<p>If \((1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n\), then the sum of the product of the coefficients taken two at a time can be represented by \(\displaystyle\sum_{i=0}^{n}\sum_{j=i+1}^{n} C_i C_j = 2^a - \dfrac{b!}{c(d!)^2}\). Then which of the following are correct?</p>
Step-by-Step Solution
Key Concept: The sum of products of binomial coefficients taken two at a time equals half the difference between (∑Cᵢ)² and ∑Cᵢ². Use (1+x)ⁿ(1+x)ⁿ = (1+x)²ⁿ to find ∑Cᵢ² = C₂ₙⁿ, and (∑Cᵢ)² = 2²ⁿ.
<p><strong>Step 1:</strong> Start with (∑Cᵢ)² = (2ⁿ)² = 2²ⁿ</p><p><strong>Step 2:</strong> Expand: (∑Cᵢ)² = ∑Cᵢ² + 2∑∑(i<j) CᵢCⱼ</p><p><strong>Step 3:</strong> Therefore: 2∑∑(i<j) CᵢCⱼ = 2²ⁿ - ∑Cᵢ²</p><p><strong>Step 4:</strong> Find ∑Cᵢ² using (1+x)ⁿ(1+x)ⁿ = (1+x)²ⁿ. Coefficient of xⁿ in RHS is C₂ₙⁿ = (2n)!/(n!)²</p><p><strong>Step 5:</strong> Thus: ∑∑(i<j) CᵢCⱼ = [2²ⁿ - (2n)!/(n!)²]/2 = 2²ⁿ⁻¹ - (1/2)·(2n)!/(n!)²</p><p><strong>Step 6:</strong> Match with 2ᵃ - b!/(c(d!)²): a = 2n-1, b = 2n, c = 2, d = n</p><p>∴ Answer: ABCD (values a=2n-1, b=2n, c=2, d=n are all correct)</p>
Correct Answer: ABCD