1. **State the problem:** We need to show that every term of the sequence defined by $U_k = 2(4^k) + 1$ is divisible by 3 for all integers $k \geq 0$. 2. **Recall the divisibility rule:** A number is divisible by 3 if it leaves a remainder of 0 when divided by 3. 3. **Express the problem in modular arithmetic:** We want to prove that $U_k \equiv 0 \pmod{3}$ for all $k$. 4. **Analyze the term $4^k$ modulo 3:** Since $4 \equiv 1 \pmod{3}$, then $$4^k \equiv 1^k \equiv 1 \pmod{3}.$$ 5. **Substitute back into $U_k$ modulo 3:** $$U_k = 2(4^k) + 1 \equiv 2(1) + 1 = 3 \equiv 0 \pmod{3}.$$ 6. **Conclusion:** Since $U_k \equiv 0 \pmod{3}$, every term $U_k$ is divisible by 3 for all integers $k \geq 0$. Thus, the sequence terms are all divisible by 3 as required.