Find the sum of the first 8 terms of the following sequence. Round to the nearest hundredth if necessary. \[ 32, \quad-16, \quad 8, \ldots \] Sum of a finite geometric series: \[ S_{n}=\frac{a_{1}-a_{1} r^{n}}{1-r} \] Answer Attempt 1 out of 3

To solve for the sum of the first 8 terms of the geometric sequence \( 32, -16, 8, \ldots \), we first identify the first term \( a_1 = 32 \) and the common ratio \( r \). The common ratio is found by dividing the second term by the first term: \[ r = \frac{-16}{32} = -\frac{1}{2} \] Now we can apply the formula for the sum of a finite geometric series: \[ S_n = \frac{a_1(1 - r^n)}{1 - r} \] for \( n = 8 \): \[ S_8 = \frac{32(1 - (-\frac{1}{2})^8)}{1 - (-\frac{1}{2})} \] Calculating \( (-\frac{1}{2})^8 = \frac{1}{256} \): \[ S_8 = \frac{32(1 - \frac{1}{256})}{1 + \frac{1}{2}} = \frac{32(\frac{255}{256})}{\frac{3}{2}} \] Now simplifying further: \[ S_8 = \frac{32 \cdot 255}{256 \cdot \frac{3}{2}} = \frac{32 \cdot 255 \cdot 2}{256 \cdot 3} = \frac{32 \cdot 255}{384} \] Calculating \( \frac{32}{384} = \frac{1}{12} \): \[ S_8 = \frac{255}{12} \approx 21.25 \] So, the sum of the first 8 terms rounded to the nearest hundredth is: \[ \boxed{21.25} \]