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Question

If ${D}_{k}=\left|\begin{array}{ccc}1& n& n\\ 2k& {n}^{2}+n+2& {n}^{2}+n\\ 2k-1& {n}^{2}& {n}^{2}+n+2\end{array}\right|\mathrm{and}\sum _{k=1}^{n}{D}_{k}=48$, then n equals (a) 4 (b) 6 (c) 8 (d) none of these

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Solution

(a) 4 ${D}_{k}=\left|1nn\phantom{\rule{0ex}{0ex}}2k{n}^{2}+n+2{n}^{2}+n\phantom{\rule{0ex}{0ex}}2k-1{n}^{2}{n}^{2}+n+2\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left|1nn\phantom{\rule{0ex}{0ex}}1n+2-2\phantom{\rule{0ex}{0ex}}2k-1{n}^{2}{n}^{2}+n+2\right|\left[\mathrm{Applying}{R}_{2}\to {R}_{2}-{R}_{3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left|1nn\phantom{\rule{0ex}{0ex}}02-2-n\phantom{\rule{0ex}{0ex}}2k-1{n}^{2}{n}^{2}+n+2\right|\left[\mathrm{Applying}{R}_{2}\to {R}_{2}-{R}_{1}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\sum }_{k=1}^{n}{D}_{k}=\left|1nn\phantom{\rule{0ex}{0ex}}02-2-n\phantom{\rule{0ex}{0ex}}1{n}^{2}{n}^{2}+n+2\right|+\left|1nn\phantom{\rule{0ex}{0ex}}02-2-n\phantom{\rule{0ex}{0ex}}3{n}^{2}{n}^{2}+n+2\right|+...+\left|1nn\phantom{\rule{0ex}{0ex}}02-2-n\phantom{\rule{0ex}{0ex}}n{n}^{2}{n}^{2}+n+2\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\sum }_{k=1}^{n}{D}_{k}=1\left(2\left({n}^{2}+n+2\right)+\left(2+n\right){n}^{2}\right)+1\left(n\left(-2-n\right)-2n\right)+1\left(2\left({n}^{2}+n+2\right)+\left(2+n\right){n}^{2}\right)+2\left(n\left(-2-n\right)-2n\right)+...+1\left(2\left({n}^{2}+n+2\right)+\left(2+n\right){n}^{2}\right)+n\left(n\left(-2-n\right)-2n\right)\phantom{\rule{0ex}{0ex}}{\sum }_{k=1}^{n}{D}_{k}=n\left(2\left({n}^{2}+n+2\right)+\left(2+n\right){n}^{2}\right)+\left(n\left(-2-n\right)-2n\right)\left(1+3+5+7+...+n\right)\phantom{\rule{0ex}{0ex}}{\sum }_{k=1}^{n}{D}_{k}=n\left(2\left({n}^{2}+n+2\right)+\left(2+n\right){n}^{2}\right)+\left(n\left(-2-n\right)-2n\right)\left({n}^{2}\right)\phantom{\rule{0ex}{0ex}}{\sum }_{k=1}^{n}{D}_{k}=2{n}^{2}+4n\phantom{\rule{0ex}{0ex}}⇒2{n}^{2}+4n=48\phantom{\rule{0ex}{0ex}}⇒\left(n-6\right)\left(n-4\right)=0\phantom{\rule{0ex}{0ex}}⇒n=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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