(f) Consider the geometric series \( 50+25+12 \frac{1}{2}+\ndots+\frac{25}{8192} \). (1) Calculate the sum of the series. (2) Determine \( m \) if \( 50+25+12 \frac{1}{2}+\ndots \) (to \( m \) terms \( )=\frac{6375}{64} \). (3) Determine the least number of terms for which the sum will be greater than 95.

To solve the problem, we can start with the geometric series given: \( 50 + 25 + 12 \frac{1}{2} + \ldots + \frac{25}{8192} \). 1. **Calculation of the sum of the series:** The first term \( a \) of the series is 50, and the common ratio \( r \) can be calculated as follows: \[ r = \frac{25}{50} = \frac{1}{2} \] To find the number of terms \( n \) in the series, we note that the last term is \( \frac{25}{8192} \). To find \( n \): \[ \frac{50}{2^{n-1}} = \frac{25}{8192} \quad \Rightarrow \quad 50 \cdot 8192 = 25 \cdot 2^{n-1} \] \[ 409600 = 25 \cdot 2^{n-1} \quad \Rightarrow \quad 16384 = 2^{n-1} \] \[ 2^{14} = 2^{n-1} \quad \Rightarrow \quad n-1 = 14 \quad \Rightarrow \quad n = 15 \] Now, we use the formula for the sum of a geometric series: \[ S_n = \frac{a(1 - r^n)}{1 - r} \] Plugging in the values: \[ S_{15} = \frac{50(1 - (\frac{1}{2})^{15})}{1 - \frac{1}{2}} = \frac{50(1 - \frac{1}{32768})}{\frac{1}{2}} = 100(1 - \frac{1}{32768}) = 100 \cdot \frac{32767}{32768} \] \[ S_{15} = \frac{3276700}{32768} \approx 99.998 \text{ (almost 100)} \] So the sum of the series is \( \frac{3276700}{32768} \). 2. **Determine \( m \) if the sum equals \( \frac{6375}{64} \):** Using the same formula \( S_m \): \[ S_m = \frac{50(1 - (\frac{1}{2})^m)}{1 - \frac{1}{2}} = 100(1 - (\frac{1}{2})^m) \] Setting this equal to \( \frac{6375}{64} \): \[ 100(1 - (\frac{1}{2})^m) = \frac{6375}{64} \quad \Rightarrow \quad 1 - (\frac{1}{2})^m = \frac{6375}{6400} \] \[ (\frac{1}{2})^m = 1 - \frac{6375}{6400} = \frac{25}{6400} = \frac{1}{256} \] From this, we can deduce: \[ (\frac{1}{2})^m = (\frac{1}{2})^8 \quad \Rightarrow \quad m = 8 \] 3. **Determine the least number of terms for which the sum will be greater than 95:** To find \( m \) such that: \[ 100(1 - (\frac{1}{2})^m) > 95 \] We solve: \[ 1 - (\frac{1}{2})^m > 0.95 \quad \Rightarrow \quad (\frac{1}{2})^m < 0.05 \] Taking logarithms: \[ m \log(\frac{1}{2}) < \log(0.05) \quad \Rightarrow \quad m > \frac{\log(0.05)}{\log(0.5)} \approx \frac{-1.301}{-0.301} \approx 4.32 \] Thus, the least integer \( m = 5 \) is required for the sum to exceed 95.