Geometric Sum Calculation

Write a function `geometric_sum(n)` that calculates the geometric sum of `n-1` using recursion. The geometric sum is defined as:

\[
S(n) = \sum_{i=0}^{n} \frac{1}{2^i}
\]

For example:

- `geometric_sum(0)` should return `1.0` (since \( \frac{1}{2^0} = 1 \)).
- `geometric_sum(1)` should return `1.5` (since \( \frac{1}{2^0} + \frac{1}{2^1} = 1.5 \)).
- `geometric_sum(2)` should return `1.75` (since \( \frac{1}{2^0} + \frac{1}{2^1} + \frac{1}{2^2} = 1.75 \)).

#### Example Usage
```python [main.nopy]
print(geometric_sum(3))  # Output: 1.875
print(geometric_sum(5))  # Output: 1.96875
```

#### Constraints
- The function should use recursion.
- Assume `n` is a non-negative integer.
def geometric_sum(n):
    """
    Calculate the geometric sum of n-1 using recursion.

    Args:
        n (int): The input non-negative integer.

    Returns:
        float: The geometric sum.
    """
    # Base case: when n is 0
    # Recursive case: calculate the sum for n-1
    pass
Intermediate recursionmathfunctions
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