5 Find the value of \( n \) if: a \( 2+4+8+\ldots \) to \( n \) terms \( =1022 \) b \( 6-18+54+\ldots \) to \( n \) terms \( =-120 \) c \( \sum_{r=1}^{n} 3 \cdot 2^{r-1}=381 \) d \( \sum_{i=1}^{n} \frac{1}{5}(3)^{r-1}=24 \frac{1}{5} \) e \( \sum_{k=1}^{n}(-8) \cdot(0,5)^{k-1}=-\frac{63}{4} \)

To find the values of \( n \) for each of the given series, let's take them one by one. **a.** The series \( 2 + 4 + 8 + \ldots \) is a geometric series where the first term \( a = 2 \) and the common ratio \( r = 2 \). The sum of the first \( n \) terms of a geometric series can be expressed as: \[ S_n = a \frac{r^n - 1}{r - 1} = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1) \] Setting this equal to 1022 gives: \[ 2(2^n - 1) = 1022 \\ 2^n - 1 = 511 \\ 2^n = 512 \\ n = 9 \] **b.** For the series \( 6 - 18 + 54 + \ldots \), the first term \( a = 6 \) and the common ratio \( r = -3 \). The sum of the first \( n \) terms is: \[ S_n = a \frac{1 - r^n}{1 - r} = 6 \frac{1 - (-3)^n}{1 - (-3)} = 6 \frac{1 - (-3)^n}{4} = \frac{3(1 - (-3)^n)}{2} \] Setting this equal to -120 gives: \[ \frac{3(1 - (-3)^n)}{2} = -120 \\ 3(1 - (-3)^n) = -240 \\ 1 - (-3)^n = -80 \\ -(-3)^n = -81 \\ (-3)^n = 81 \\ n = 4 \] **c.** For the series \( \sum_{r=1}^{n} 3 \cdot 2^{r-1} \), we can factor out the 3: \[ 3 \sum_{r=1}^{n} 2^{r-1} = 3 \cdot \frac{2^n - 1}{2 - 1} = 3(2^n - 1) = 381 \] Thus, \[ 2^n - 1 = 127 \\ 2^n = 128 \\ n = 7 \] **d.** For the series \( \sum_{i=1}^{n} \frac{1}{5}(3)^{r-1} \): \[ \frac{1}{5} \sum_{r=1}^{n} 3^{r-1} = \frac{1}{5} \cdot \frac{3^n - 1}{3 - 1} = \frac{1}{5} \cdot \frac{3^n - 1}{2} = \frac{3^n - 1}{10} \] Setting this equal to \( 24 \frac{1}{5} = \frac{121}{5} \) gives: \[ \frac{3^n - 1}{10} = \frac{121}{5} \\ 3^n - 1 = 242 \\ 3^n = 243 \\ n = 5 \] **e.** For the series \( \sum_{k=1}^{n}(-8)(0.5)^{k-1} \): \[ \sum_{k=1}^{n} -8 (0.5)^{k-1} = -8 \cdot \frac{1 - (0.5)^n}{1 - 0.5} = -8 \cdot 2(1 - (0.5)^n) = -16(1 - (0.5)^n) \] Setting this equal to \(-\frac{63}{4}\) gives: \[ -16(1 - (0.5)^n) = -\frac{63}{4}\\ 16(1 - (0.5)^n) = \frac{63}{4}\\ 64(1 - (0.5)^n) = 63\\ 1 - (0.5)^n = \frac{63}{64}\\ (0.5)^n = \frac{1}{64}\\ 0.5^n = 0.5^6\\ n = 6 \] Summarized results: a) \( n = 9 \) b) \( n = 4 \) c) \( n = 7 \)