<p>The sum of first nine 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: 1³+2³+...+n³ = [n(n+1)/2]², and the denominator is the sum of first n odd numbers = n². This simplifies each term dramatically.
<p><strong>Step 1:</strong> Identify the general term pattern.</p><p>Numerator of nth term: 1³ + 2³ + 3³ + ... + n³ = [n(n+1)/2]²</p><p>Denominator of nth term: 1 + 3 + 5 + ... + (2n-1) = n²</p><p><strong>Step 2:</strong> Simplify the nth term.</p><p>T_n = [n(n+1)/2]² / n² = [n²(n+1)²/4] / n² = (n+1)²/4</p><p><strong>Step 3:</strong> Calculate 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)[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]</p><p>= (1/4)(384)</p><p>= 96</p><p>∴ Answer: D</p>
Correct Answer: D