Question

Let D_k=\(\begin{vmatrix} 1 & 2k & 2k-1\\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix} \) If \(∑^n_{k=1} D_k=96,\) The n is equal to

Answer 1

Correct Answer: 6

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: \( n = 6 \)