If S_{r} = left|begin{array}{ccc} 2r & x & n( | vedclass

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(D) We have, $S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \ 6r^{2}-1 & y & n^{2}(2n+3) \ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}ight|$.Applyi

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$x, y, z$ and $n$

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(D) We have, $S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \ 6r^{2}-1 & y & n^{2}(2n+3) \ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}ight|$.Applying the summation $\sum_{r=1}^{n}$ to the determinant, we get:$\sum_{r=1}^{n} S_{r} = \left|\begin{array}{ccc} 2 \sum_{r=1}^{n} r & x & n(n+1) \ \sum_{r=1}^{n} (6r^{2}-1) & y & n^{2}(2n+3) \ \sum_{r=1}^{n} (4r^{3}-2nr) & z & n^{3}(n+1) \end{array}ight|$.Using the standard summation formulas $\sum r = \frac{n(n+1)}{2}$, $\sum r^{2} = \frac{n(n+1)(2n+1)}{6}$, and $\sum r^{3} = \frac{n^{2}(n+1)^{2}}{4}$, we calculate the entries of the first column:$C_{11} = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$.$C_{21} = 6 \cdot \frac{n(n+1)(2n+1)}{6} - n = n(n+1)(2n+1) - n = n(2n^{2}+3n+1-1) = n^{2}(2n+3)$.$C_{31} = 4 \cdot \frac{n^{2}(n+1)^{2}}{4} - 2n \cdot \frac{n(n+1)}{2} = n^{2}(n+1)^{2} - n^{2}(n+1) = n^{2}(n+1)(n+1-1) = n^{3}(n+1)$.Substituting these back, we get:$\sum_{r=1}^{n} S_{r} = \left|\begin{array}{ccc} n(n+1) & x & n(n+1) \ n^{2}(2n+3) & y & n^{2}(2n+3) \ n^{3}(n+1) & z & n^{3}(n+1) \end{array}ight|$.Since column $C_{1}$ and column $C_{3}$ are identical, the value of the determinant is $0$.Thus, the sum is $0$, which is independent of $x, y, z$ and $n$.

If $S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \\ 6r^{2}-1 & y & n^{2}(2n+3) \\ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}\right|$, then the value of $\sum_{r=1}^{n} S_{r}$ is independent of

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