Calculate the sum of each of the following geometric series: 1) \( 6+12+24+\ldots+6144 \) (2) \( 4-8+16-\ldots+16384 \) \[ -5+15-45+\ldots-3645 \] (4) \( 100-10+1-\ldots+\frac{1}{10000} \) \[ 200\left(1+(1,01)+(1.01)^{2}+\ldots+(1,01)^{2 R}\right) \] (6) \( -12-6-3-\ldots-\frac{3}{256} \)

To sum a geometric series, you can use the formula for the sum of the first \( n \) terms of a geometric series: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \( S_n \) is the sum of the series, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Let's calculate the sums for the given series. 1) **\( 6 + 12 + 24 + \ldots + 6144 \)** This series has a first term \( a = 6 \) and a common ratio \( r = 2 \). To find \( n \): If \( 6144 = 6 \cdot 2^{n-1} \Longrightarrow 1024 = 2^{n-1} \Longrightarrow n - 1 = 10 \Longrightarrow n = 11 \). Now using the formula: \[ S_{11} = 6 \frac{1 - 2^{11}}{1 - 2} = 6 \frac{1 - 2048}{-1} = 6 \cdot 2047 = 12282 \] 2) **\( 4 - 8 + 16 - \ldots + 16384 \)** Here, \( a = 4 \) and \( r = -2 \). To find \( n \): If \( 16384 = 4 \cdot (-2)^{n-1} \Longrightarrow 4096 = (-2)^{n-1} \Longrightarrow n - 1 = 12 \Longrightarrow n = 13 \). Now applying the formula: \[ S_{13} = 4 \frac{1 - (-2)^{13}}{1 - (-2)} = 4 \frac{1 + 8192}{3} = 4 \cdot \frac{8193}{3} = 4 \cdot 2731 = 10924 \] 3) **\(-5 + 15 - 45 + \ldots - 3645\)** Here, \( a = -5 \) and \( r = -3 \). To find \( n \): If \( -3645 = -5 \cdot (-3)^{n-1} \Longrightarrow 729 = (-3)^{n-1} \Longrightarrow n - 1 = 6 \Longrightarrow n = 7 \). Using the formula: \[ S_{7} = -5 \frac{1 - (-3)^{7}}{1 - (-3)} = -5 \frac{1 + 2187}{4} = -5 \cdot \frac{2188}{4} = -5 \cdot 547 = -2735 \] 4) **\( 100 - 10 + 1 - \ldots + \frac{1}{10000} \)** Here, \( a = 100 \) and \( r = -0.1 \). To find \( n \): If \( \frac{1}{10000} = 100 \cdot (-0.1)^{n-1} \Longrightarrow 0.00001 = (-0.1)^{n-1} \Longrightarrow n - 1 = 5 \Longrightarrow n = 6 \). Now applying the formula: \[ S_{6} = 100 \frac{1 - (-0.1)^{6}}{1 - (-0.1)} = 100 \frac{1 - 0.000001}{1.1} = 100 \cdot \frac{0.999999}{1.1} \approx 90.909 \] 5) **\( 200\left(1 + (1.01) + (1.01)^{2} + \ldots + (1.01)^{2R}\right) \)** Here, \( a = 200 \) and \( r = 1.01 \). To apply the geometric series sum formula, we'd need to clarify \( R \), but generally: \[ S = 200 \cdot \frac{1 - (1.01)^{2(R+1)}}{1 - 1.01} = 200 \cdot \frac{(1.01)^{2(R+1)} - 1}{-0.01} = -20000 \cdot \left( (1.01)^{2(R+1)} - 1 \right). \] 6) **\( -12 - 6 - 3 - \ldots - \frac{3