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Answer:
f(N) = 2+ N/2 + 6N² units of time.
Step-by-step explanation:
Assigning 0 to the variable sum takes one unit of time.
Each time you increment sum by B, you need to call the value of sum, sum it to B and assign it to sum, which takes three units of time in total. You are repeating this process for each value of B which ranges from 1 to 4n and for each value of A which ranges from 1 to n/2. Opening the FOR takes also another unit of time, so, as a result, we have
f(N) 1 + 1 (open the FOR in A)+ N/2*(1 (open the FOR in B) + 4N*3) = 2+ N/2 + 6N² units of time. It has order complexity O(N²).
