Let's break down the two parts of the problem:

Part (a): Recursive function $f(n+1) = f(n) + 2$

We are given the initial condition $f(0) = 1$ and the recursive relation $f(n+1) = f(n) + 2$. Now, let's find $f(1), f(2), f(3),$ and $f(4)$.

1. $f(1) = f(0) + 2 = 1 + 2 = 3$
2. $f(2) = f(1) + 2 = 3 + 2 = 5$
3. $f(3) = f(2) + 2 = 5 + 2 = 7$
4. $f(4) = f(3) + 2 = 7 + 2 = 9$

So, the values are:

• $f(1) = 3$
• $f(2) = 5$
• $f(3) = 7$
• $f(4) = 9$

Part (b): Recursive function $f(n+1) = 3f(n)$

We are given the initial condition $f(0) = 1$ and the recursive relation $f(n+1) = 3f(n)$. Now, let's find $f(1), f(2), f(3),$ and $f(4)$.

1. $f(1) = 3f(0) = 3 \times 1 = 3$
2. $f(2) = 3f(1) = 3 \times 3 = 9$
3. $f(3) = 3f(2) = 3 \times 9 = 27$
4. $f(4) = 3f(3) = 3 \times 27 = 81$

So, the values are:

• $f(1) = 3$
• $f(2) = 9$
• $f(3) = 27$
• $f(4) = 81$

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Let me know if you'd like more details or have questions.

Here are 5 related questions to deepen understanding:

1. What would $f(5)$ be for both part (a) and part (b)?
2. How would the function change in part (a) if the recursive relation was $f(n+1) = f(n) + 3$?
3. What is the general formula for $f(n)$ for part (a)?
4. Can you find the general formula for $f(n)$ in part (b)?
5. How does changing the initial condition $f(0)$ affect the results in both parts?

Tip: Recursion is often simplified by identifying patterns, which can sometimes lead to discovering a closed-form solution!