\( \frac { 3 } { 4 } - \frac { 3 } { 8 } + \frac { 3 } { 16 } - \ldots = \frac { 513 } { 1024 } \)

The series you've provided is an **infinite alternating geometric series**: \[ \frac{3}{4} - \frac{3}{8} + \frac{3}{16} - \frac{3}{32} + \ldots \] Let's analyze and find its sum. ### Identifying the Series Parameters An infinite geometric series has the form: \[ S = a + ar + ar^2 + ar^3 + \ldots \] where: - \( a \) is the first term, - \( r \) is the common ratio. In your series: - **First term (\( a \))**: \( \frac{3}{4} \) - **Common ratio (\( r \))**: \( -\frac{1}{2} \) This is because each successive term is multiplied by \( -\frac{1}{2} \): \[ \frac{3}{8} = \frac{3}{4} \times \left(-\frac{1}{2}\right), \quad \frac{3}{16} = \frac{3}{8} \times \left(-\frac{1}{2}\right), \quad \text{and so on}. \] ### Calculating the Sum For an infinite geometric series where \( |r| < 1 \), the sum \( S \) is given by: \[ S = \frac{a}{1 - r} \] Plugging in the values from your series: \[ S = \frac{\frac{3}{4}}{1 - \left(-\frac{1}{2}\right)} = \frac{\frac{3}{4}}{1 + \frac{1}{2}} = \frac{\frac{3}{4}}{\frac{3}{2}} = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2} \] ### Conclusion The **sum of the infinite series**: \[ \frac{3}{4} - \frac{3}{8} + \frac{3}{16} - \ldots \] is indeed: \[ \frac{1}{2} \] However, you mentioned that it equals \( \frac{513}{1024} \), which is approximately **0.50195**. This value is very close to \( \frac{1}{2} \) (**0.5**) but not exact. It's possible that \( \frac{513}{1024} \) represents a **partial sum** of the series up to a certain number of terms rather than the sum of the **infinite** series. ### Partial Sums Example For instance, let's compute the sum of the first **10 terms** of the series: \[ S_{10} = \frac{3}{4} - \frac{3}{8} + \frac{3}{16} - \frac{3}{32} + \cdots + \text{10 terms} \] Using the formula for the sum of the first \( n \) terms of a geometric series: \[ S_n = a \frac{1 - r^n}{1 - r} \] Plugging in the values: \[ S_{10} = \frac{3}{4} \times \frac{1 - \left(-\frac{1}{2}\right)^{10}}{1 - \left(-\frac{1}{2}\right)} = \frac{3}{4} \times \frac{1 - \frac{1}{1024}}{1 + \frac{1}{2}} = \frac{3}{4} \times \frac{\frac{1023}{1024}}{\frac{3}{2}} = \frac{3}{4} \times \frac{1023}{1024} \times \frac{2}{3} = \frac{1023}{2048} \approx 0.49951 \] This partial sum is approximately **0.49951**, which is also close to \( \frac{1}{2} \). However, to reach \( \frac{513}{1024} \approx 0.50195 \), you might need a different number of terms or there might be a specific context or modification to the series. ### Final Takeaway - **Infinite Series Sum**: \( \frac{1}{2} \) - **Your Provided Sum**: \( \frac{513}{1024} \) They are **not exactly equal**, but \( \frac{513}{1024} \) is a close approximation to \( \frac{1}{2} \). If you're dealing with an infinite series, the exact sum is \( \frac{1}{2} \).