1. **State the problem:** We want to show that every term of the sequence $U_k = 2(4^k) + 1$ 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 mathematically:** We want to prove that $3 \mid U_k$, i.e., $U_k \equiv 0 \pmod{3}$. 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 $U_k \equiv 0 \pmod{3}$ for all $k$, every term of the sequence is divisible by 3.