1. **Problem a:** Prove by mathematical induction that $$2 + 7 + 12 + \cdots + (5n - 3) = \frac{n}{2}(5n - 1)$$ for all positive integers $n$. 2. **Base case:** For $n=1$, the left side is $2$ and the right side is $$\frac{1}{2}(5(1) - 1) = \frac{1}{2}(5 - 1) = \frac{1}{2} \times 4 = 2.$$ Both sides are equal, so the base case holds. 3. **Inductive hypothesis:** Assume the formula holds for some positive integer $k$, i.e., $$2 + 7 + 12 + \cdots + (5k - 3) = \frac{k}{2}(5k - 1).$$ 4. **Inductive step:** We need to prove it holds for $k+1$: $$2 + 7 + 12 + \cdots + (5k - 3) + (5(k+1) - 3) = \frac{k+1}{2}(5(k+1) - 1).$$ 5. Substitute the inductive hypothesis into the left side: $$\frac{k}{2}(5k - 1) + (5(k+1) - 3).$$ 6. Simplify the added term: $$5(k+1) - 3 = 5k + 5 - 3 = 5k + 2.$$ 7. So the left side becomes: $$\frac{k}{2}(5k - 1) + 5k + 2 = \frac{k}{2}(5k - 1) + \frac{2(5k + 2)}{2} = \frac{k}{2}(5k - 1) + \frac{10k + 4}{2}.$$ 8. Combine over common denominator 2: $$\frac{k(5k - 1) + 10k + 4}{2} = \frac{5k^2 - k + 10k + 4}{2} = \frac{5k^2 + 9k + 4}{2}.$$ 9. Now simplify the right side: $$\frac{k+1}{2}(5(k+1) - 1) = \frac{k+1}{2}(5k + 5 - 1) = \frac{k+1}{2}(5k + 4).$$ 10. Expand: $$\frac{(k+1)(5k + 4)}{2} = \frac{5k^2 + 4k + 5k + 4}{2} = \frac{5k^2 + 9k + 4}{2}.$$ 11. Both sides are equal, so the formula holds for $k+1$. By mathematical induction, the formula is true for all positive integers $n$. --- 12. **Problem b:** Show that for all positive integers $n$, $9^n + 7$ is divisible by 8. 13. **Base case:** For $n=1$, $$9^1 + 7 = 9 + 7 = 16,$$ which is divisible by 8. 14. **Inductive hypothesis:** Assume for some $k \geq 1$, $9^k + 7$ is divisible by 8. 15. **Inductive step:** Consider $9^{k+1} + 7 = 9 \times 9^k + 7.$ 16. Since $9 \equiv 1 \pmod{8}$, we have $$9^{k+1} + 7 \equiv 1 \times 9^k + 7 = 9^k + 7 \pmod{8}.$$ 17. By the inductive hypothesis, $9^k + 7$ is divisible by 8, so $9^{k+1} + 7$ is also divisible by 8. 18. Therefore, by induction, $9^n + 7$ is divisible by 8 for all positive integers $n$.