<p>Find the coefficient of \(x^{98}\) in the continued product \((x+1)(x+2)(x+3)\cdots(x+100)\).</p>
Step-by-Step Solution
Key Concept: The coefficient of x^98 in (x+1)(x+2)...(x+100) equals the sum of all products of 2 distinct numbers chosen from {1,2,...,100}, which is the elementary symmetric polynomial e₂(1,2,...,100).
<p><strong>Step 1:</strong> Recognize that (x+1)(x+2)...(x+100) is a monic polynomial of degree 100. When expanded, the coefficient of x^k comes from elementary symmetric polynomials in {1,2,...,100}.</p><p><strong>Step 2:</strong> The coefficient of x^98 corresponds to e₂, the sum of all products of pairs: we need to choose 2 numbers from {1,2,...,100} and multiply them, then sum all such products.</p><p><strong>Step 3:</strong> Use the identity: e₂ = ½[(∑aᵢ)² - ∑aᵢ²]</p><p>where ∑aᵢ = 1+2+...+100 = 100(101)/2 = 5050</p><p><strong>Step 4:</strong> Calculate ∑aᵢ² = 1² + 2² + ... + 100² = 100(101)(201)/6 = 338350</p><p><strong>Step 5:</strong> Compute e₂ = ½[(5050)² - 338350] = ½[25502500 - 338350] = ½(25164150) = 12582075</p><p>∴ Answer: 12582075</p>
Correct Answer: 12582075