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If $\Delta(r)=\left|\begin{array}{cc}r & r^{3} \\ 1 & n(n+1)\end{array}\right|$, then $\sum_{r=1}^{n} \Delta(r)$ is equal to
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The correct answer is:
$\sum_{r=1}^{n} r^{3}$
$\Delta(\mathrm{r})=\left|\begin{array}{cc}\mathrm{r} & \mathrm{r}^{3} \\ 1 & \mathrm{n}(\mathrm{n}+1)\end{array}\right|$
$\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}} \Delta(\mathrm{r})=\left|\begin{array}{cc}\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} & \sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{3} \\ 1 & \mathrm{n}(\mathrm{n}+1)\end{array}\right|$
$=\left|\begin{array}{ll}\frac{n(n+1)}{2} & \frac{[n(n+1)]^{2}}{2} \\ 1 & n(n+1)\end{array}\right|$
$=\frac{[\mathrm{n}(\mathrm{n}+1)]^{2}}{2}-\frac{[\mathrm{n}(\mathrm{n}+1)]^{2}}{4}$
$=\frac{[\mathrm{n}(\mathrm{n}+1)]^{2}}{2}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{3}$
$\Rightarrow \sum_{\mathrm{r}=1}^{\mathrm{n}} \Delta(\mathrm{r})=\left|\begin{array}{cc}\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} & \sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{3} \\ 1 & \mathrm{n}(\mathrm{n}+1)\end{array}\right|$
$=\left|\begin{array}{ll}\frac{n(n+1)}{2} & \frac{[n(n+1)]^{2}}{2} \\ 1 & n(n+1)\end{array}\right|$
$=\frac{[\mathrm{n}(\mathrm{n}+1)]^{2}}{2}-\frac{[\mathrm{n}(\mathrm{n}+1)]^{2}}{4}$
$=\frac{[\mathrm{n}(\mathrm{n}+1)]^{2}}{2}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^{3}$
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