Let Dk=\(\begin{vmatrix} 1 & 2k & 2k-1\\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix} \) If \(∑^n_{k=1} D_k=96,\) The n is equal to
Correct Answer: 6
Solution and Explanation
The given determinant \( D_k \) is:
\[ D_k = \begin{vmatrix} 1 & 2k & 2k - 1 \\ n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \end{vmatrix} \]
The sum of determinants is:
\[ \sum_{k=1}^{n} D_k = 96 \]
Expanding the determinant for the summation:
\[ \begin{vmatrix} \sum_{k=1}^{n} 1 & \sum_{k=1}^{n} 2k & \sum_{k=1}^{n} (2k-1) \\ n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \end{vmatrix} = 96 \]
Simplify the summations:
\( \sum_{k=1}^{n} 1 = n \), \( \sum_{k=1}^{n} 2k = n(n+1) \), \( \sum_{k=1}^{n} (2k-1) = n^2 \)
Substitute back:
\[ \begin{vmatrix} n & n(n+1) & n^2 \\ n & n^2 + n + 2 & n^2 \\ n & n^2 + n & n^2 + n + 2 \end{vmatrix} = 96 \]
Perform row operations to simplify:
\( R_2 \rightarrow R_2 - R_1, \quad R_3 \rightarrow R_3 - R_1 \)
This gives:
\[ \begin{vmatrix} n & n^2 + n & n^2 \\ 0 & 2 & 0 \\ 0 & 0 & n+2 \end{vmatrix} = 96 \]
Calculate the determinant:
\( n \cdot 2 \cdot (n+2) = 96 \)
Simplify:
\( 2n(n+2) = 96 \Rightarrow n(n+2) = 48 \)
Solve for \( n \):
\( n^2 + 2n - 48 = 0 \)
Factorize:
\( (n - 6)(n + 8) = 0 \Rightarrow n = 6 \) (as \( n > 0 \))
Final Answer: \( n = 6 \)
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