<p>Let <br/> \(\sum_{r=1}^{N} A_r = \begin{vmatrix} N(N+1) & x & N(N+1) \\ 2N^3+3N^2 & y & N^2(2N+3) \\ N^3(N+1) & z & N^3(N+1) \end{vmatrix}\)<br/> then \(\sum_{r=1}^{N} A_r\) equals:</p>
<p>(a) 0</p>
<p>(b) depends on x, y, z</p>
<p>(c) independent of N</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Recognize that columns 1 and 3 are identical, making the determinant zero regardless of N, x, y, z values. The sum telescopes to a determinant where two columns collapse to the same expression.
<p><strong>Step 1:</strong> Observe the given determinant structure:</p><p>Column 1: N(N+1), 2N³+3N², N³(N+1)</p><p>Column 3: N(N+1), N²(2N+3), N³(N+1)</p><p><strong>Step 2:</strong> Check if Column 1 = Column 3:</p><p>• First element: N(N+1) = N(N+1) ✓</p><p>• Second element: 2N³+3N² = N²(2N+3) ✓</p><p>• Third element: N³(N+1) = N³(N+1) ✓</p><p><strong>Step 3:</strong> Apply the fundamental property: <em>A determinant with two identical columns equals zero.</em></p><p>Since Column 1 ≡ Column 3, the determinant = 0</p><p><strong>Step 4:</strong> This holds for all values of N, therefore:</p><p>∑(r=1 to N) A<sub>r</sub> = <strong>0</strong></p><p>∴ Answer: <strong>a</strong> (which represents 0)</p>
Correct Answer: a