<p>The sum of the series \(\dfrac{1^3}{1} + \dfrac{1^3+2^3}{1+3} + \dfrac{1^3+2^3+3^3}{1+3+5} + \cdots\) up to 9 terms is:</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator is the sum of cubes (which equals the square of sum of natural numbers) and the denominator is the sum of first n odd numbers (which equals n²). This creates a telescoping or simplifiable pattern.
<p><strong>Step 1:</strong> Identify the general term. For the nth term:</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 nth 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> Sum the series up to 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)[2² + 3² + 4² + ... + 10²]</p><p>= (1/4)[Σ(k=1 to 10) k² - 1²]</p><p>= (1/4)[10·11·21/6 - 1]</p><p>= (1/4)[385 - 1] = (1/4)·384 = 96</p><p>∴ Answer: A (96)</p>
Correct Answer: A