1. The given expression is \( \frac{1-r^k}{1-r} \).\n2. This expression is a common formula in algebra called the sum of a geometric series.\n3. It represents the sum of the series \(1 + r + r^2 + \dots + r^{k-1}\) for \(r \neq 1\).\n4. To understand it, you can multiply the sum by \(1-r\):\n$$ (1-r) \times \sum_{n=0}^{k-1} r^n = (1-r)(1 + r + r^2 + \dots + r^{k-1}) $$\n5. Expanding the right side using distributive property:\n$$ (1 + r + r^2 + \dots + r^{k-1}) - (r + r^2 + \dots + r^k) $$\n6. Observe that all intermediate terms cancel out, leaving:\n$$ 1 - r^k $$\n7. Dividing both sides by \(1-r\) gives:\n$$ \sum_{n=0}^{k-1} r^n = \frac{1-r^k}{1-r} $$\n8. This formula helps quickly compute the sum of the geometric series without adding each term separately.\n\nFinal answer: \n\[ \sum_{n=0}^{k-1} r^n = \frac{1-r^k}{1-r} \]