f(1) = 3, f(n) = f(n-1) + 9for n ≥ 2
In this problem, we are given a function defined recursively
In this problem, we are given a function defined recursively. The function is denoted as f(n), where n is an integer greater than or equal to 1.
The base case for the recursion is given as f(1) = 3. This is the initial value of the function.
For n ≥ 2, the function is defined as f(n) = f(n-1) + 9. This means that to find the value of f(n), we need to find the value of f(n-1) (the previous term) and add 9 to it.
Let’s use this definition to find the values of f(n) for some specific values of n:
1. f(1) = 3 (given base case)
2. f(2) = f(1) + 9 = 3 + 9 = 12
3. f(3) = f(2) + 9 = 12 + 9 = 21
4. f(4) = f(3) + 9 = 21 + 9 = 30
5. f(5) = f(4) + 9 = 30 + 9 = 39
Continuing this pattern, we can find the values of f(n) for any positive integer n.
To summarize, the function f(n) is defined recursively as f(1) = 3, and for n ≥ 2, it is given by f(n) = f(n-1) + 9. Recursive functions are defined in terms of their previous terms, and you can find the value of f(n) by repeatedly using the given definition.
More Answers:
[next_post_link]