If $D_k=\begin{vmatrix}1&n&n\\2k&n^2+n+2&n^2+n\\2k-1&n^2&n^2+n+2\end{vmatrix}$ and $\sum\limits_{k=1}^{n}D_k=48$ then n equals

4

We have,

$\sum\limits_{k=1}^{n}D_k=48$

$⇒\begin{vmatrix}\sum\limits_{k=1}^{n}1&n&n\\\sum\limits_{k=1}^{n}2k&n^2+n+2&n^2+n\\\sum\limits_{k=1}^{n}2k-1&n^2&n^2+n+2\end{vmatrix}$

$⇒\begin{vmatrix}n&n&n\\n(n+1)&n^2+n+2&n^2+n\\n^2&n^2&n^2+n+2\end{vmatrix}$

$⇒\begin{vmatrix}n&0&0\\n^2+n&2&0\\n^2&0&n+2\end{vmatrix}=48$  [Applying $C_2→C_2-C_1,C_3→C_3-C_1$]

$⇒n(2n+4)=48$

$⇒n^2+2n-24=0$

$⇒(n+6)(n-4)=0⇒n=4$   $[∵n+6≠0]$