1. The problem is to simplify the expression $$\frac{1-r^{k-1}}{1-r}$$. 2. Recognize that the denominator is a difference of terms involving $r$, commonly appearing as the denominator of a geometric series sum. 3. The expression resembles the sum of a geometric series from the first term up to $r^{k-2}$. 4. Recall that the sum of a geometric series is given by: $$\sum_{i=0}^{n} r^i=\frac{1-r^{n+1}}{1-r}$$ 5. Comparing with our expression, note that: $$\frac{1-r^{k-1}}{1-r} = \sum_{i=0}^{k-2} r^i$$ 6. Therefore, the expression is the sum of the geometric series with $k-1$ terms starting from 1 (when $i=0$) and increasing the exponent by 1 each time up to $r^{k-2}$. Final answer: $$\boxed{\sum_{i=0}^{k-2} r^i = \frac{1-r^{k-1}}{1-r}}$$