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 with respect to 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 be independent of n.
The determinant $S_r$ can be written as a function of $r$ in the first column. Since the determinant is a linear function of its columns, $\sum_{r=1}^n S_r$ is equal to the determinant where the first column is replaced by $\sum_{r=1}^n (2r, 6r^2-1, 4r^3-2nr)^T$. Calculating these sums: $\sum 2r = n(n+1)$, $\sum (6r^2-1) = 6\frac{n(n+1)(2n+1)}{6} - n = n(n+1)(2n+1) - n = 2n^3+3n^2+n-n = 2n^3+3n^2 = n^2(2n+3)$, and $\sum (4r^3-2nr) = 4\frac{n^2(n+1)^2}{4} - 2n\frac{n(n+1)}{2} = n^2(n+1)^2 - n^2(n+1) = n^2(n+1)(n+1-1) = n^3(n+1)$. The first column becomes identical to the third column, so the determinant is 0, which is independent of $n$.
Correct Answer: 3