<p>Let \(a = \displaystyle\sum_{r=1}^{\infty} \dfrac{1}{r^2}\) and \(b = \displaystyle\sum_{r=1}^{\infty} \dfrac{1}{(2r-1)^2}\). Then the value of \(\dfrac{3a}{b}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Separate the series 'a' into odd and even terms: a = b + (sum of 1/(2r)² terms). The even terms sum equals a/4, giving you b = 3a/4, which directly yields 3a/b = 4.
<p><strong>Step 1:</strong> Write out series a by separating odd and even indexed terms:</p><p>a = ∑(r=1 to ∞) 1/r² = ∑(r=1 to ∞) 1/(2r-1)² + ∑(r=1 to ∞) 1/(2r)²</p><p><strong>Step 2:</strong> Recognize that the first sum on the right is exactly b:</p><p>a = b + ∑(r=1 to ∞) 1/(4r²)</p><p><strong>Step 3:</strong> Factor out from the second sum:</p><p>a = b + (1/4)∑(r=1 to ∞) 1/r² = b + (1/4)a</p><p><strong>Step 4:</strong> Solve for b in terms of a:</p><p>a - (1/4)a = b</p><p>(3/4)a = b</p><p><strong>Step 5:</strong> Calculate 3a/b:</p><p>3a/b = 3a/[(3/4)a] = 3a × (4/3a) = 4</p><p>∴ Answer: C (which equals 4)</p>
Correct Answer: C