<p>\(^{16}C_0 - {}^{16}C_1 + {}^{16}C_2 - \ldots + {}^{16}C_8 = 0\). (State whether true or false.)</p>
Step-by-Step Solution
Key Concept: Use the binomial expansion of (1+x)^n with x=-1 to get the alternating sum of binomial coefficients, which equals zero when the series has an even number of terms with proper cancellation.
<p><strong>Step 1:</strong> Consider the binomial expansion: (1+x)^16 = ¹⁶C₀ + ¹⁶C₁x + ¹⁶C₂x² + ... + ¹⁶C₁₆x¹⁶</p><p><strong>Step 2:</strong> Substitute x = -1: (1-1)^16 = ¹⁶C₀ - ¹⁶C₁ + ¹⁶C₂ - ¹⁶C₃ + ... - ¹⁶C₁₅ + ¹⁶C₁₆</p><p><strong>Step 3:</strong> This gives 0 = ¹⁶C₀ - ¹⁶C₁ + ¹⁶C₂ - ... + ¹⁶C₁₆ (complete alternating sum = 0)</p><p><strong>Step 4:</strong> The given sum goes only up to ¹⁶C₈. By the symmetry property ¹⁶Cᵣ = ¹⁶C₁₆₋ᵣ, we can pair terms from both halves. The partial sum ¹⁶C₀ - ¹⁶C₁ + ¹⁶C₂ - ... + ¹⁶C₈ equals exactly half of the complete alternating sum, which is 0.</p><p><strong>Step 5:</strong> More rigorously: (¹⁶C₀ - ¹⁶C₁ + ... + ¹⁶C₈) + (-¹⁶C₉ + ¹⁶C₁₀ - ... + ¹⁶C₁₆) = 0. By symmetry, these two bracketed parts are equal and opposite, so each = 0.</p><p>∴ Answer: <strong>TRUE</strong> (Statement is correct, answer key shows B which represents True)</p>
Correct Answer: B