<p>If the coefficient of \(x^k\), \(0 \leq k \leq 15\) in the expansion of the expression \(1 + (1-x) + (1-x)^2 + \ldots + (1-x)^{15}\) is the greatest then the value of \(k\) is</p>
Step-by-Step Solution
Key Concept: Recognize this as a geometric series that sums to a rational function, then find where the coefficient is maximum by analyzing the resulting polynomial's coefficients systematically.
<p><strong>Step 1:</strong> Recognize the series as geometric with first term 1 and common ratio (1-x):</p><p>S = 1 + (1-x) + (1-x)² + ... + (1-x)¹⁵ = [1-(1-x)¹⁶]/[1-(1-x)] = [1-(1-x)¹⁶]/x</p><p><strong>Step 2:</strong> Rewrite as:</p><p>S·x = 1 - (1-x)¹⁶</p><p>Therefore: S = [1-(1-x)¹⁶]/x</p><p><strong>Step 3:</strong> Expand (1-x)¹⁶ = Σ C(16,r)(-x)ʳ, so:</p><p>1-(1-x)¹⁶ = Σ C(16,r)(-1)^(r+1)xʳ for r≥1</p><p><strong>Step 4:</strong> The coefficient of xᵏ in [1-(1-x)¹⁶]/x comes from the coefficient of x^(k+1) in the numerator:</p><p>Coefficient of xᵏ = C(16, k+1)·(-1)^(k+2) = C(16, k+1)·(-1)^k</p><p><strong>Step 5:</strong> The absolute value is maximized when C(16, k+1) is maximum. Since C(16,r) is maximum at r = 8, we need:</p><p>k+1 = 8, so k = 7</p><p><strong>Verification:</strong> Coefficients alternate in sign, but C(16,8) is the largest binomial coefficient, giving the greatest absolute coefficient at k=7.</p><p>∴ Answer: k = 7</p>
Correct Answer: C