Question
Let $D_{k}=\left|\begin{array}{ccc}2 k-1 & n^{2} & n^{2} \\ 2 k & n^{2}+n+1 & n^{2}+n \\ 6 k^{2} & 2 n^{3}+3 n^{2}+n & 2 n^{3}+6 n^{2}-2 n\end{array}\right|$ Then the value of $n$, if $\sum_{k=1}^{n} D_{k}$ equals 10752, is(a) 8(b) 0(c) 4(d) 6
Step 1
Step 1: We start by calculating the determinant \( D_k \) given by the matrix: \[ D_k = \begin{vmatrix} 2k-1 & n^2 & n^2 \\ 2k & n^2+n+1 & n^2+n \\ 6k^2 & 2n^3+3n^2+n & 2n^3+6n^2-2n \end{vmatrix} \] Show more…
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