1. We are asked to prove that for all $n \in \mathbb{N}$, the sum $\sum_{k=0}^{n-1} (2k+1)$ equals $n^2$. 2. Rewrite the sum explicitly to understand it better: $$\sum_{k=0}^{n-1} (2k+1) = (2\cdot0 + 1) + (2\cdot1 + 1) + \cdots + (2(n-1) + 1).$$ 3. Notice that the terms are the first $n$ odd numbers: $1, 3, 5, ..., 2n-1$. 4. One way to prove this is by induction: **Base case:** For $n=1$, $$\sum_{k=0}^0 (2k+1) = 1,$$ and $$1^2 = 1.$$ So the base case holds. 5. **Inductive step:** Assume the formula holds for some $n = m$, that is, $$\sum_{k=0}^{m-1} (2k+1) = m^2.$$ We want to show it holds for $n = m+1$: $$\sum_{k=0}^m (2k+1) = \left( \sum_{k=0}^{m-1} (2k+1) \right) + (2m+1).$$ 6. By the inductive hypothesis, substitute: $$= m^2 + (2m + 1).$$ 7. Simplify: $$m^2 + 2m + 1 = (m+1)^2.$$ 8. Hence, $$\sum_{k=0}^m (2k+1) = (m+1)^2,$$ completing the induction step. 9. By induction, the formula $$\sum_{k=0}^{n-1} (2k+1) = n^2$$ holds for all natural numbers $n$. Final answer: $$\boxed{\sum_{k=0}^{n-1} (2k+1) = n^2}.$$