Answer:
Step-by-step explanation:
From the given information:
Consider the first positive integer power of 3 as 3¹ = 3
SO; 3 ∈ S
Thus, for every positive integer power of 3, it is set to be the previous integer of 3 multiplied by 3.
i.e. 3s ∈ S if and only if s ∈ S
Now for the first iteration, the recursive definition on s = 3
In the second iteration;
3s = 3 × 3 = 9 ∈ S
Thus, we use the recursive definition on s = 9 in the second iteration.
In the third iteration;
3s = 3 × 9 = 27 ∈ S
Thus, we use the recursive definition on s = 27 in the third iteration.
Then; 3s = 3 × 27 = 81 ∈ S .... and so on.
Therefore, the set is {3,9,27,81,...}