Recursive Math Function | Calculating f(n) using f(n-1) + 6

f(1)=3, f(n)= f(n-1) +6 for n ≥ 2

In this problem, we are given a recursive function f(n), where the value of f(n) is defined based on the previous value of f(n-1)

In this problem, we are given a recursive function f(n), where the value of f(n) is defined based on the previous value of f(n-1). Specifically, we are told that f(1) = 3 and for any n greater than or equal to 2, f(n) is equal to f(n-1) plus 6.

To find the value of f(2), we can use the given recursive definition. Since f(1) is already known to be 3, we can find f(2) by substituting n=2 into the recursive formula:

f(2) = f(2-1) + 6
= f(1) + 6
= 3 + 6
= 9

So, f(2) is equal to 9.

Similarly, we can find the values of f(3), f(4), f(5), and so on, by repeatedly applying the recursive definition. Let’s evaluate f(3) as an example:

f(3) = f(3-1) + 6
= f(2) + 6 (using the previously calculated value of f(2))
= 9 + 6
= 15

So, f(3) is equal to 15.

Using the same method, we can calculate values for any positive integer n. The pattern is that each value of f(n) is equal to the previous value of f(n-1) plus 6.

In summary, the recursive function f(n) is defined as f(n) = f(n-1) + 6 for n ≥ 2, with an initial value of f(1) = 3. We can calculate the value of f(n) for any positive integer n by repeatedly applying the recursive formula.

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