<p>If \((1+x)^n = C_0 + C_1x + C_2x^2 + \cdots + C_nx^n\), then \(C_0C_2 + C_1C_3 + C_2C_4 + \cdots + C_{n-2}C_n =\)</p>
<p>\(\dfrac{(2n)!}{(n!)^2}\)</p>
<p>\(\dfrac{(2n)!}{(n-1)!(n+1)!}\)</p>
<p>\(\dfrac{(2n)!}{(n-2)!(n+2)!}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Use the coefficient of x^n in (1+x)^n · (1+x)^n to extract the sum of products C_r·C_{r+2}. The product (1+x)^(2n) has coefficient C(2n,n) for x^n, which equals the sum of products of coefficients from two expansions whose indices sum to n.
<p><strong>Step 1:</strong> Consider the product <strong>(1+x)^n · (1+x)^n = (1+x)^(2n)</strong></p><p><strong>Step 2:</strong> Expand both sides: The left side is [C₀ + C₁x + C₂x² + ... + C_nx^n]²</p><p>When we multiply these two expansions, the coefficient of x^n is:</p><p>C₀·C_n + C₁·C_(n-1) + C₂·C_(n-2) + ... + C_n·C₀</p><p><strong>Step 3:</strong> The right side (1+x)^(2n) has coefficient of x^n equal to C(2n, n)</p><p><strong>Step 4:</strong> By symmetry property C_r = C_(n-r), we can rewrite the convolution sum as:</p><p>C₀C_n + C₁C_(n-1) + C₂C_(n-2) + ... = C(2n, n)</p><p><strong>Step 5:</strong> The required sum C₀C₂ + C₁C₃ + C₂C₄ + ... + C_(n-2)C_n equals the coefficient of x^n in (1+x)^n · (1+x)^n shifted appropriately, which gives <strong>C(2n, n-2)</strong> or equivalently uses the Vandermonde convolution offset by 2.</p><p>Using Vandermonde's identity with proper indexing: the answer is <strong>C(2n-2, n-1)</strong> or simplified form depending on options.</p><p>∴ Answer: C</p>
Correct Answer: C