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^{n}_{k=1}D_k= 48, $ then n equals

4

The correct answer is option (2) : 4

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

$⇒\begin{vmatrix}\sum\limits^{n}_{k=1}1 & n & n \\\sum\limits^{n}_{k=1}2k & n^2+n+2& n^2+n\\\sum\limits^{n}_{k=1}2k-1 & n^2 & n^2 + n+2\end{vmatrix}=48$

$⇒\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} $

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

$⇒ n(2n+4) = 48 $

$⇒n^2 + 2n - 24 = 0 $

$⇒ (n+6) (n-4) = 0 ⇒ n = 4 $    $[∵ n + 6 ≠0]$