<p>The sum of first 9 terms of the series \(\dfrac{1^3}{1} + \dfrac{1^3+2^3}{1+3} + \dfrac{1^3+2^3+3^3}{1+3+5} + \cdots\) is</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator is the sum of cubes formula [n(n+1)/2]² and the denominator is the sum of first n odd numbers = n². This simplifies each term to [(n(n+1)/2)²]/n² = [n²(n+1)²/4]/n² = (n+1)²/4.
<p><strong>Step 1:</strong> Identify the general term. The nth term has:</p><p>Numerator: 1³ + 2³ + 3³ + ... + n³ = [n(n+1)/2]²</p><p>Denominator: 1 + 3 + 5 + ... + (2n-1) = n²</p><p><strong>Step 2:</strong> Simplify the general term:</p><p>aₙ = [n(n+1)/2]²/n² = [n²(n+1)²/4]/n² = (n+1)²/4</p><p><strong>Step 3:</strong> Find the sum of first 9 terms:</p><p>S₉ = Σ(n=1 to 9) (n+1)²/4 = (1/4)Σ(n=1 to 9)(n+1)²</p><p>= (1/4)Σ(m=2 to 10) m² = (1/4)[Σ(m=1 to 10) m² - 1]</p><p><strong>Step 4:</strong> Apply formula Σm² = n(n+1)(2n+1)/6:</p><p>Σ(m=1 to 10) m² = 10(11)(21)/6 = 385</p><p>S₉ = (1/4)(385 - 1) = 384/4 = 96</p><p>∴ Answer: <strong>96</strong></p>
Correct Answer: A