<p>If the sum to infinity of the series \(1 + 2r + 3r^2 + 4r^3 + \cdots\) is 9/4, then value of <em>r</em> is</p>
Step-by-Step Solution
Key Concept: Recognize this as a derivative of geometric series: if S = 1 + r + r² + r³ + ... = 1/(1-r), then dS/dr gives 1 + 2r + 3r² + 4r³ + ... = 1/(1-r)². Use this formula with the given sum to find r.
<p><strong>Step 1:</strong> Identify the series pattern. The series 1 + 2r + 3r² + 4r³ + ... has coefficients that are natural numbers.</p><p><strong>Step 2:</strong> Recognize this comes from differentiating the geometric series. Starting with:</p><p>1 + r + r² + r³ + ... = 1/(1-r) for |r| < 1</p><p><strong>Step 3:</strong> Differentiate both sides with respect to r:</p><p>d/dr[1 + r + r² + r³ + ...] = d/dr[1/(1-r)]</p><p>1 + 2r + 3r² + 4r³ + ... = 1/(1-r)²</p><p><strong>Step 4:</strong> Use the given condition that the sum equals 9/4:</p><p>1/(1-r)² = 9/4</p><p><strong>Step 5:</strong> Solve for r:</p><p>(1-r)² = 4/9</p><p>1-r = ±2/3</p><p><strong>Step 6:</strong> Consider convergence condition |r| < 1:</p><p>If 1-r = 2/3, then r = 1/3 ✓ (satisfies |r| < 1)</p><p>If 1-r = -2/3, then r = 5/3 ✗ (violates |r| < 1)</p><p><strong>Verification:</strong> 1/(1-1/3)² = 1/(2/3)² = 9/4 ✓</p><p>∴ Answer: B (r = 1/3)</p>
Correct Answer: B