Question

If $△\left(r\right)=\left|\begin{array}{cc}r& r^3\\ 1& n(n+1)\end{array}\right|$, then $\sum _{r=1}^n△\left(r\right)$ is equal 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 1

The correct option is B $\sum _{r=1}^n{r}^3$

$△\left(r\right)=\left|\begin{array}{cc}r& r^3\\ 1& n(n+1)\end{array}\right|$
$⟹\sum _{r=1}n△\left(r\right)=\left|\begin{array}{cc}\sum _{r=1}^{n}r& \sum _{r=1}^n{r}^3\\ 1& n(n+1)\end{array}\right|$

$=\left|\begin{array}{cc}\frac{n(n+1)}{2}& \frac{\left[n\right(n+1){]}^2}{4}\\ 1& n(n+1)\end{array}\right|$

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

$=\frac{\left[n\right(n+1){]}^2}{4}=\sum _{r=1}^n{r}^3$