1. **State the problem:** Prove by mathematical induction that for all $n \geq 1$, the sum of the sequence $1 + 4 + 7 + \cdots + (3n - 2)$ equals $\frac{n(3n-1)}{2}$. 2. **Base case:** For $n=1$, the left side is just the first term $1$. Calculate the right side: $$\frac{1(3\times1 - 1)}{2} = \frac{1 \times 2}{2} = 1$$ Since both sides equal 1, the base case holds. 3. **Inductive hypothesis:** Assume the formula holds for some $k \geq 1$: $$1 + 4 + 7 + \cdots + (3k - 2) = \frac{k(3k - 1)}{2}$$ 4. **Inductive step:** Prove it holds for $k+1$: Consider the sum up to $k+1$ terms: $$1 + 4 + 7 + \cdots + (3k - 2) + [3(k+1) - 2]$$ Using the inductive hypothesis, this is: $$\frac{k(3k - 1)}{2} + (3k + 3 - 2) = \frac{k(3k - 1)}{2} + (3k + 1)$$ 5. **Simplify the right side:** $$\frac{k(3k - 1)}{2} + \frac{2(3k + 1)}{2} = \frac{3k^2 - k + 6k + 2}{2} = \frac{3k^2 + 5k + 2}{2}$$ 6. **Factor the numerator:** $$3k^2 + 5k + 2 = (3k + 2)(k + 1)$$ So the sum is: $$\frac{(3k + 2)(k + 1)}{2}$$ 7. **Rewrite the right side formula for $k+1$:** $$\frac{(k+1)(3(k+1) - 1)}{2} = \frac{(k+1)(3k + 3 - 1)}{2} = \frac{(k+1)(3k + 2)}{2}$$ 8. **Conclusion:** The expression matches the simplified sum, so the formula holds for $k+1$. By the principle of mathematical induction, the formula $$1 + 4 + 7 + \cdots + (3n - 2) = \frac{n(3n - 1)}{2}$$ holds for all $n \geq 1$. **Final answer:** The formula is proven by induction.