Let D_k=vmatrix 1 2k 2k-1 n n^2+n+2 n^2 n n^2+n n^2+n+2 vmatrix.

Question

Let ${D}_{k}=\begin{vmatrix} 1 \quad  2k \quad  2k-1 \ n \quad  {n}^{2}+n+2 \quad  {n}^{2} \ n \quad  {n}^{2}+n \quad  {n}^{2}+n+2 \end{vmatrix}$. If $\sum _{k=1}^{n}{D}_{k}=96$, then $n$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 6

Accepted Answer

$\begin{gathered} \sum _{k=1}^{n}{D}_{k}=|\begin{array}{ccc}\sum 1 \quad  2\sum k \quad  2\sum k-\sum 1 \ n \quad  {n}^{2}+n+2 \quad  {n}^{2} \ n \quad  {n}^{2}+n \quad  {n}^{2}+n+2\end{array}| \ =|\begin{array}{ccc}n \quad  n(n+1) \quad  {n}^{2} \ n \quad  {n}^{2}+n+2 \quad  {n}^{2} \ n \quad  {n}^{2}+n \quad  {n}^{2}+n+2\end{array}| \ =|\begin{array}{ccc}0 \quad  -2 \quad  0 \ 0 \quad  2 \quad  -n-2 \ n \quad  {n}^{2}+n \quad  {n}^{2}+n+2\end{array}| \ =2((-n)(-n-2))=96 \ \Rightarrow {n}^{2}+2n=48 \ \Rightarrow n=6,-8 \ \Rightarrow n=6 \end{gathered}$

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