Problem: Prove that for all $ n,k \in \mathbb{N}, k > 1$, we have $$\sum \limits_{i=0}^{n} k^i = \frac{k^{n+1}-1}{k-1}$$ Solution: Representing the numbers in base $ k$, we have that each term of the sum is all 0’s except for a 1 in the $ i$th place. Hence, the sum of all terms is the $ n$-digit number comprised of all 1’s. Multiplying by $ k-1$ gives us the $ n$-digit number where every digit is $ k-1$.