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If $S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|$, then the
value of $\sum_{r=1}^{n} S_{r}$ is independent of
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value of $\sum_{r=1}^{n} S_{r}$ is independent of
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The correct answer is:
$x y, z$ and $n$
We have, $S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|$
$\Rightarrow \quad \sum_{r=1}^{n} S_{r}=\left|\begin{array}{ccc}2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n}\left(6 r^{2}-1\right) & y & n^{2}(2 n+3) \\ \sum_{r=1}^{n}\left(4 r^{3}-2 n r\right) & z & n^{3}(n+1)\end{array}\right|$
$=\left|\begin{array}{ccc}n(n+1) & x & n(n+1) \\ n^{2}(2 n+3) & y & n^{2}(2 n+3) \\ n^{3}(n+1) & z & n^{3}(n+1)\end{array}\right|$
$=0 \quad\left(\because C_{1}\right.$ and $C_{3}$ are identical $)$
Hence, $\sum_{r=1}^{n} S_{r}$ is independent of $x, y, z$ and $n$.
$\Rightarrow \quad \sum_{r=1}^{n} S_{r}=\left|\begin{array}{ccc}2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n}\left(6 r^{2}-1\right) & y & n^{2}(2 n+3) \\ \sum_{r=1}^{n}\left(4 r^{3}-2 n r\right) & z & n^{3}(n+1)\end{array}\right|$
$=\left|\begin{array}{ccc}n(n+1) & x & n(n+1) \\ n^{2}(2 n+3) & y & n^{2}(2 n+3) \\ n^{3}(n+1) & z & n^{3}(n+1)\end{array}\right|$
$=0 \quad\left(\because C_{1}\right.$ and $C_{3}$ are identical $)$
Hence, $\sum_{r=1}^{n} S_{r}$ is independent of $x, y, z$ and $n$.
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