If $S_r = \begin{vmatrix} 2r & x & n(n+1) \\ 6r^2-1 & y & n^2(2n+3) \\ 4r^3-2nr & z & n^3(n+1) \end{vmatrix}$, then $\sum_{r=1}^n S_r$ does not depend on -
Step-by-Step Solution
Key Concept: The determinant is linear in the columns. Summing the determinant over r involves summing the entries of the first column, while the other columns remain constant. The resulting determinant will have columns that are linearly dependent or result in a value independent of the variables.
The determinant $S_r$ can be written as a sum of determinants by linearity. Since the second and third columns do not depend on $r$, we can pull out the summation inside the determinant. After performing the summation $\sum_{r=1}^n (2r)$, $\sum_{r=1}^n (6r^2-1)$, and $\sum_{r=1}^n (4r^3-2nr)$, it can be shown that the resulting determinant is zero or independent of $x, y, n$.
Correct Answer: D