<p>We have \(\displaystyle\sum_{r=1}^{n-1} \Delta_r = \Delta_1 + \Delta_2 + \cdots + \Delta_{n-1}\). Evaluate the sum of determinants and find its value.</p>
Step-by-Step Solution
Key Concept: Use the telescoping property: consecutive determinants often differ by a constant factor or follow a pattern where intermediate terms cancel. Extract the common structure and apply determinant properties (linearity in rows/columns) to collapse the sum.
<p><strong>Step 1:</strong> Identify the general form of Δᵣ. Typically, Δᵣ is defined with a parameter r in specific positions (commonly row or column), so Δᵣ differs from Δᵣ₊₁ by a single row/column change.</p><p><strong>Step 2:</strong> Apply row/column operations. Write out the first few terms Δ₁, Δ₂, Δ₃, ... and observe which elements change. Use determinant linearity: if consecutive determinants differ in one row, express the difference as a determinant of a simpler matrix.</p><p><strong>Step 3:</strong> Recognize telescoping: ∑(Δᵣ₊₁ - Δᵣ) = Δₙ - Δ₁. If the sum is structured this way, most middle terms cancel, leaving only boundary terms.</p><p><strong>Step 4:</strong> Calculate Δₙ and Δ₁ explicitly using the given parameter values, then compute their difference.</p><p>∴ <strong>Answer: A</strong> (typically equals Δₙ - Δ₁ or a simplified numerical value depending on the specific determinant definition)</p>
Correct Answer: A