Answer:
[tex]S_9=-18.703[/tex].
Step-by-step explanation:
The given series is,
[tex]\sum_{i=1}^9(-\frac{1}{2})^{i-1}[/tex]
When we substitute [tex]i=1[/tex], we get the first term, which is [tex]a_1=-28(-\frac{1}{2})^{1-1}[/tex]
This implies that,
[tex]a_1=-28(-\frac{1}{2})^{0}[/tex]
[tex]a_1=-28(1)=-28[/tex].
The common ratio is
[tex]r=-\frac{1}{2}[/tex]
The finite geometric sum is given by the formula,
[tex]S_n=\frac{a_1(r^n-1)}{r-1} , -1\:<\:r<\:1[/tex].
Since there are 9 terms, we find the sum of the first nine terms by putting [tex]n=9[/tex] in to the formula to get,
[tex]S_9=\frac{-28((-\frac{1}{2})^9-1)}{-\frac{1}{2}-1}[/tex].
[tex]S_9=\frac{-28((-\frac{1}{2})^9-1)}{-\frac{1}{2}-1}[/tex].
[tex]S_9=\frac{-28((-\frac{1}{512})-1)}{-\frac{3}{2}}[/tex].
[tex]S_9=\frac{-28(-\frac{513}{512})}{-\frac{3}{2}}[/tex].
[tex]S_9=-28(\frac{171}{256})[/tex].
[tex]S_9=-\frac{1197}{64}[/tex].
[tex]S_9=-18.703[/tex].
The correct answer is B
