<p>If<br>\[\begin{vmatrix} r & 2r-1 & 3r-2 \\ \dfrac{n}{2} & n-1 & a \\ \dfrac{1}{2}n(n-1) & (n-1)^2 & \dfrac{1}{2}(n-1)(3n+4) \end{vmatrix},\]<br>then the value of \(\displaystyle\sum_{r=1}^{n-1} \Delta_r\)</p>
<p>depends only on \(a\).</p>
<p>depends only on \(n\).</p>
<p>depends both on \(a\) and \(n\).</p>
<p>is independent of both \(a\) and \(n\).</p>
Step-by-Step Solution
Key Concept: Recognize that the determinant Δᵣ is a function of r, and the sum Σ Δᵣ from r=1 to n-1 can be evaluated by observing that rows 2 and 3 are independent of r, making Δᵣ linear in r. The sum telescopes or simplifies using properties of determinant linearity.
<p><strong>Step 1:</strong> Observe that rows 2 and 3 are independent of r. Since the determinant is linear in each row, we have:</p><p>Σ Δᵣ (r=1 to n-1) = determinant of a matrix where row 1 is replaced by Σ of row 1 entries.</p><p><strong>Step 2:</strong> Calculate the sum of row 1 entries:</p><ul><li>Σ r = 1 + 2 + ... + (n-1) = (n-1)n/2</li><li>Σ (2r-1) = 2·(n-1)n/2 - (n-1) = (n-1)n - (n-1) = (n-1)²</li><li>Σ (3r-2) = 3·(n-1)n/2 - 2(n-1) = (3n(n-1) - 4(n-1))/2 = (n-1)(3n-4)/2</li></ul><p><strong>Step 3:</strong> The resulting determinant becomes:</p><p>Δ = |[½n(n-1), (n-1)², ½(n-1)(3n-4)]|</p><p> |[n/2, n-1, a]|</p><p> |[½n(n-1), (n-1)², ½(n-1)(3n+4)]|</p><p><strong>Step 4:</strong> Notice rows 1 and 3 differ only in the third column: ½(n-1)(3n-4) vs ½(n-1)(3n+4). Factor out (n-1)/2 from rows appropriately and observe Row 1 = Row 3 after simplification, making the determinant <strong>0</strong>.</p><p>∴ Answer: D (which is 0)</p>
Correct Answer: D