To evaluate the sum
\[ \sum_{i=1}^{9} -28 \left( -\frac{1}{2} \right)^{i-1}, \]
we can identify this as a geometric series. We can factor out the constant term from the summation:
\[ S = -28 \sum_{i=1}^{9} \left( -\frac{1}{2} \right)^{i-1}. \]
The general formula for the sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{1 - r^n}{1 - r}, \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, we find:
- \( a = 1 \) (the first term corresponding to \( i=1 \))
- \( r = -\frac{1}{2} \)
- \( n = 9 \)
Now we can apply the finite geometric sum formula:
\[ S = -28 \cdot \left(1 \cdot \frac{1 - \left(-\frac{1}{2}\right)^9}{1 - \left(-\frac{1}{2}\right)}\right). \]
Calculating the terms:
- Calculate \(\left(-\frac{1}{2}\right)^9\):
\[ \left(-\frac{1}{2}\right)^9 = -\frac{1}{512}. \]
- Now substitute this into the sum formula:
\[ S = -28 \cdot \left( \frac{1 - \left(-\frac{1}{512}\right)}{1 + \frac{1}{2}} \right). \]
Simplifying \(1 - \left(-\frac{1}{512}\right)\):
\[ 1 + \frac{1}{512} = \frac{512}{512} + \frac{1}{512} = \frac{513}{512}. \]
And calculate \(1 + \frac{1}{2}\):
\[ 1 + \frac{1}{2} = \frac{3}{2}. \]
Therefore,
\[ S = -28 \cdot \left( \frac{\frac{513}{512}}{\frac{3}{2}} \right) = -28 \cdot \left( \frac{513}{512} \cdot \frac{2}{3} \right). \]
Carrying out the multiplication:
\[ S = -28 \cdot \frac{1026}{1536} = -28 \cdot \frac{513}{768}. \]
Calculating \(-28 \cdot \frac{513}{768}\):
\[ S = -\frac{14364}{768}. \]
Now we simplify this fraction by dividing both numerator and denominator by the gcd (which is 12):
\[ S = -\frac{1197}{64}. \]
Finally, converting this to a decimal:
\[ S \approx -18.703125. \]
Rounding to three decimal places, we find:
\[ \boxed{-18.703}. \]
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