If S_r = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mn>2</mn><mi>r</mi></mtd><mtd><mi>x</mi></mtd><mtd><mi>n</mi><mo>(</mo><mi>n</mi><mo>+</mo><mn>1</mn><mo>)</mo></mtd></mtr><mtr><mtd><mn>6</mn><msup><mi>r</mi><mn>2</mn></msup><mo>-</mo><mn>1</mn></mtd><mtd><mi>y</mi></mtd><mtd><msup><mi>n</mi><mn>2</mn></msup><mo>(</mo><mn>2</mn><mi>n</mi><mo>+</mo><mn>3</mn><mo>)</mo></mtd></mtr><mtr><mtd><mn>4</mn><msup><mi>r</mi><mn>3</mn></msup><mo>-</mo><mn>2</mn><mi>n</mi><mi>r</mi></mtd><mtd><mi>z</mi></mtd><mtd><msup><mi>n</mi><mn>3</mn></msup><mo>(</mo><mi>n</mi><mo>+</mo><mn>1</mn><mo>)</mo></mtd></mtr></mtable></mfenced></math>, then <math xmlns="http://www.w3.org/1998/Math/MathML"><munderover><mo>∑</mo><mrow><mi>r</mi><mo>=</mo><mn>1</mn></mrow><mi>n</mi></munderover><msub><mi>S</mi><mi>r</mi></msub></math> does not depend on -
Step-by-Step Solution
Key Concept: The determinant is linear with respect to its columns. Summing the determinant over r involves summing the elements of the first column. The resulting determinant will have columns that are independent of n, making the sum independent of n.
The determinant S_r is linear in its first column. When we compute the sum of the determinants, we can sum the first column elements: sum(2r) = 2*n(n+1)/2 = n(n+1), sum(6r^2-1) = 6*n(n+1)(2n+1)/6 - n = n(2n^2+3n+1-1) = n^2(2n+3), sum(4r^3-2nr) = 4*(n(n+1)/2)^2 - 2n*n(n+1)/2 = n^2(n+1)^2 - n^2(n+1) = n^2(n+1)(n+1-1) = n^3(n+1). The resulting determinant has the first column equal to the third column, so the determinant is 0, which is independent of n.
Correct Answer: C