(b) Calculate the sum of each of the following geometric series: \( \begin{array}{ll}\text { (1) } 6+12+24+\ldots+6144 & \text { (2) } 4-8+16 \ldots+16384 \\ \text { (3) }-5+15-45+\ldots-3645 & \text { (4) } 100-10+1-\ldots+\frac{1}{10000} \\ \text { (5) } 200\left(1+(1,01)+(1,01)^{2}+\ldots+(1,01)^{28}\right) & \text { (6) }-12-6-3-\ldots-\frac{3}{256} \\ \text { (7) } x y^{2}+x y+x+\ldots+\frac{x}{y^{12}} & \text { (8) } x^{5}-x^{3}+x-\ldots+\frac{1}{x^{11}}\end{array} \)

To find the sum of each of these geometric series, we can apply the formula for the sum of a finite geometric series, which is \( S_n = a \frac{1 - r^n}{1 - r} \), where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. (1) For the series \( 6 + 12 + 24 + \ldots + 6144 \): - Here, \( a = 6 \), \( r = 2 \), and \( n = 13 \) (since \( 6144 = 6 \cdot 2^{12} \)). - The sum is \( S_{13} = 6 \frac{1 - 2^{13}}{1 - 2} = 6 \cdot (8192 - 1) = 6 \cdot 8191 = 49146 \). (2) For \( 4 - 8 + 16 + \ldots + 16384 \): - In this case, \( a = 4 \), \( r = -2 \), and \( n = 14 \) (since \( 16384 = 4 \cdot (-2)^{13} \)). - The sum is \( S_{14} = 4 \frac{1 - (-2)^{14}}{1 - (-2)} = 4 \cdot \frac{1 - 16384}{3} = 4 \cdot \frac{-16383}{3} = -21844 \). (3) For \( -5 + 15 - 45 + \ldots - 3645 \): - Here, \( a = -5 \), \( r = -3 \), and \( n = 8 \) (since \( -3645 = -5 \cdot (-3)^{7} \)). - The sum is \( S_{8} = -5 \frac{1 - (-3)^{8}}{1 - (-3)} = -5 \cdot \frac{1 - 6561}{4} = -5 \cdot \frac{-6560}{4} = 8200 \). (4) For \( 100 - 10 + 1 - \ldots + \frac{1}{10000} \): - In this series, \( a = 100 \), \( r = -0.1 \), and \( n = 5 \) (since \( \frac{1}{10000} = 100 \cdot (-0.1)^{5} \)). - The sum is \( S_{5} = 100 \frac{1 - (-0.1)^{5}}{1 - (-0.1)} = 100 \cdot \frac{1 - (-0.00001)}{1.1} = 100 \cdot \frac{1.00001}{1.1} = 90.9091 \). (5) For \( 200\left(1 + (1.01) + (1.01)^{2} + \ldots + (1.01)^{28}\right) \): - Here, \( a = 1 \), \( r = 1.01 \), and \( n = 29 \). - The sum inside the parentheses is \( S_{29} = 1 \frac{1 - (1.01)^{29}}{1 - 1.01} \approx 1 \cdot \frac{1 - 1.34885}{-0.01} = 34.885 \), so \( 200 \cdot 34.885 \approx 6977 \). (6) For \( -12 - 6 - 3 - \ldots - \frac{3}{256} \): - Here, \( a = -12 \), \( r = \frac{1}{2} \), and \( n = 9 \) (since \( -\frac{3}{256} = -12 \cdot \left(\frac{1}{2}\right)^{8} \)). - The sum is \( S_{9} = -12 \frac{1 - \left(\frac{1}{2}\right)^{9}}{1 - \frac{1}{2}} = -12 \cdot \frac{1 - \frac{1}{512}}{\frac{1}{2}} = -12 \cdot 2 \cdot \frac{511}{512} = -12 \cdot \frac{1022}{512} = -24.149 \). (7) For \( xy^{2} + xy + x +