1. **Problem:** Prove by mathematical induction that $$4 + 7 + 10 + \cdots + (3n + 1) = \frac{n(3n+5)}{2}$$ for all positive integers $n$. 2. **Base Case:** For $n=1$, the left side is $4$ and the right side is $\frac{1(3(1)+5)}{2} = \frac{1 \times 8}{2} = 4$. Both sides are equal, so the base case holds. 3. **Inductive Hypothesis:** Assume the formula holds for some $k \geq 1$, i.e., $$4 + 7 + 10 + \cdots + (3k + 1) = \frac{k(3k+5)}{2}$$ 4. **Inductive Step:** We need to prove it holds for $k+1$: $$4 + 7 + 10 + \cdots + (3k + 1) + [3(k+1) + 1] = \frac{(k+1)(3(k+1)+5)}{2}$$ 5. Substitute the inductive hypothesis into the left side: $$\frac{k(3k+5)}{2} + (3k + 3 + 1) = \frac{k(3k+5)}{2} + (3k + 4)$$ 6. Simplify the right side of the inductive step: $$\frac{(k+1)(3k + 3 + 5)}{2} = \frac{(k+1)(3k + 8)}{2}$$ 7. Now simplify the left side: $$\frac{k(3k+5)}{2} + (3k + 4) = \frac{k(3k+5)}{2} + \frac{2(3k + 4)}{2} = \frac{k(3k+5) + 2(3k + 4)}{2}$$ 8. Expand numerator: $$k(3k+5) + 2(3k + 4) = 3k^2 + 5k + 6k + 8 = 3k^2 + 11k + 8$$ 9. So left side is: $$\frac{3k^2 + 11k + 8}{2}$$ 10. Expand right side: $$(k+1)(3k + 8) = 3k^2 + 8k + 3k + 8 = 3k^2 + 11k + 8$$ 11. Both sides equal: $$\frac{3k^2 + 11k + 8}{2}$$ 12. Therefore, the formula holds for $k+1$. 13. **Conclusion:** By mathematical induction, the formula is true for all positive integers $n$. **Final answer:** $$4 + 7 + 10 + \cdots + (3n + 1) = \frac{n(3n+5)}{2}$$