1)

If  $\triangle (r)= \begin{bmatrix}r & r^{3} \\1 & n(n+1),  \end{bmatrix}$  , then  $\sum_{r=1}^{n}\triangle (r)$ eqal to


A) $\sum_{r=1}^{n}r^{2}$

B) $\sum_{r=1}^{n}r^{3}$

C) $\sum_{r=1}^{n}r^{}$

D) $\sum_{r=1}^{n}r^{4}$

Answer:

Option B

Explanation:

 $\triangle (r)= \begin{bmatrix}r & r^{3} \\1 & n(n+1),  \end{bmatrix}$ 

$\Rightarrow \sum_{r=1}^{n} \triangle (r)= \begin{bmatrix}\sum_{r=1}^{n}r & \sum_{r=1}^{n} r^{3} \\1 & n(n+1),  \end{bmatrix}$

 $= \begin{bmatrix}\frac{n(n+1)}{2} & \frac{[n(n+1)]^{2}}{2} \\1 & n(n+1),  \end{bmatrix}$

=$\frac{[n(n+1)]^{2}}{2}-\frac{[n(n+1)]^{2}}{4}$

=$\frac{[n(n+1)]^{2}}{2}=\sum_{r=1}^{n}r^{3}$