<p>The sum of the series \(^{20}C_0 + ^{20}C_1 + ^{20}C_2 + \ldots + ^{20}C_{10}\) is</p>
<p>(a) \(2^{20}\)</p>
<p>(b) \(2^{19}\)</p>
<p>(c) \(2^{19} + \dfrac{1}{2} \cdot {}^{20}C_{10}\)</p>
<p>(d) \(2^{19} - \dfrac{1}{2} \cdot {}^{20}C_{10}\)</p>
Step-by-Step Solution
Key Concept: The sum of all binomial coefficients in row 20 equals 2^20. By symmetry, the sum of coefficients from C_0 to C_10 equals exactly half of this total, since C_r = C_(n-r).
<p><strong>Step 1:</strong> Apply the binomial theorem. We know that (1+1)^20 = Σ(k=0 to 20) C(20,k) = 2^20</p><p><strong>Step 2:</strong> Use symmetry property: C(20,r) = C(20,20-r). This means C_0 = C_20, C_1 = C_19, ..., C_10 = C_10</p><p><strong>Step 3:</strong> The sum splits equally: C_0 + C_1 + ... + C_10 + C_11 + ... + C_20 = 2^20</p><p><strong>Step 4:</strong> By symmetry, C_0 + C_1 + ... + C_10 = C_10 + C_11 + ... + C_20</p><p><strong>Step 5:</strong> Note that C_10 appears in both halves. Therefore: 2(C_0 + C_1 + ... + C_10) - C_10 = 2^20</p><p><strong>Step 6:</strong> Solving: C_0 + C_1 + ... + C_10 = (2^20 + C_10)/2 = 2^19 + C(20,10)/2</p><p><strong>Alternative approach (cleaner):</strong> Since there are 21 terms total (C_0 through C_20) and by perfect symmetry around the middle term C_10, the sum from C_0 to C_9 equals the sum from C_11 to C_20. Thus: C_0 + C_1 + ... + C_9 + C_10 + C_11 + ... + C_20 = 2^20, which gives 2(C_0 + ... + C_9) + C_10 = 2^20</p><p>∴ Answer: <strong>2^19 + C(20,10)/2</strong> or equivalently <strong>2^19 + 92378</strong> if C(20,10) is evaluated, but the standard form is <strong>2^19</strong> when C_10 is excluded from one side, or the answer is <strong>(2^20 + C(20,10))/2</strong></p>
Correct Answer: C