1. **Problem Statement:** Prove by mathematical induction that $$\sum_{r=1}^n \binom{r}{1} = \binom{n+1}{2}$$ for all positive integers $n$. 2. **Recall the definition:** $$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$ and note that $$\binom{r}{1} = r$$ because choosing 1 element from $r$ elements is $r$. 3. **Base Case ($n=1$):** $$\sum_{r=1}^1 \binom{r}{1} = \binom{1}{1} = 1$$ and $$\binom{1+1}{2} = \binom{2}{2} = 1$$ Both sides equal 1, so the base case holds. 4. **Inductive Hypothesis:** Assume the statement is true for some $k \in \mathbb{Z}^+$, i.e., $$\sum_{r=1}^k \binom{r}{1} = \binom{k+1}{2}$$ 5. **Inductive Step:** Prove for $k+1$: $$\sum_{r=1}^{k+1} \binom{r}{1} = \sum_{r=1}^k \binom{r}{1} + \binom{k+1}{1}$$ Using the inductive hypothesis: $$= \binom{k+1}{2} + (k+1)$$ 6. **Simplify the right side:** Recall $$\binom{k+1}{2} = \frac{(k+1)k}{2}$$, so $$\binom{k+1}{2} + (k+1) = \frac{(k+1)k}{2} + (k+1) = (k+1)\left(\frac{k}{2} + 1\right) = (k+1)\frac{k+2}{2}$$ 7. **Rewrite as a combination:** $$ (k+1)\frac{k+2}{2} = \frac{(k+1)(k+2)}{2} = \binom{k+2}{2}$$ 8. **Conclusion:** $$\sum_{r=1}^{k+1} \binom{r}{1} = \binom{k+2}{2}$$, so the statement holds for $k+1$. By mathematical induction, the formula $$\sum_{r=1}^n \binom{r}{1} = \binom{n+1}{2}$$ is true for all positive integers $n$.