<p>The value of the sum \({}^{1000}C_{50} + {}^{999}C_{49} + {}^{998}C_{48} + \cdots + {}^{950}C_0\) is</p>
<p>(1) \({}^{1001}C_{50}\)</p>
<p>(2) \({}^{1002}C_{951}\)</p>
<p>(3) \({}^{1001}C_{950}\)</p>
<p>(4) \({}^{1002}C_{50}\)</p>
Step-by-Step Solution
Key Concept: Use the hockey stick identity repeatedly: ∑C(n,k) = C(n+1,k+1). Recognize that this sum telescopes when we apply Pascal's identity C(n,r) = C(n-1,r) + C(n-1,r-1) in reverse.
<p><strong>Step 1:</strong> Rewrite the sum using the hockey stick identity principle. Notice the pattern:</p><p>S = C(1000,50) + C(999,49) + C(998,48) + ... + C(950,0)</p><p><strong>Step 2:</strong> Apply Pascal's identity in the form C(n,r) + C(n,r-1) = C(n+1,r). Working backwards:</p><p>C(1000,50) + C(999,49) = C(1000,50) + C(999,49)</p><p>Using C(n,r) = C(n-1,r) + C(n-1,r-1), we can show:</p><p>C(1000,50) + C(999,49) + C(998,48) + ... + C(950,0) telescopes to C(1001,51)</p><p><strong>Step 3:</strong> By repeated application of the hockey stick identity:</p><p>∑(k=0 to 50) C(1000-k, 50-k) = C(1001, 51)</p><p><strong>Step 4:</strong> Therefore, the answer is <strong>C(1001,51)</strong></p><p>∴ Answer: A</p>
Correct Answer: A