def F(n):
if n == 1: return 1
elif n == 2: return 1
else: return F(n-1) + F(n-2),so f(5) equal to what

Question

def F(n):
if n == 1: return 1
elif n == 2: return 1
else: return F(n-1) + F(n-2),so f(5) equal to what

Accepted Answer

```python # define a recursive function named F to calculate the nth Fibonacci number def F(n): # base case: if n is 1, return 1 if n == 1: return 1 # base case: if n is 2, return 1 elif n == 2: return 1 # recursive case: return the sum of the two previous Fibonacci numbers else: return F(n-1) + F(n-2) ``` ```python # main function that tests the F function def main(): # test the F function # use case 1: F(5) should return 5 assert F(5) == 5 # use case 2: F(6) should return 8 assert F(6) == 8 # call the main function to run the tests if __name__ == "__main__": main() ``` ∻Explanation∻ ⚹The function F is defined to calculate the nth Fibonacci number recursively. The base cases return 1 for the first two numbers in the Fibonacci sequence. For all other cases, the function calls itself with the previous two numbers until it reaches the base cases. The main function tests the F function with two use cases to ensure it works correctly.⚹ ∻Step-by-step Instruction∻ ‖ ▪ ⋮ Define the recursive function F to calculate the nth Fibonacci number. ‖ ‖ ▪ ⋮ Include base cases for when n is 1 or 2, returning 1 as per the Fibonacci sequence. ‖ ‖ ▪ ⋮ For other cases, recursively call F with (n-1) and (n-2) and return their sum. ‖ ‖ ▪ ⋮ Define the main function to test the F function with specific use cases. ‖ ‖ ▪ ⋮ Use assert statements to verify the correctness of the F function for known Fibonacci numbers. ‖ ∻Time Complexity∻ ⚹The time complexity of the recursive Fibonacci function is exponential, specifically O(2^n), because it makes two recursive calls for each non-base case.⚹ ∻Space Complexity∻ ⚹The space complexity is O(n) due to the maximum depth of the recursion call stack.⚹

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