Suppose that f(n)=f(n/5)+3n when n is a positive integer divisible by 5, and f(1)=4.

By means of recurrence formulas and a given initial value, we find the following three results: f(5) = 19, f(125) = 469, f(3125) = 11719.
Sequences are sets of values defined by at least one condition. In this case, we have three conditions to generate the required values:
f(n) = f(n/5) + 3 · n
f(1) = 4
[tex]n \in \mathbb{N}[/tex]
Now we proceed to find the elements by recurrence:
f(5) = f(1) + 3 · 5
f(5) = 4 + 15
f(5) = 19
f(25) = f(5) + 3 · 25
f(25) = 19 + 75
f(25) = 94
f(125) = f(25) + 3 · 125
f(125) = 94 + 375
f(125) = 469
f(625) = f(125) + 3 · 625
f(625) = 469 + 3 · 625
f(625) = 469 + 1875
f(625) = 2344
f(3125) = f(625) + 3 · 3125
f(3125) = 2344 + 9375
f(3125) = 11719
By means of recurrence formulas and a given initial value, we find the following three results: f(5) = 19, f(125) = 469, f(3125) = 11719.
To learn more on recurrence formulas: https://brainly.com/question/13391306
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