We are tasked with finding $f(5)$, given:

1. $f(1) = 10$
2. $f(n) = -2f(n-1)$ for $n \geq 2$

Step-by-step Solution:

Using the recurrence relation $f(n) = -2f(n-1)$, we compute the values of $f(2)$, $f(3)$, $f(4)$, and $f(5)$:

1. Base case: $f(1) = 10$
2. $f(2) = -2f(1) = -2 \cdot 10 = -20$
3. $f(3) = -2f(2) = -2 \cdot (-20) = 40$
4. $f(4) = -2f(3) = -2 \cdot 40 = -80$
5. $f(5) = -2f(4) = -2 \cdot (-80) = 160$

Final Answer:

$f(5) = 160$

Let me know if you'd like further clarification or additional details!

Follow-up Questions:

1. How does a recurrence relation define a sequence?
2. What happens if the initial value $f(1)$ changes?
3. Can this recurrence relation be expressed explicitly as a closed-form formula?
4. What are some examples of recurrence relations in real-world applications?
5. How do negative coefficients affect the behavior of sequences?

Tip: When solving recurrence relations, always compute several terms to ensure your understanding of the sequence's behavior.