<p>The value of \(({}^{21}C_1 - {}^{10}C_1) + ({}^{21}C_2 - {}^{10}C_2) + ({}^{21}C_3 - {}^{10}C_3) + ({}^{21}C_4 - {}^{10}C_4) + \ldots + ({}^{21}C_{10} - {}^{10}C_{10})\) is</p>
<p>\(2^{20} - 2^{10}\)</p>
<p>\(2^{21} - 2^{11}\)</p>
<p>\(2^{21} - 2^{10}\)</p>
<p>\(2^{20} - 2^9\)</p>
Step-by-Step Solution
Key Concept: Use the binomial theorem identity: ∑(r=0 to n) C(n,r) = 2^n. Separate the sum into two parts and apply this identity to each, recognizing that C(n,r) = 0 when r > n.
<p><strong>Step 1:</strong> Separate the given sum into two parts:</p><p>∑(r=1 to 10) [C(21,r) - C(10,r)] = ∑(r=1 to 10) C(21,r) - ∑(r=1 to 10) C(10,r)</p><p><strong>Step 2:</strong> Apply the binomial theorem. We know that:</p><p>(1+1)^n = ∑(r=0 to n) C(n,r) = 2^n</p><p>Therefore: ∑(r=0 to 21) C(21,r) = 2^21 and ∑(r=0 to 10) C(10,r) = 2^10</p><p><strong>Step 3:</strong> Rewrite the first sum by excluding r=0:</p><p>∑(r=1 to 10) C(21,r) = ∑(r=0 to 21) C(21,r) - C(21,0) - ∑(r=11 to 21) C(21,r)</p><p>Since ∑(r=0 to 21) C(21,r) = 2^21 and C(21,0) = 1, and by symmetry ∑(r=11 to 21) C(21,r) = 2^20:</p><p>∑(r=1 to 10) C(21,r) = 2^21 - 1 - 2^20 = 2^20 - 1</p><p><strong>Step 4:</strong> For the second sum:</p><p>∑(r=1 to 10) C(10,r) = 2^10 - C(10,0) = 2^10 - 1</p><p><strong>Step 5:</strong> Combine results:</p><p>(2^20 - 1) - (2^10 - 1) = 2^20 - 2^10 = 2^10(2^10 - 1) = 1024 × 1023</p><p>∴ Answer: <strong>2^20 - 2^10</strong> or <strong>1024 × 1023 = 1,047,552</strong></p>
Correct Answer: A