<p>If \(a = \displaystyle\sum_{r=0}^{n} \dfrac{1}{{}^nC_r}\), then the value of \(\displaystyle\sum_{0 \le i < j \le n} \left(\dfrac{i}{{}^nC_i} + \dfrac{j}{{}^nC_j}\right)\) is</p>
Step-by-Step Solution
Key Concept: Use the symmetry property of binomial coefficients (C(n,r) = C(n,n-r)) to pair terms and recognize that the sum involves products of reciprocals of binomial coefficients arranged symmetrically.
<p><strong>Step 1:</strong> Note that <strong>a</strong> = Σ(r=0 to n) 1/C(n,r) is given.</p><p><strong>Step 2:</strong> For the double sum, use the symmetry C(n,r) = C(n,n-r), so 1/C(n,r) = 1/C(n,n-r).</p><p><strong>Step 3:</strong> Rewrite the double sum: Σ(0≤i<j≤n) 1/(C(n,i)·C(n,j)) = Σ(0≤i<j≤n) [1/C(n,i)·1/C(n,j)]</p><p><strong>Step 4:</strong> Recognize that [Σ(r=0 to n) 1/C(n,r)]² = Σ(all pairs) 1/(C(n,i)·C(n,j)), which includes:</p><p>• Diagonal terms: Σ(r=0 to n) 1/C(n,r)² = a² - 2·Σ(0≤i<j≤n) 1/(C(n,i)·C(n,j))</p><p><strong>Step 5:</strong> Therefore: Σ(0≤i<j≤n) 1/(C(n,i)·C(n,j)) = <strong>(a² - Σ 1/C(n,r)²)/2</strong></p><p>∴ Answer: D</p>
Correct Answer: D