Question

${\sum }_{r=1}^n{r}^2-{\sum }_{m=1}^n{\sum }_{r=1}^{m}r$ is equal to

• A. 0
• B. ${\sum }_{r=1}^n{r}^2+{\sum }_{r=1}^{n}r$
• C. $\frac{1}{2}({\sum }_{r=1}^n{r}^2-{\sum }_{r=1}^{n}r)$
• D. none of these

Answer 1

The correct option is C $\frac{1}{2}({\sum }_{r=1}^n{r}^2-{\sum }_{r=1}^{n}r)$

${\sum }_{r=1}^n{r}^2-{\sum }_{m=1}^n{\sum }_{r=1}^{m}r={\sum }_{r=1}^n{r}^2-{\sum }_{m=1}^n\frac{m(m+1)}{2}={\sum }_{r=1}^n{r}^2-{\sum }_{m=1}^n\frac{m^2}{2}-{\sum }_{m=1}^n\frac{m}{2}={\sum }_{r=1}^n{r}^2-\frac{1}{2}{\sum }_{r=1}^n{r}^2-\frac{1}{2}{\sum }_{r=1}^{n}r=\frac{1}{2}{\sum }_{r=1}^n{r}^2-\frac{1}{2}{\sum }_{r=1}^{n}r=\frac{1}{2}[{\sum }_{r=1}^n{r}^2-{\sum }_{r=1}^{n}r]$