Order the parts of the proof for:\\
$\sum_{i=0}^{n} r^i = \frac{r^{n+1}-1}{r-1}, \forall n \ge 0$\\
$\frac{r^{k+1}-1 + r^{k+1} - r^{k+1}}{r-1} = \frac{r^{k+2}-1}{r-1}$ \\
Therefore, if $\sum_{i=0}^{k} r^i = \frac{r^{k+1}-1}{r-1}$ for some $k \ge 0$, then\\
$\sum_{i=0}^{k+1} r^i = \frac{r^{k+2}-1}{r-1}$ \\
$\sum_{i=0}^{k+1} r^i = (\sum_{i=0}^{k} r^i) + r^{k+1} = (\frac{r^{k+1}-1}{r-1}) + \frac{(r-1)r^{k+1}}{r-1}$ \\
Let n=0,\\
By definition, $\sum_{i=0}^{0} r^i = r^0 = 1$ \\
By conjecture, $\frac{r^{0+1}-1}{r-1} = \frac{r-1}{r-1} = 1$ \\
Which confirms the conjecture for n=0.\\
Assume $\sum_{i=0}^{k} r^i = \frac{r^{k+1}-1}{r-1}$ for some $k \ge 0$ \\
Therefore, $\sum_{i=0}^{n} r^i = \frac{r^{n+1}-1}{r-1}, \forall n \ge 0$
