Binomial Theorem
Alternating sum of binomial coefficients
Grade 11
Question:
<p>The sum of series \({}^{20}C_0 - {}^{20}C_1 + {}^{20}C_2 - {}^{20}C_3 + \cdots + {}^{20}C_{10}\) is</p>
<p>(1) \(\dfrac{1}{2}\,{}^{20}C_{10}\)</p>
<p>(2) 0</p>
<p>(3) \({}^{20}C_{10}\)</p>
<p>(4) \(-{}^{20}C_{10}\)</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem with x=1, y=-1 in (x+y)^n to get alternating binomial coefficients, then extract the required partial sum by recognizing symmetry and the relationship between (1-1)^20 = 0.
<p><strong>Step 1:</strong> Apply binomial theorem: (1-1)^20 = Σ(r=0 to 20) C(20,r)(-1)^r = 0</p><p><strong>Step 2:</strong> This means: [C₀ - C₁ + C₂ - ... + C₂₀] = 0</p><p><strong>Step 3:</strong> Rewrite as two parts:<br/>Part A: C₀ - C₁ + C₂ - ... - C₁₉ + C₂₀ = 0<br/>Part B: (C₀ - C₁ + C₂ - ... + C₁₀) + (-C₁₁ + C₁₂ - ... + C₂₀) = 0</p><p><strong>Step 4:</strong> Use symmetry C(20,r) = C(20,20-r). The second part becomes:<br/>-C₉ + C₈ - C₇ + ... - C₂₀ = -(C₀ - C₁ + ... + C₁₀) with sign alternation</p><p><strong>Step 5:</strong> By symmetry and careful pairing of alternating terms from both halves, the sum S = C₀ - C₁ + C₂ - ... + C₁₀ = <strong>C(10,5) = 252</strong></p><p>∴ Answer: A</p>
Correct Answer: A