<p>\(\displaystyle\sum_{r=0}^{300} a_r x^r = (1+x+x^2+x^3)^{100}\). If \(a = \displaystyle\sum_{r=0}^{300} a_r\), then \(\displaystyle\sum_{r=0}^{300} r\, a_r\) is equal to</p>
Step-by-Step Solution
Key Concept: To find ∑r·aᵣ, differentiate the generating function (1+x+x²+x³)¹⁰⁰ and evaluate at x=1. This converts the coefficient problem into a derivative evaluation problem.
<p><strong>Step 1:</strong> Recognize that ∑ᵣ₌₀³⁰⁰ r·aᵣ = x·(d/dx)[(1+x+x²+x³)¹⁰⁰]|ₓ₌₁</p><p><strong>Step 2:</strong> Simplify 1+x+x²+x³ = (1-x⁴)/(1-x), so (1+x+x²+x³)¹⁰⁰ = [(1-x⁴)/(1-x)]¹⁰⁰</p><p><strong>Step 3:</strong> Differentiate using chain and quotient rules:<br/>d/dx[(1-x⁴)/(1-x)]¹⁰⁰ = 100[(1-x⁴)/(1-x)]⁹⁹ · d/dx[(1-x⁴)/(1-x)]</p><p><strong>Step 4:</strong> d/dx[(1-x⁴)/(1-x)] = [(-4x³)(1-x) - (1-x⁴)(-1)]/(1-x)² = [-4x³ + 4x⁴ + 1 - x⁴]/(1-x)² = (1 - 4x³ + 3x⁴)/(1-x)²</p><p><strong>Step 5:</strong> At x=1: (1+x+x²+x³)¹⁰⁰|ₓ₌₁ = 4¹⁰⁰. For the derivative at x=1, use L'Hôpital's rule or direct substitution in factored form. The derivative term at x=1 evaluates using the limiting behavior: (1 - 4x³ + 3x⁴)/(1-x)² → need careful analysis.</p><p><strong>Step 6:</strong> More directly: ∑ᵣ₌₀³⁰⁰ r·aᵣ = 100·4⁹⁹·3 = 300·4⁹⁹ (by evaluating d/dx at x=1 where numerator gives 1-4+3=0, requiring derivative of numerator and denominator)</p><p>∴ Answer: <strong>300·4⁹⁹</strong> or equivalent form (C)</p>
Correct Answer: C