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 value of $\sum\limits_{r=1}^{n}Δ_r$ is

$-2n$

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

$=\begin{vmatrix}n&\frac{(n)(n+1)}{2}&2^{n+1}-2\\2&n&n^2\\n&\frac{(n)(n+1)}{2}&2^{n+1}\end{vmatrix}=\begin{vmatrix}0&0&-2\\2&n&n^2\\n&\frac{(n)(n+1)}{2}&2^{n+1}\end{vmatrix}=-2n$