<p>The value of \({}^nC_1 x(1-x)^{n-1} + 2 \cdot {}^nC_2 x^2(1-x)^{n-2} + 3 \cdot {}^nC_3 x^3(1-x)^{n-3} + \ldots + n \cdot {}^nC_n x^n\) is</p>
Step-by-Step Solution
Key Concept: Recognize this sum as the derivative of the binomial expansion (1-x+x)^n with respect to x, then apply the binomial theorem strategically.
<p><strong>Step 1:</strong> Recognize the sum pattern. Note that the coefficient r·ⁿCᵣ appears before xʳ(1-x)ⁿ⁻ʳ. This suggests we consider the binomial expansion:</p><p>(1-x+x)ⁿ = 1 (trivially)</p><p><strong>Step 2:</strong> Instead, rewrite: Let S = Σ r·ⁿCᵣ xʳ(1-x)ⁿ⁻ʳ. Consider the binomial expansion:</p><p>(y + x)ⁿ = Σ ⁿCᵣ yⁿ⁻ʳ xʳ where y = (1-x)</p><p><strong>Step 3:</strong> Differentiate both sides of (1-x+x)ⁿ = 1 with respect to x is trivial. Instead, differentiate (y+x)ⁿ where y=(1-x):</p><p>d/dx[(1-x+x)ⁿ] = d/dx[1ⁿ] = 0, but this approach fails.</p><p><strong>Step 4:</strong> Take f(x) = Σ ⁿCᵣ xʳ(1-x)ⁿ⁻ʳ = [(1-x)+x]ⁿ = 1</p><p>Differentiate: f'(x) = Σ r·ⁿCᵣ xʳ⁻¹(1-x)ⁿ⁻ʳ - Σ (n-r)·ⁿCᵣ xʳ(1-x)ⁿ⁻ʳ⁻¹ = 0</p><p><strong>Step 5:</strong> Multiply f'(x) by x and (1-x): xf'(x) = Σ r·ⁿCᵣ xʳ(1-x)ⁿ⁻ʳ = n·x·(1-x+x)ⁿ⁻¹ = nx</p><p>∴ Answer: <strong>nx</strong> (or Option C if nx is provided)</p>
Correct Answer: C