<p>Consider \(f(k) = \dbinom{2k+1}{2k}C - \dbinom{2^k}{2^{k-1}}C\), which of the following is true about \(f(k)\)</p>
<p>\(f(k)\) divides \(2^{3k}\) but does not divide \(2^{3k+1}\) for all k</p>
<p>\(f(k)\) divides \(2^{3k}\) but does not divide \(2^{3k+1}\) only for certain values of k</p>
<p>\(f(k)\) divides \(2^{3k}\) but does not divide \(2^{3k+1}\) only if k < N here N is a 3 digit natural number.</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that C(n,r) = C(n, n-r), so C(2k+1, 2k) = C(2k+1, 1) = 2k+1, and simplify using binomial coefficient properties to evaluate f(k) for specific values.
<p><strong>Step 1:</strong> Simplify C(2k+1, 2k) using the symmetry property of binomial coefficients.</p><p>C(2k+1, 2k) = C(2k+1, 2k+1-2k) = C(2k+1, 1) = 2k+1</p><p><strong>Step 2:</strong> Analyze C(2^k, 2^(k-1)). For k ≥ 1, we have 2^(k-1) < 2^k, so this is well-defined. However, note that 2^(k-1) is exactly half of 2^k, making this the central binomial coefficient when k ≥ 1.</p><p><strong>Step 3:</strong> For k = 0: f(0) = C(1,0) - C(1, 0.5) = 1 - undefined. For k = 1: f(1) = C(3,2) - C(2,1) = 3 - 2 = 1</p><p><strong>Step 4:</strong> For k ≥ 1, f(k) = (2k+1) - C(2^k, 2^(k-1)). As k increases, C(2^k, 2^(k-1)) grows exponentially, making f(k) negative for large k.</p><p>∴ Answer: A</p>
Correct Answer: A