Question

If \[D_k = \begin{vmatrix}1 & n & n \\ 2k & n^2 + n + 2 & n^2 + n \\ 2k - 1 & n^2 & n^2 + n + 2\end{vmatrix} and \sum^n_{k = 1} D_k = 48\], then n equals

 

Options

• 4

• 6

• 8

•  none of these

Answer 1

(a) 4
\[D_k = \begin{vmatrix} 1 & n & n\\ 2k & n^2 + n + 2 & n^2 + n\\2k - 1 & n^2 & n^2 + n + 2 \end{vmatrix}\]
\[ = \begin{vmatrix} 1 & n & n\\ 1 & n + 2 & - 2\\2k - 1 & n^2 & n^2 + n + 2 \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_3 \right]\]
\[ = \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\2k - 1 & n^2 & n^2 + n + 2 \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ \sum\nolimits_{k = 1}^n D_k = \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\ 1 & n^2 & n^2 + n + 2 \end{vmatrix} + \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\ 3 & n^2 & n^2 + n + 2 \end{vmatrix} + . . . + \begin{vmatrix} 1 & n & n\\ 0 & 2 & - 2 - n\\ n & n^2 & n^2 + n + 2 \end{vmatrix}\]
\[ \sum\nolimits_{k = 1}^n D_k = 1\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + 1\left( n\left( - 2 - n \right) - 2n \right) + 1\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + 2\left( n\left( - 2 - n \right) - 2n \right) + . . . + 1\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + n\left( n\left( - 2 - n \right) - 2n \right)\]
\[ \sum\nolimits_{k = 1}^n D_k = n\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + \left( n\left( - 2 - n \right) - 2n \right)\left( 1 + 3 + 5 + 7 + . . . + n \right)\]
\[ \sum\nolimits_{k = 1}^n D_k = n\left( 2\left( n^2 + n + 2 \right) + \left( 2 + n \right) n^2 \right) + \left( n\left( - 2 - n \right) - 2n \right)\left( n^2 \right)\]
\[ \sum\nolimits_{k = 1}^n D_k = 2 n^2 + 4n\]
\[ \Rightarrow 2 n^2 + 4n = 48\]
\[ \Rightarrow \left( n - 6 \right)\left( n - 4 \right) = 0\]
\[ \Rightarrow n = 4\]