(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}} \text { How many terms of each the following geometric series must be added to obtain the given } \\ \text { sum? } \\ \text { (1) } 2+12+72+\ldots=18662 & \text { (2) } 8+\frac{8}{3}+\frac{8}{9}+\ldots=\frac{8744}{729} \\ \text { (3) }-4+16-64+\ldots=52428 & \text { (4) } \frac{3}{4}-\frac{3}{8}+\frac{3}{16}-\ldots=\frac{513}{1024}\end{array} \
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(1) The series \(6 + 12 + 24 + \ldots + 6144\) is a geometric series where the first term \(a = 6\) and the common ratio \(r = 2\). The number of terms \(n\) can be found using the formula for the nth term of a geometric series, which yields \(n = 13\). The sum \(S_n\) is calculated as \(S_n = a \frac{r^n - 1}{r - 1} = 6 \frac{2^{13} - 1}{2 - 1} = 6(8191) = 49146\). (2) The second series \(4 - 8 + 16 - \ldots + 16384\) features an alternating sign with \(a = 4\) and \(r = -2\). The number of terms can be derived in a similar fashion; here, \(n = 14\). For the sum, apply the geometric sum formula: \(S_n = 4 \frac{(-2)^{14} - 1}{-2 - 1} = 4 \frac{16384 - 1}{-3} = \frac{-65512}{-3} = 21837\). (3) For the series \(-5 + 15 - 45 + \ldots - 3645\), where \(a = -5\) and \(r = -3\), the number of terms is determined to be \(n = 8\). The sum calculated will reveal that \(S_n = -5 \frac{(-3)^8 - 1}{-3 - 1} = -5 \frac{6561 - 1}{-4} = 5(1640) = 8200\). (4) In the series \(100 - 10 + 1 - \ldots + \frac{1}{10000}\), the initial term is \(a = 100\) and the common ratio is \(-\frac{1}{10}\). Identify \(n = 5\) through analysis, leading to a sum \(S_n = 100 \frac{(-\frac{1}{10})^5 - 1}{- \frac{1}{10} - 1} = 100 \frac{-0.00001 - 1}{-1.1} = 90.9090\). (5) Explore \(200(1 + (1.01) + (1.01)^2 + \ldots + (1.01)^{28})\), with \(a = 1\) and \(r = 1.01\) leading to \(n = 29\). The sum involves using the geometric series formula: \(S_n = 200 \cdot \frac{1.01^{29} - 1}{1.01 - 1}\), approximately resulting in \(200 \cdot 298.847 = 59769.4\). (6) Analyzing \(-12 - 6 - 3 - \ldots - \frac{3}{256}\), the series where \(a = -12\) and \(r = \frac{1}{2}\) gives \(n = 9\). Here, the sum becomes \(S_n = -12 \frac{(1/2)^9 - 1}{1/2 - 1} = -12 \cdot 0.001953 - 1.7775\). (7) In the expression \(xy^2 + xy + x + \ldots + \frac{x}{y^{12}}\), one can isolate common factors leading to identification of \(a = xy^2\) and \(r = \frac{1}{y}\) with the series length \(n = 13\), summing as \(xy^2 \frac{1 - \frac{1}{y^{13}}}{1 - \frac{1}{y}} = xy^2(y^{13} - 1)/(y^{13} - y)\). (8) For \(x^5 - x^3 + x - \ldots + \frac{1}{x^{11}}\), where \(a=x^5\) and \(r = -\frac{1}{x^2}\) provides a structure to compute \(n = 12\) as it unfolds as the familiar geometric series, used to discover that each term leads toward \(x^5 \cdot \frac{1 - (-\frac{1}{x^2})^{-12}}{1 + \frac{1}{x^2}}\). So, geometry isn't just about shapes; it's about finding
