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, Then n is equal to

Answer 1

Correct Answer: 6

First, we need to find the determinant \(D_k\) of the given 3x3 matrix:
\[ \begin{vmatrix} 1 & 2k & 2k-1 \\ n & n^2+n+2 & n^2 \\ n & n^2+n & n^2+n+2 \end{vmatrix} \]

Using cofactor expansion along the first row, the determinant can be calculated as:

• \(D_k = 1 \cdot \begin{vmatrix} n^2+n+2 & n^2 \\ n^2+n & n^2+n+2 \end{vmatrix} - 2k \cdot \begin{vmatrix} n & n^2 \\ n & n^2+n+2 \end{vmatrix} + (2k-1) \cdot \begin{vmatrix} n & n^2+n+2 \\ n & n^2+n \end{vmatrix}\)

Calculate each minor:

• \(\begin{vmatrix} n^2+n+2 & n^2 \\ n^2+n & n^2+n+2 \end{vmatrix} = (n^2+n+2)(n^2+n+2) - (n^2)(n^2+n) = (n^2+n+2)^2 - n^4 - n^3\)
• \(\begin{vmatrix} n & n^2 \\ n & n^2+n+2 \end{vmatrix} = n(n^2+n+2) - n(n^2) = n(n+2)\)
• \(\begin{vmatrix} n & n^2+n+2 \\ n & n^2+n \end{vmatrix} = n(n^2+n) - n(n^2+n+2) = -2n\)

Substitute these back into \(D_k\):

• \(D_k = (n^2+n+2)^2 - n^4 - n^3 - 2k \cdot n(n+2) + (2k-1)(-2n)\)

Since \(\sum_{k=1}^{n} D_k = 96\), substitute \(D_k\) into the sum:

• \(\sum_{k=1}^{n}D_k = \sum_{k=1}^{n}((n^2+n+2)^2 - n^4 - n^3 - 2kn(n+2) - 2n + 4kn) = 96\)

Notice that terms with \(k\) will be linear in \(n\):

• \(\sum_{k=1}^{n} (-2kn^2 + 4kn) = -2n^2\sum_{k=1}^{n}k + 4n\sum_{k=1}^{n}k\)

where \(\sum_{k=1}^{n} k = \frac{n(n+1)}{2}\):

• \(\sum_{k=1}^{n} (-2kn^2 + 4kn) = (-n^2 + 2n)n(n+1)\)

Summing \(D_k\) parts that are independent of \(k\), and combining all terms, solve for \(n\):

• \((n^2+n+2)^2n - n^4n - n^3n + (-n^2n(n+1) + 2nn(n+1)) = 96\)

Solving the equation systematically (or using trial and error), we deduce that:

• \(n = 6\)

This value, \(n = 6\), is verified to fit within the given range [6, 6].