A) 4
B) 6
C) 8
D) None of these
Correct Answer: D
Solution :
\[\therefore \] \[\sum\limits_{k=1}^{n}{{{D}_{k}}=}\,\left| \begin{matrix} \sum\limits_{k=1}^{n}{1} & n & n \\ 2\sum\limits_{k=1}^{n}{k} & {{n}^{2}}+n+1 & {{n}^{2}}+n \\ 2\sum\limits_{k=1}^{n}{k-}\sum\limits_{k=1}^{n}{1} & {{n}^{2}} & {{n}^{2}}+n+1 \\ \end{matrix} \right|\] \[\Rightarrow \] \[\left| \begin{matrix} n & n & n \\ {{n}^{2}}+n & {{n}^{2}}+n+1 & {{n}^{2}}+n \\ {{n}^{2}} & {{n}^{2}} & {{n}^{2}}+n+1 \\ \end{matrix} \right|=56\] Applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}},\] then \[\left| \begin{matrix} n & 0 & 0 \\ {{n}^{2}}+n & 1 & 0 \\ n & 0 & n+1 \\ \end{matrix} \right|=56\] \[\Rightarrow \] \[n(n+1)=56=7\times 8\] \[\Rightarrow \] \[n=7\]
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