<p>The coefficient of \(x^n\) in the expansion of \((1 + x + x^2 + x^3 + \cdots)^2\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the infinite geometric series 1 + x + x² + x³ + ... = 1/(1-x) for |x| < 1, so we need the coefficient of x^n in [1/(1-x)]². Use the binomial series expansion of (1-x)^(-2).
<p><strong>Step 1:</strong> Recognize the infinite geometric series:</p><p>1 + x + x² + x³ + ... = 1/(1-x) for |x| < 1</p><p><strong>Step 2:</strong> Rewrite the problem:</p><p>(1 + x + x² + ...)² = [1/(1-x)]² = (1-x)^(-2)</p><p><strong>Step 3:</strong> Apply the binomial series expansion for (1-x)^(-2):</p><p>(1-x)^(-2) = Σ C(n+1, 1) x^n = Σ (n+1)x^n</p><p>Using the generalized binomial coefficient: C(-2, n) = (-2)(-3)...(-2-n+1)/n! = (-1)^n(n+1)(-1)^n/n! = (n+1)</p><p><strong>Step 4:</strong> The coefficient of x^n is:</p><p>Coefficient = n + 1</p><p>∴ Answer: <strong>n + 1</strong></p>
Correct Answer: D