<p>Let \(\binom{n}{k}\) represent the combination of 'n' things taken 'k' at a time, then the value of the sum \(\binom{99}{97} + \binom{98}{96} + \binom{97}{95} + \ldots + \binom{3}{1} + \binom{2}{0}\) equals:</p>
<p>(a) \(\binom{99}{97}\)</p>
<p>(b) \(\binom{100}{98}\)</p>
<p>(c) \(\binom{99}{98}\)</p>
<p>(d) \(\binom{100}{97}\)</p>
Step-by-Step Solution
Key Concept: Use the symmetry property of combinations and apply Pascal's identity (hockey-stick identity) to telescope the sum. The sum can be rewritten by recognizing that consecutive binomial coefficients with specific patterns can be combined using the recurrence relation C(n,k) + C(n,k-1) = C(n+1,k).
<p><strong>Step 1:</strong> Rewrite the sum explicitly. We have:</p><p>S = C(99,97) + C(98,96) + C(97,95) + ... + C(3,1) + C(2,0)</p><p><strong>Step 2:</strong> Use the symmetry property C(n,k) = C(n,n-k). Rewrite each term:</p><p>C(99,97) = C(99,2), C(98,96) = C(98,2), C(97,95) = C(97,2), ..., C(3,1) = C(3,2), C(2,0) = C(2,2)</p><p>So: S = C(99,2) + C(98,2) + C(97,2) + ... + C(3,2) + C(2,2)</p><p><strong>Step 3:</strong> Apply the Hockey-Stick Identity: Σ(i=2 to 99) C(i,2) = C(100,3)</p><p>This is because the hockey-stick identity states: Σ(i=r to n) C(i,r) = C(n+1,r+1)</p><p>With r=2 and n=99: Σ(i=2 to 99) C(i,2) = C(100,3)</p><p><strong>Step 4:</strong> Simplify C(100,3) using symmetry:</p><p>C(100,3) = C(100, 100-3) = C(100,97)</p><p>∴ Answer: D</p>
Correct Answer: D