If $I_n = \begin{vmatrix}1 & k & k \\ 2n & k^2+k+1 & k^2 +k\\2n-1 & k^2 & k^2 +k+1\end{vmatrix} $ and $\sum\limits^{k}_{n=1} I_n = 72, $ then k =

8

The correct answer is option (1) : 8

$\sum\limits^{k}_{n=1}I_n= 72 $

$⇒\sum\limits^{k}_{n=1}I_n= \begin{vmatrix}1 & k & k \\ 2n & k^2+k+1 & k^2 +k\\2n-1 & k^2 & k^2 +k+1\end{vmatrix}=72 $

$⇒\begin{vmatrix}\sum\limits^{k}_{n=1}1 & k & k \\ 2\sum\limits^{k}_{n=1}n & k^2+k+1 & k^2 +k\\\sum\limits^{k}_{n=1}2n-1 & k^2 & k^2 +k+1\end{vmatrix}=72 $

$⇒ \begin{vmatrix}k & k & k \\ k^2 +k & k^2+k+1 & k^2 +k\\k^2& k^2 & k^2 +k+1\end{vmatrix}=72 $

$⇒ \begin{vmatrix}0 & 0 & k \\ 0  & 1& k^2 +k\\-(k+1)& -(k+1) & k^2 +k+1\end{vmatrix}=72 $

[Applying $C_1→C_1-C_3, C_2 →C_2-C_3$]

$⇒k(k+1) = 72 $

$⇒k(k+1) = 8 ×9 ⇒k = 8 $