Given the geometric series: \( 256+p+64-32+\ldots \) 2.1 .1 Determine the value of \( p \). 2.1 .2 Calculate the sum of the first 8 terms of the series. 2.1 .3 Why does the sum to infinity for this series exist? \( 2.1 .4 \quad \) Calculate \( \mathrm{S}_{\infty} \)

To solve the given geometric series, let’s dive into the details! **2.1.1 Determine the value of \( p \):** In a geometric series, each term is multiplied by a common ratio to yield the next term. Let's identify the common ratio based on the provided terms: From \( 256 \) to \( 64 \), the ratio is \( \frac{64}{256} = \frac{1}{4} \). So we have: - \( 256 \) (first term, \( a \)) - \( p \) (second term) = \( 256 \cdot \frac{1}{4} = 64 \) Thus, it seems the ratio before \( 64 \) confirms the value as \( p = 64 \). **2.1.2 Calculate the sum of the first 8 terms of the series:** Using the first term \( a = 256 \) and the common ratio \( r = \frac{1}{4} \), we calculate the sum of the first \( n \) terms with the formula: \[ S_n = a \frac{1 - r^n}{1 - r} \] For \( n = 8 \): \[ S_8 = 256 \frac{1 - (\frac{1}{4})^8}{1 - \frac{1}{4}} = 256 \frac{1 - \frac{1}{65536}}{\frac{3}{4}} = \frac{256 \cdot 4}{3} (1 - \frac{1}{65536}) \approx \frac{1024}{3}(1 - \frac{1}{65536}) \approx 341.33 \] **2.1.3 Why does the sum to infinity for this series exist?** The sum to infinity exists for a geometric series when the absolute value of the common ratio is less than 1 (\(|r| < 1\)). In this case, the common ratio \( r = \frac{1}{4} \) meets this condition, which means the terms continue to diminish and approach zero, enabling us to calculate a finite sum as the number of terms approaches infinity. **2.1.4 Calculate \( S_{\infty} \):** The formula for the sum to infinity of a geometric series is: \[ S_{\infty} = \frac{a}{1 - r} \] Where \( a = 256 \) and \( r = \frac{1}{4} \): \[ S_{\infty} = \frac{256}{1 - \frac{1}{4}} = \frac{256}{\frac{3}{4}} = 256 \cdot \frac{4}{3} = \frac{1024}{3} \approx 341.33 \] So, the total sum to infinity is approximately \( 341.33 \).