If

$Δ_r=\begin{vmatrix}1 & r & 2^r\\2 & n & n^2\\n & \frac{n(n+1)}{2} & 2^{n+1}\end{vmatrix},$ then the value of $\sum\limits^{n}_{r=1} Δ_r$ is

$-2n$

The correct answer is option (3) : $-2n$

$\sum\limits^{n}_{r=1} Δ_r=\begin{vmatrix}\sum\limits^{n}_{r=1}1 & \sum\limits^{n}_{r=1}r & \sum\limits^{n}_{r=1}2^r\\2 & n & n^2\\n & \frac{n(n+1)}{2} & 2^{n+1}\end{vmatrix},$

$⇒\sum\limits^{n}_{r=1} Δ_r=\begin{vmatrix}n & \frac{n(n+1)}{2} & 2^{n+1}-2\\2 & n & n^2\\n & \frac{n(n+1)}{2} & 2^{n+1}\end{vmatrix}$

$⇒\sum\limits^{n}_{r=1} Δ_r=\begin{vmatrix}n & \frac{n(n+1)}{2} & 2^{n+1}-2\\2 & n & n^2\\0 & 0 &2\end{vmatrix}$    [Applying $R_3→R_3-R_1$]

$⇒\sum\limits^{n}_{r=1} Δ_r=2\begin{Bmatrix}n^2-n(n+1)\end{Bmatrix}=-2n$