To evaluate the geometric series [tex]\(\sum_{i=1}^8 2^{i-1}\)[/tex], we need to recognize that it is a geometric series with the first term [tex]\(a = 2^{1-1} = 1\)[/tex] and the common ratio [tex]\(r = 2\)[/tex]. The series is given by:
[tex]\[ S_n = \sum_{i=1}^n ar^{i-1} = a (1 + r + r^2 + \cdots + r^{n-1}) \][/tex]
For the series [tex]\(\sum_{i=1}^8 2^{i-1}\)[/tex]:
- The number of terms [tex]\(n = 8\)[/tex]
- The first term [tex]\(a = 1\)[/tex]
- The common ratio [tex]\(r = 2\)[/tex]
The sum of the first [tex]\(n\)[/tex] terms of a geometric series can be calculated using the formula:
[tex]\[ S_n = a \frac{r^n - 1}{r - 1} \][/tex]
Plugging in the values we have:
[tex]\[ S_8 = 1 \times \frac{2^8 - 1}{2 - 1} \][/tex]
Let's simplify inside the fraction:
[tex]\[ 2^8 = 256 \][/tex]
So:
[tex]\[ S_8 = \frac{256 - 1}{1} = \frac{255}{1} = 255 \][/tex]
Thus, the correct answer is:
B. 255
