The expression you provided is:

$\frac{3^{n-1} + 3^{n-2}}{3^{n-3}}$

Let’s simplify this step by step:

Step 1: Factor the numerator

The numerator consists of two terms, $3^{n-1} + 3^{n-2}$. We can factor out the common factor, which is $3^{n-2}$:

$3^{n-1} + 3^{n-2} = 3^{n-2}(3^1 + 1) = 3^{n-2}(3 + 1) = 3^{n-2} \cdot 4$

Step 2: Divide by the denominator

Now, divide the factored numerator by $3^{n-3}$:

$\frac{3^{n-2} \cdot 4}{3^{n-3}} = 4 \cdot 3^{(n-2) - (n-3)} = 4 \cdot 3^1 = 4 \cdot 3 = 12$

Final Answer:

The expression simplifies to $12$.

Let me know if you need further explanation!

Here are five related questions for further exploration:

1. How would this change if we had powers of 5 instead of 3?
2. Can you express this in logarithmic form?
3. What happens if you add more terms with decreasing powers of 3?
4. How do you apply the same approach to expressions with different bases?
5. How can this be extended to solve similar problems in calculus?

Tip: Always look for common factors in the numerator before diving into complex simplifications.