Binomial Theorem
Summation of Binomial Coefficients
Grade 11
Question:
<p>If <span>\((1+x)^{20} = {}^{20}C_0 + {}^{20}C_1 x + \cdots + {}^{20}C_{20}x^{20}\)</span>, then <span>\({}^{20}C_0 - {}^{20}C_1 + \cdots - {}^{20}C_9 + {}^{20}C_{10}\)</span> equals:</p>
<p>\(\dfrac{1}{2}\,{}^{20}C_{10}\)</p>
<p>\({}^{20}C_{10}\)</p>
<p>\(2\,{}^{20}C_{10}\)</p>
<p>0</p>
Step-by-Step Solution
Key Concept: Substitute x = -1 in the binomial expansion to get alternating sum, then recognize that the sum up to the middle term equals half of the resulting value due to symmetry of binomial coefficients.
<p><strong>Step 1:</strong> Use the binomial expansion with x = -1:</p><p>(1-1)^20 = C₀ - C₁ + C₂ - C₃ + ... - C₁₉ + C₂₀ = 0</p><p>where C_r denotes ²⁰C_r.</p><p><strong>Step 2:</strong> Pair up symmetric terms using C_r = C_(20-r):</p><p>(C₀ - C₁ + ... + C₁₀) + (C₁₁ - C₁₂ + ... + C₂₀) = 0</p><p><strong>Step 3:</strong> In the second group, substitute C_(20-k) for C_k:</p><p>(C₁₀ - C₉ + C₈ - ... + C₀) = -(C₀ - C₁ + ... + C₁₀)</p><p><strong>Step 4:</strong> Since both groups are equal and opposite, and their sum is 0:</p><p>2(C₀ - C₁ + C₂ - ... + C₁₀) = 0</p><p>∴ C₀ - C₁ + C₂ - ... - C₉ + C₁₀ = <strong>0</strong></p>
Correct Answer: A