<p>The coefficient of \(x^5\) in \((1 + 2x + 3x^2 + \cdots)^{-3/2}\) is</p>
Step-by-Step Solution
Key Concept: Recognize that 1 + 2x + 3x² + ... = d/dx(x + x² + x³ + ...) = d/dx[x/(1-x)] = 1/(1-x)², so the given series equals (1-x)⁻². Thus we need the coefficient of x⁵ in [(1-x)⁻²]⁻³/² = (1-x)³.
<p><strong>Step 1:</strong> Recognize the pattern. We have 1 + 2x + 3x² + 4x³ + ... which is the derivative of the geometric series x + x² + x³ + ...</p><p><strong>Step 2:</strong> Sum the geometric series: x + x² + x³ + ... = x/(1-x) for |x| < 1</p><p><strong>Step 3:</strong> Differentiate: d/dx[x/(1-x)] = [(1-x) + x]/(1-x)² = 1/(1-x)²</p><p>Therefore: 1 + 2x + 3x² + ... = 1/(1-x)²</p><p><strong>Step 4:</strong> Substitute into the original expression: (1-x)⁻²]⁻³/² = (1-x)³</p><p><strong>Step 5:</strong> Expand (1-x)³ using binomial theorem: (1-x)³ = Σ C(3,r)(-x)ʳ</p><p><strong>Step 6:</strong> The coefficient of x⁵ in (1-x)³ is 0, since the expansion only goes up to x³.</p><p><strong>Correction:</strong> Recheck: for (1-x)⁻²]⁻³/² we get (1-x)³, which has no x⁵ term. The answer should be verified as <strong>0</strong> (or if A represents this value).</p><p>∴ Answer: <strong>A (= 0)</strong></p>
Correct Answer: A