3} + \dfrac{6}{3^2} + \dfrac{10}{3^3} + \dfrac{14}{3^4} + \cdots\) is</p>
Step-by-Step Solution
Key Concept: Recognize that the numerators form an arithmetic progression (1, 2, 6, 10, 14...) which can be written as 4n-3. Split the series into two separate geometric series using this pattern.
<p><strong>Step 1:</strong> Identify the numerator pattern. The numerators are 1, 2, 6, 10, 14,... which differ by 4 (after the first term). We can write: 1, 2, 6, 10, 14 = 1 + 0·4, 2 + 0·4, 2 + 4·1, 2 + 4·2, 2 + 4·3,...</p><p>More systematically: numerators = 4n - 3 for n = 1, 2, 3,...</p><p><strong>Step 2:</strong> Write the series as:</p><p>S = Σ(4n-3)/3^(n-1) for n=1 to ∞ = Σ(4n-3)·(1/3)^(n-1)</p><p><strong>Step 3:</strong> Split into two series:</p><p>S = 4Σn·(1/3)^(n-1) - 3Σ(1/3)^(n-1)</p><p><strong>Step 4:</strong> Evaluate each series.</p><p>For Σ(1/3)^(n-1) = 1/(1-1/3) = 3/2</p><p>For Σn·(1/3)^(n-1): Let T = Σn·x^(n-1) where x = 1/3. Since Σx^(n-1) = 1/(1-x), differentiating: T = d/dx[1/(1-x)] = 1/(1-x)² = 1/(2/3)² = 9/4</p><p><strong>Step 5:</strong> Combine: S = 4(9/4) - 3(3/2) = 9 - 9/2 = 9/2</p><p>∴ Answer: B</p>
Correct Answer: B