<p>The coefficient of \(x^{50}\) in the expansion of \((1+x)^{1000} + 2x(1+x)^{999} + 3x^2(1+x)^{998} + \cdots + 1001 x^{1000}\) is:</p>
Step-by-Step Solution
Key Concept: Recognize the series as the derivative of a geometric series: the sum equals d/dx[(1+x)^1001 - x^1001]/(1-x), which simplifies to (1001(1+x)^1000 - 1001x^1000)/(1-x)^2. Extract the coefficient of x^50 using binomial expansion.
<p><strong>Step 1:</strong> Rewrite the series as ∑(k=0 to 1000) (k+1)x^k(1+x)^(1000-k).</p><p><strong>Step 2:</strong> Factor out (1+x)^1000 from a shifted index perspective. Recognize this as the derivative: d/dx[∑(k=0 to 1000) x^(k+1)(1+x)^(1000-k)] = d/dx[x(1+x)^1001 - x^1001)/(1-x)].</p><p><strong>Step 3:</strong> Simplify using the quotient rule or recognize the closed form is (1001(1+x)^1000 - 1001x^1000)/(1-x)^2.</p><p><strong>Step 4:</strong> The coefficient of x^50 comes from: (1001)[coeff of x^50 in (1+x)^1000 - 0] × [coeff of x^0 in 1/(1-x)^2] + (1001)[coeff of x^49 in (1+x)^1000] × [coeff of x^1 in 1/(1-x)^2].</p><p><strong>Step 5:</strong> This equals 1001·C(1000,50)·1 + 1001·C(1000,49)·2 = 1001[C(1000,50) + 2C(1000,49)] = 1001·C(1001,50).</p><p>∴ Answer: D</p>
Correct Answer: D