1. **Problem Statement:** Prove by the principle of mathematical induction that $$3^{2n+1} + (-1)^n 2 \equiv 0 \pmod{5}$$ for all positive integers $n$. 2. **Principle of Mathematical Induction:** To prove a statement $P(n)$ for all positive integers $n$: 1. **Base Case:** Show $P(1)$ is true. 2. **Inductive Step:** Assume $P(k)$ is true for some $k \geq 1$ (inductive hypothesis). 3. Prove $P(k+1)$ is true using the assumption. 3. **Base Case ($n=1$):** Calculate $$3^{2(1)+1} + (-1)^1 2 = 3^3 + (-1) \times 2 = 27 - 2 = 25.$$ Since $25 \equiv 0 \pmod{5}$, the base case holds. 4. **Inductive Hypothesis:** Assume for $n=k$, $$3^{2k+1} + (-1)^k 2 \equiv 0 \pmod{5}.$$ This means there exists an integer $m$ such that $$3^{2k+1} + (-1)^k 2 = 5m.$$ 5. **Inductive Step:** Prove for $n=k+1$: Consider $$3^{2(k+1)+1} + (-1)^{k+1} 2 = 3^{2k+3} + (-1)^{k+1} 2 = 3^{2k+1} \times 3^2 + (-1)^{k+1} 2 = 9 \times 3^{2k+1} - (-1)^k 2.$$ Rewrite as $$9 \times 3^{2k+1} - (-1)^k 2 = 9 \times 3^{2k+1} - (-1)^k 2.$$ Using the inductive hypothesis, $$3^{2k+1} = 5m - (-1)^k 2.$$ Substitute: $$9(5m - (-1)^k 2) - (-1)^k 2 = 45m - 9(-1)^k 2 - (-1)^k 2 = 45m - 10(-1)^k 2.$$ Factor out 5: $$= 5(9m - 2(-1)^k).$$ Since $9m - 2(-1)^k$ is an integer, the expression is divisible by 5. 6. **Conclusion:** By induction, $$3^{2n+1} + (-1)^n 2 \equiv 0 \pmod{5}$$ for all positive integers $n$. **Final answer:** The statement is true for all positive integers $n$ by mathematical induction.