Answer:
Option B
Explanation:
$\triangle (r)= \begin{bmatrix}r & r^{3} \\1 & n(n+1), \end{bmatrix}$
$\Rightarrow \sum_{r=1}^{n} \triangle (r)= \begin{bmatrix}\sum_{r=1}^{n}r & \sum_{r=1}^{n} r^{3} \\1 & n(n+1), \end{bmatrix}$
$= \begin{bmatrix}\frac{n(n+1)}{2} & \frac{[n(n+1)]^{2}}{2} \\1 & n(n+1), \end{bmatrix}$
=$\frac{[n(n+1)]^{2}}{2}-\frac{[n(n+1)]^{2}}{4}$
=$\frac{[n(n+1)]^{2}}{2}=\sum_{r=1}^{n}r^{3}$