1. **Problem statement:** Find a formula for the sum of the first $n$ even positive integers. 2. **Understanding the problem:** The first $n$ even positive integers are $2, 4, 6, \ldots, 2n$. 3. **Formula for the sum:** The sum $S$ of the first $n$ even positive integers is $$S = 2 + 4 + 6 + \cdots + 2n$$ 4. **Factor out 2:** $$S = 2(1 + 2 + 3 + \cdots + n)$$ 5. **Use the formula for the sum of the first $n$ natural numbers:** $$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$ 6. **Substitute this into the sum:** $$S = 2 \times \frac{n(n+1)}{2}$$ 7. **Simplify by canceling 2:** $$S = \cancel{2} \times \frac{n(n+1)}{\cancel{2}} = n(n+1)$$ 8. **Final formula:** $$\boxed{S = n(n+1)}$$ This means the sum of the first $n$ even positive integers is $n(n+1)$. --- **Slug:** sum even integers **Subject:** algebra **desmos:** {"latex":"y=n(n+1)","features":{"intercepts":true,"extrema":true}} **q_count:** 17