To evaluate the expression \((\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + ...) - 1\), we first need to recognize the series involved in the expression. ### Step 1: Identify the Series The series can be rewritten as: \[ S = \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + ... \] This is an infinite geometric series where the first term \(a = \frac{1}{2}\) and the common ratio \(r = -\frac{1}{2}\). ### Step 2: Formula for the Sum of an Infinite Geometric Series The sum \(S\) of an infinite geometric series can be calculated using the formula: \[ S = \frac{a}{1 - r} \] This formula is valid when the absolute value of the common ratio \(|r| < 1\). ### Step 3: Apply the Formula In our case: - \(a = \frac{1}{2}\) - \(r = -\frac{1}{2}\) Now, substituting these values into the formula: \[ S = \frac{\frac{1}{2}}{1 - (-\frac{1}{2})} = \frac{\frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \] ### Step 4: Subtract 1 from the Sum Now that we have the sum \(S = \frac{1}{3}\), we need to evaluate the entire expression: \[ S - 1 = \frac{1}{3} - 1 \] To perform this subtraction, we convert 1 into a fraction with a denominator of 3: \[ 1 = \frac{3}{3} \] Now we can subtract: \[ \frac{1}{3} - \frac{3}{3} = \frac{1 - 3}{3} = \frac{-2}{3} \] ### Final Answer Thus, the final answer is: \[ \frac{-2}{3} \] ### Explanation of Other Options - **Option A: \(\frac{2}{3}\)** - This option is incorrect because it suggests a positive value, while our calculation shows a negative result. - **Option B: zero** - This option is incorrect as the sum of the series is not zero; it is \(\frac{1}{3}\), which when subtracted from 1 gives a negative result. - **Option D: -1** - This option is incorrect because the sum of the series is \(\frac{1}{3}\), and subtracting 1 from \(\frac{1}{3}\) does not yield -1. ### Revision Summary - The series \(\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + ...\) is an infinite geometric series with first term \(\frac{1}{2}\) and common ratio \(-\frac{1}{2}\). - The sum of the series is calculated using the formula \(S = \frac{a}{1 - r}\), resulting in \(S = \frac{1}{3}\). - Subtracting 1 from the sum gives \(\frac{-2}{3}\). - The correct answer is \(\frac{-2}{3}\) (Option C).