1. **Problem statement:** Prove by induction that $$\sum_{r=1}^n r = \frac{n(n+1)}{2}$$ for all natural numbers $n$. 2. **Base case:** For $n=1$, the left side is $1$ and the right side is $\frac{1(1+1)}{2} = 1$. Both sides are equal, so the base case holds. 3. **Inductive hypothesis:** Assume the formula holds for some $k \geq 1$, i.e., assume $$\sum_{r=1}^k r = \frac{k(k+1)}{2}.$$ 4. **Inductive step:** We need to prove the formula holds for $k+1$: $$\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + (k+1).$$ Using the inductive hypothesis: $$= \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}.$$ 5. This matches the formula with $n = k+1$, so the statement holds for $k+1$. 6. **Conclusion:** By the principle of mathematical induction, the formula $$\sum_{r=1}^n r = \frac{n(n+1)}{2}$$ holds for all natural numbers $n$. Final answer: $$\boxed{\sum_{r=1}^n r = \frac{n(n+1)}{2}}$$