Find f(1), f(2), f(3), and, f(4) if f(n) is defined recursively by f(0) = 1 and for n = 0, 1, 2, . . .
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ICS 141: Discrete Mathematics I (Fall 2014)
a) f(n+1)=f(n)+2
f(1) = f(0 + 1) = f(0) + 2 = 1 + 2 = 3 f(2) = f(1 + 1) = f(1) + 2 = 3 + 2 = 5 f(3) = f(2 + 1) = f(2) + 2 = 5 + 2 = 7 f(4) = f(3 + 1) = f(3) + 2 = 7 + 2 = 9
b) f(n+1)=3f(n)
f(1) = f(0 + 1) = 3f(0) = 3(1) = 3 f(2) = f(1 + 1) = 3f(1) = 3(3) = 9 f(3) = f(2 + 1) = 3f(2) = 3(9) = 27 f(4) = f(3 + 1) = 3f(3) = 3(27) = 81

Given the recursive function definitions for both cases (a) and (b), we can now compute \( f(1) \), \( f(2) \), \( f(3) \), and \( f(4) \).

### Case (a) \( f(n+1) = f(n) + 2 \) and \( f(0) = 1 \):

We can compute the values step by step as follows:

- \( f(0) = 1 \)
- \( f(1) = f(0) + 2 = 1 + 2 = 3 \)
- \( f(2) = f(1) + 2 = 3 + 2 = 5 \)
- \( f(3) = f(2) + 2 = 5 + 2 = 7 \)
- \( f(4) = f(3) + 2 = 7 + 2 = 9 \)

Thus, for case (a), we have:

- \( f(1) = 3 \)
- \( f(2) = 5 \)
- \( f(3) = 7 \)
- \( f(4) = 9 \)

### Case (b) \( f(n+1) = 3f(n) \) and \( f(0) = 1 \):

Similarly, compute the values step by step:

- \( f(0) = 1 \)
- \( f(1) = 3 \cdot f(0) = 3 \cdot 1 = 3 \)
- \( f(2) = 3 \cdot f(1) = 3 \cdot 3 = 9 \)
- \( f(3) = 3 \cdot f(2) = 3 \cdot 9 = 27 \)
- \( f(4) = 3 \cdot f(3) = 3 \cdot 27 = 81 \)

Thus, for case (b), we have:

- \( f(1) = 3 \)
- \( f(2) = 9 \)
- \( f(3) = 27 \)
- \( f(4) = 81 \)

Would you like any more details or explanations?

### Related questions:
1. What is the general solution for \( f(n) \) in case (a)?
2. How would you solve a recursive function where \( f(n+1) = f(n) + n \)?
3. Can you express case (b) in terms of a closed-form equation?
4. How would the solution change if \( f(0) \neq 1 \) in either case?
5. How do you approach finding recursive sequences in general?

**Tip:** When dealing with recursive functions, it's helpful to compute the first few terms to observe any patterns, which may help in formulating a closed form.

Find f(1), f(2), f(3), and f(4) if f(n) is defined recursively by f(0) = 1 and for n = 0, 1, 2, ... (a) f(n+1) = f(n) + 2, (b) f(n+1) = 3f(n).

Solve the recursive function f(n) with two cases: (a) f(n+1) = f(n) + 2 and (b) f(n+1) = 3f(n). Step-by-step solutions include the arithmetic and geometric progressions used to find f(1), f(2), f(3), and f(4). Suitable for Grades 9-12.

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