Problem 1: $F(0) = 1$, $F(n) = 3 \cdot F(n-1)$

This recurrence defines a geometric progression. Let's solve it step by step.

Step 1: Base case

$F(0) = 1$

Step 2: Recurrence relation

For $n = 1$: $F(1) = 3 \cdot F(0) = 3 \cdot 1 = 3$

For $n = 2$: $F(2) = 3 \cdot F(1) = 3 \cdot 3 = 9$

For $n = 3$: $F(3) = 3 \cdot F(2) = 3 \cdot 9 = 27$

The solution is: $F(n) = 3^n$

Problem 2: $F(0) = 5$, $F(n) = 5 + F(n-1)$

This recurrence defines an arithmetic progression. Let's solve it step by step.

Step 1: Base case

$F(0) = 5$

Step 2: Recurrence relation

For $n = 1$: $F(1) = 5 + F(0) = 5 + 5 = 10$

For $n = 2$: $F(2) = 5 + F(1) = 5 + 10 = 15$

For $n = 3$: $F(3) = 5 + F(2) = 5 + 15 = 20$

The solution is: $F(n) = 5n + 5$

Final Results:

1. $F(n) = 3^n$
2. $F(n) = 5n + 5$

Let me know if you want further explanations!
Here are 5 related questions to deepen understanding:

1. What happens if the multiplier in problem 1 changes to a different number?
2. How would the solution change in problem 2 if the base case $F(0)$ was different?
3. Can you write the explicit formula for other recurrence relations?
4. How do these solutions relate to common sequences like Fibonacci or arithmetic progressions?
5. Can you derive the formulas for these problems using summation notation?

Tip: Recurrences can often be simplified by looking for patterns in their initial terms!