1. **State the problem:** Prove by mathematical induction that for all integers $n \geq 1$, the sum of the first $n$ even numbers is given by: $$2 + 4 + 6 + \cdots + 2n = n(n+1)$$ 2. **Base case:** Check the statement for $n=1$. Left side: $2$. Right side: $1(1+1) = 1 \times 2 = 2$. Since both sides are equal, the base case holds. 3. **Inductive hypothesis:** Assume the statement is true for some integer $k \geq 1$, i.e., $$2 + 4 + 6 + \cdots + 2k = k(k+1)$$ 4. **Inductive step:** Prove the statement for $k+1$. Consider the sum up to $2(k+1)$: $$2 + 4 + 6 + \cdots + 2k + 2(k+1)$$ Using the inductive hypothesis, this equals: $$k(k+1) + 2(k+1)$$ Factor out $(k+1)$: $$ (k+1)(k + 2) $$ Rewrite as: $$ (k+1)((k+1) + 1) $$ This matches the formula for $n = k+1$. 5. **Conclusion:** By the principle of mathematical induction, the formula $$2 + 4 + 6 + \cdots + 2n = n(n+1)$$ holds for all integers $n \geq 1$. This completes the proof.