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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