1. The problem states that the sequence $a_n = 3^n$ for $n \geq 2$ satisfies the recurrence relation: $$2a_{n-1} - a_{n-2} = a_n$$ 2. We want to verify this by substituting $a_n = 3^n$ into the recurrence. 3. Substitute $a_{n-1} = 3^{n-1}$ and $a_{n-2} = 3^{n-2}$: $$2a_{n-1} - a_{n-2} = 2 \cdot 3^{n-1} - 3^{n-2}$$ 4. Factor out $3^{n-2}$: $$= 3^{n-2} (2 \cdot 3 - 1) = 3^{n-2} (6 - 1) = 3^{n-2} \cdot 5$$ 5. But the problem states the expression equals $3^n$, so let's re-examine the original substitution carefully: Actually, the problem's step is: $$2a_{n-1} - a_{n-2} = 2 \cdot 3^{n-1} - 3^{n-2} = 6 \cdot 3^{n-2} - 3^{n-2} = (6 - 1) 3^{n-2} = 5 \cdot 3^{n-2}$$ This does not equal $3^n = 3^{n-2} \cdot 3^2 = 9 \cdot 3^{n-2}$. 6. The problem's original statement has a typo in the simplification step. The correct simplification is: $$2 \cdot 3^{n-1} - 3^{n-2} = 2 \cdot 3 \cdot 3^{n-2} - 3^{n-2} = (6 - 1) 3^{n-2} = 5 \cdot 3^{n-2}$$ 7. Since $5 \cdot 3^{n-2} \neq 3^n$, the sequence $a_n = 3^n$ does not satisfy the recurrence relation $2a_{n-1} - a_{n-2} = a_n$. 8. However, if the recurrence relation is $2a_{n-1} - a_{n-2} = 3a_{n-1}$, then substituting $a_n = 3^n$ works: $$2 \cdot 3^{n-1} - 3^{n-2} = 3 \cdot 3^{n-1}$$ $$6 \cdot 3^{n-2} - 3^{n-2} = 3^{n}$$ $$5 \cdot 3^{n-2} = 3^{n}$$ This is false unless $5 = 9$, so no. 9. Conclusion: The original problem's verification is incorrect as stated. Final answer: The sequence $a_n = 3^n$ does not satisfy the recurrence relation $2a_{n-1} - a_{n-2} = a_n$ for $n \geq 2$ as shown by substitution and simplification.