<p>Let \(a = \displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2}\) and \(b = \displaystyle\sum_{r=1}^{\infty} \frac{1}{(2r-1)^2}\). Then the value of \(\dfrac{3a}{b}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Split the series a into odd and even indexed terms: a = Σ(1/r²) = Σ(1/(2k-1)²) + Σ(1/(2k)²) = b + (1/4)a, which gives the relationship between a and b.
<p><strong>Step 1:</strong> Recognize that the infinite series a contains all positive integers, while b contains only odd-indexed denominators.</p><p><strong>Step 2:</strong> Decompose a into odd and even terms:</p><p>a = Σ(r=1 to ∞) 1/r² = Σ(odd r) 1/r² + Σ(even r) 1/r²</p><p>a = b + Σ(k=1 to ∞) 1/(2k)²</p><p><strong>Step 3:</strong> Simplify the even terms:</p><p>Σ(k=1 to ∞) 1/(2k)² = Σ(k=1 to ∞) 1/(4k²) = (1/4)Σ(k=1 to ∞) 1/k² = (1/4)a</p><p><strong>Step 4:</strong> Substitute back:</p><p>a = b + (1/4)a</p><p>a - (1/4)a = b</p><p>(3/4)a = b</p><p><strong>Step 5:</strong> Find 3a/b:</p><p>3a/b = 3a/[(3/4)a] = 3a · (4/3a) = 4</p><p>∴ Answer: C (which equals <strong>4</strong>)</p>
Correct Answer: C