Mathematics | VITEEE 2015 | Discussion Page

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

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

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

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

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

Option B

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

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

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

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

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

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