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 $\sum\limits_{n=1}^{k}I_n=72$, then k = 

8

We have,

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

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

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

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

$⇒\begin{vmatrix}0&0&k\\0&1&k^2+k\\-(k+1)&-(k+1)&k^2+k+1\end{vmatrix}=72$ [Applying $C_1→C_1-C_3, C_2 →C_2-C_3$]

$⇒k (k+1)=72$

$⇒k (k+1)=8×9⇒ k=8$.