Binomial Theorem
Vieta's formulas and binomial coefficients
Grade 11
Question:
<p><strong>For Problems 7–9:</strong> An equation \(a_0 + a_1 x + a_2 x^2 + \cdots + a_{99} x^{99} + x^{100} = 0\) has roots \({}^{99}C_0, {}^{99}C_1, {}^{99}C_2, \ldots, {}^{99}C_{99}\).</p><p><strong>7.</strong> The value of \(a_{99}\) is equal to</p>
<p>(1) \(2^{98}\)</p>
<p>(2) \(2^{99}\)</p>
<p>(3) \(-2^{99}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: By Vieta's formulas, the coefficient of x in a monic polynomial equals the negative sum of all roots. Here, a₉₉ = -(sum of all binomial coefficients), which equals -2⁹⁹ by the binomial theorem.
<p><strong>Step 1:</strong> The polynomial is monic with leading coefficient 1 and roots r₁ = ⁹⁹C₀, r₂ = ⁹⁹C₁, ..., r₁₀₀ = ⁹⁹C₉₉.</p><p><strong>Step 2:</strong> By Vieta's formulas for a monic polynomial of degree 100, the coefficient of x⁹⁹ (which is a₉₉) equals the negative sum of all roots:</p><p>a₉₉ = -(⁹⁹C₀ + ⁹⁹C₁ + ⁹⁹C₂ + ... + ⁹⁹C₉₉)</p><p><strong>Step 3:</strong> By the binomial theorem, (1 + 1)⁹⁹ = Σ ⁹⁹Cₖ = 2⁹⁹</p><p><strong>Step 4:</strong> Therefore, a₉₉ = -2⁹⁹</p><p><strong>Note:</strong> If the answer stated is 3, this likely refers to a simplified form or different parameterization of the problem. The standard answer is a₉₉ = -2⁹⁹.</p>
Correct Answer: 3