Question

If $D_{k}=\begin{vmatrix} 1 & n & n \\ 2k & n^{2}+n+1 & n^{2}+n \\ 2k-1 & n^{2} & n^{2}+n+1 \end{vmatrix}$ and $\displaystyle \sum _{k = 1}^{n}D_{k}=56$ , then $n$ equals to

• A. $4$
• B. $8$
• C. $6$
• D. None of these

Answer 1

Answer: (D)

It is given that $ \Sigma _{k = 1}^{ n}D_{k}=56$
where, $D_{k}=\begin{vmatrix} 1 & n & n \\ 2k & n^{2}+n+1 & n^{2}+n \\ 2k-1 & n^{2} & n^{2}+n+1 \end{vmatrix}$
$\Rightarrow \begin{vmatrix} \Sigma _{k = 1}^{n}1 & n & n \\ \Sigma _{k = 1}^{n}2k & n^{2}+n+1 & n^{2}+n \\ \Sigma _{k = 1}^{n}\left(2 k - 1\right) & n^{2} & n^{2}+n+1 \end{vmatrix}=56$
$\Rightarrow \begin{vmatrix} n & n & n \\ n\left(n + 1\right) & n^{2}+n+1 & n^{2}+n \\ n^{2} & n^{2} & n^{2}+n+1 \end{vmatrix}=56$
$\text{C}_{2} \rightarrow \text{C}_{2} - \text{C}_{1}$
$\text{C}_{3} \rightarrow \text{C}_{3} - \text{C}_{1}$
$\begin{vmatrix} n & 0 & 0 \\ n^{2}+n & 1 & 0 \\ n^{2} & 0 & n+1 \end{vmatrix}=56$
$\Rightarrow n\left(n + 1\right)=56$
$\Rightarrow n=7$