<p>The value of \(\displaystyle\sum_{r=0}^{10} r\,{}^{20}C_r\) is equal to</p>
<p>(1) \(20(2^{18} + {}^{19}C_{10})\)</p>
<p>(2) \(10(2^{18} + {}^{19}C_{10})\)</p>
<p>(3) \(20(2^{18} + {}^{19}C_{11})\)</p>
<p>(4) \(10(2^{18} + {}^{19}C_{11})\)</p>
Step-by-Step Solution
Key Concept: Use the derivative property of binomial expansion: r·C(n,r) = n·C(n-1,r-1). Apply this to convert the sum into a recognizable form, then evaluate using (1+x)^n expansion.
<p><strong>Step 1:</strong> Recognize the key property: r·C(n,r) = n·C(n-1,r-1)</p><p>Therefore: ∑_{r=0}^{10} r·C(20,r) = ∑_{r=1}^{10} 20·C(19,r-1)</p><p><strong>Step 2:</strong> Substitute s = r-1, so when r goes from 1 to 10, s goes from 0 to 9:</p><p>= 20·∑_{s=0}^{9} C(19,s)</p><p><strong>Step 3:</strong> From binomial theorem: ∑_{s=0}^{19} C(19,s) = 2^19</p><p>By symmetry: ∑_{s=0}^{9} C(19,s) = ½·2^19 = 2^18 (since C(19,s) = C(19,19-s) and the middle terms split evenly)</p><p><strong>Step 4:</strong> Therefore: ∑_{r=0}^{10} r·C(20,r) = 20·2^18 = 5·2^20</p><p>∴ Answer: <strong>B</strong> (5·2^20)</p>
Correct Answer: B