1. **Statement of the problem:** Prove by induction that for all natural numbers $n$, the expression $$4^n + 6n - 1$$ is divisible by 9. 2. **Formula and induction principle:** We want to prove that $$9 \mid (4^n + 6n - 1)$$ for all $n \in \mathbb{N}$. We use mathematical induction: - Base case: check for $n=0$ or $n=1$. - Inductive step: assume true for $n=k$, prove for $n=k+1$. 3. **Base case ($n=0$):** Calculate $$4^0 + 6\cdot0 - 1 = 1 + 0 - 1 = 0$$. Since 0 is divisible by 9, base case holds. 4. **Inductive hypothesis:** Assume for some $k \geq 0$, $$4^k + 6k - 1 = 9m$$ for some integer $m$. 5. **Inductive step:** We want to prove: $$4^{k+1} + 6(k+1) - 1$$ is divisible by 9. Rewrite: $$4^{k+1} + 6k + 6 - 1 = 4 \cdot 4^k + 6k + 5$$ Using the hypothesis, write $4^k = 9m - 6k + 1$: $$4 \cdot (9m - 6k + 1) + 6k + 5 = 36m - 24k + 4 + 6k + 5 = 36m - 18k + 9$$ Factor out 9: $$9(4m - 2k + 1)$$ This is divisible by 9. 6. **Conclusion:** By induction, $$4^n + 6n - 1$$ is divisible by 9 for all $n \in \mathbb{N}$. **Final answer:** The expression $$4^n + 6n - 1$$ is a multiple of 9 for every natural number $n$.