1. **Problem:** Show whether $a_n = 2^{n+1} - 1$, $n \geq 1$, is a solution to the recurrence relation $a_n = 3a_{n-1} - 2a_{n-2}$. 2. **Formula:** The recurrence relation is given by $$a_n = 3a_{n-1} - 2a_{n-2}$$ 3. **Step 1:** Calculate $a_{n-1}$ and $a_{n-2}$ using the given formula for $a_n$: $$a_{n-1} = 2^n - 1$$ $$a_{n-2} = 2^{n-1} - 1$$ 4. **Step 2:** Substitute into the right side of the recurrence: $$3a_{n-1} - 2a_{n-2} = 3(2^n - 1) - 2(2^{n-1} - 1) = 3 \cdot 2^n - 3 - 2 \cdot 2^{n-1} + 2$$ 5. **Step 3:** Simplify the expression: $$3 \cdot 2^n - 2 \cdot 2^{n-1} - 1 = 3 \cdot 2^n - 2^n - 1 = 2 \cdot 2^n - 1 = 2^{n+1} - 1$$ 6. **Step 4:** This matches the left side $a_n = 2^{n+1} - 1$, so the given $a_n$ satisfies the recurrence. **Final answer:** Yes, $a_n = 2^{n+1} - 1$ is a solution to the recurrence relation $a_n = 3a_{n-1} - 2a_{n-2}$.