Question
If $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}=55,$ find $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3$

Answer

$\sum\limits^{\text{n}}_{\text{r}=1}\text{r}=55$
$\Rightarrow\frac{\text{n}(\text{n}+1)}{2}=55\ ...(\text{i})$
Now,
$\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3=\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$
$=\big[55\big]^2$ [Using equation (i), we get]
$=3025$
Hence, $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3=3025$

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