b) Calculate the sum of each of the following geometric series (1) \( 6+12+24+\ldots+6144 \) (3) \( -5+15-45+\ldots-3645 \) (5) \( 200\left(1+(1.01)+(1.01)^{2}+\ldots+(1.01)^{25}\right) \) (7) \( 13^{2}+x y+x+\ldots+\frac{x}{y^{12}} \)

To calculate the sum of each of the given geometric series, we will use the formula for the sum of a geometric series: \[ S_n = a \frac{1 - r^n}{1 - r} \] where: - \( S_n \) is the sum of the first \( n \) terms, - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. Let's solve each series step by step. ### (1) \( 6 + 12 + 24 + \ldots + 6144 \) **Step 1: Identify \( a \), \( r \), and \( n \)** - First term \( a = 6 \) - Common ratio \( r = \frac{12}{6} = 2 \) **Step 2: Find \( n \)** The last term is \( 6144 \). We can express the \( n \)-th term of the series as: \[ a_n = a \cdot r^{n-1} \] Setting \( a_n = 6144 \): \[ 6144 = 6 \cdot 2^{n-1} \] Dividing both sides by 6: \[ 1024 = 2^{n-1} \] Since \( 1024 = 2^{10} \), we have: \[ n - 1 = 10 \implies n = 11 \] **Step 3: Calculate the sum \( S_n \)** Now we can calculate the sum: \[ S_{11} = 6 \frac{1 - 2^{11}}{1 - 2} = 6 \frac{1 - 2048}{-1} = 6 \cdot 2047 = 12282 \] ### (3) \( -5 + 15 - 45 + \ldots - 3645 \) **Step 1: Identify \( a \), \( r \), and \( n \)** - First term \( a = -5 \) - Common ratio \( r = \frac{15}{-5} = -3 \) **Step 2: Find \( n \)** The last term is \( -3645 \). Using the \( n \)-th term formula: \[ -3645 = -5 \cdot (-3)^{n-1} \] Dividing both sides by -5: \[ 729 = (-3)^{n-1} \] Since \( 729 = 3^6 \), we have: \[ (-3)^{n-1} = 3^6 \implies n - 1 = 6 \implies n = 7 \] **Step 3: Calculate the sum \( S_n \)** Now we can calculate the sum: \[ S_7 = -5 \frac{1 - (-3)^7}{1 - (-3)} = -5 \frac{1 + 2187}{4} = -5 \cdot \frac{2188}{4} = -5 \cdot 547 = -2735 \] ### (5) \( 200\left(1 + (1.01) + (1.01)^{2} + \ldots + (1.01)^{25}\right) \) **Step 1: Identify \( a \), \( r \), and \( n \)** - First term \( a = 1 \) - Common ratio \( r = 1.01 \) - Number of terms \( n = 26 \) (from \( 0 \) to \( 25 \)) **Step 2: Calculate the sum \( S_n \)** Now we can calculate the sum: \[ S_{26} = 1 \frac{1 - (1.01)^{26}}{1 - 1.01} = \frac{1 - (1.01)^{26}}{-0.01} \] Calculating \( (1.01)^{26} \): \[ (1.01)^{26} \approx 1.268241795 \] Now substituting back: \[ S_{26} = \frac{1 - 1.268241795}{-0.01} = \frac{-0.268241795}{-0.01} = 26.8241795 \] Finally, multiplying by 200: \[ 200 \cdot 26.8241795 \approx 5364.8359 \] ### (7) \( 13^{2} + x y + x + \ldots + \frac{x}{y^{12}} \) This series appears to be a geometric series where the first term is \( 13^2 \) and the common ratio is \( \frac{1}{y} \). **Step 1: Identify \( a \), \( r \), and \( n \)** - First term \( a = 13^2 = 169 \) - Common ratio \( r = \frac{1}{y} \) - The last term is \( \frac{x}{y^{12}} \), which can be expressed as \( a \cdot r^{n-1} \). **Step 2: Find \( n \)** Setting the last term: \[ \frac{x}{y^{12}} = 169 \cdot \left(\frac{1}{y}\right)^{n-1} \] This implies: \[ x = 169 \cdot \frac{1}{y^{n-1}} \cdot y^{12} = 169 \cdot y^{12 - (n-1)} = 169 \cdot y^{13 - n} \] Thus, \( n = 13 \). **Step 3: Calculate the sum \( S_n \)** Now we can calculate the sum: \[ S_{13} = 169 \frac{1 - \left(\frac{1}{y}\right)^{13}}{1 - \frac{1}{y}} = 169 \frac{1 - \frac{1}{y^{13}}}{1 - \frac{1}{y}} = 169 \frac{y^{13} - 1}{y^{13}(y - 1)} \] ### Final Results 1. \( S_1 = 12282 \) 2. \( S_3 = -2735 \) 3. \( S_5 \approx 5364.84 \) 4. \( S_7 = 169 \frac{y^{13} - 1}{y^{13}(y - 1)} \)