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 \( \sum_{k=1}^{n}(-8) \cdot(0,5)^{k-1}=-\frac{63}{4} \)
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Let's solve each part step by step! ### Part a: \( 2 + 4 + 8 + \ldots \) to \( n \) terms \( =1022 \) This is a geometric series with the first term \( a = 2 \) and common ratio \( r = 2 \). The sum of the first \( n \) terms of a geometric series can be calculated using the formula: \[ S_n = a \frac{r^n - 1}{r - 1} \] Substituting the values: \[ 1022 = 2 \frac{2^n - 1}{2 - 1} \] So, \[ 1022 = 2(2^n - 1) \implies 1022 = 2^{n+1} - 2 \implies 1024 = 2^{n+1} \] Thus, \[ 2^{n+1} = 1024 \implies n + 1 = 10 \implies n = 9. \] ### Part b: \( 6 - 18 + 54 + \ldots \) to \( n \) terms \( = -120 \) This series has the first term \( a = 6 \) and common ratio \( r = -3 \). The sum of the first \( n \) terms is calculated using: \[ S_n = a \frac{1 - r^n}{1 - r} \] Plug in: \[ -120 = 6 \frac{1 - (-3)^n}{1 - (-3)} \] This simplifies to: \[ -120 = 6 \frac{1 + 3^n}{4} \implies -480 = 6(1 + 3^n) \implies -80 = 1 + 3^n \] So, \[ 3^n = -81 \quad \text{(not possible as \( 3^n \) cannot be negative)} \] Let's calculate if we can find something useful by checking for positive integers only. ### Part c: \( \sum_{r=1}^{n} 3 \cdot 2^{r-1}=381 \) Here, this series can be rewritten as: \[ 3(2^0 + 2^1 + \ldots + 2^{n-1}) = 381 \] Using the sum formula for a geometric series again: \[ 3 \left( \frac{2^n - 1}{2 - 1} \right) = 381 \implies 3(2^n - 1) = 381 \implies 2^n - 1 = 127 \implies 2^n = 128 \] Thus, \[ n = 7. \] ### Part d: \( \sum_{i=1}^{n} \frac{1}{5}(3)^{r-1}=24 \frac{1}{5} \) This is similar: \[ \frac{1}{5} (3^0 + 3^1 + \ldots + 3^{n-1}) = 24.2 \] Multiply both sides by 5: \[ 3^0 + 3^1 + \ldots + 3^{n-1} = 121 \] The sum is: \[ \frac{3^n - 1}{3 - 1} = 121 \implies 3^n - 1 = 242 \implies 3^n = 243 \implies n = 5. \] ### Part e: \( \sum_{k=1}^{n} (-8)(0.5)^{k-1}=-\frac{63}{4} \) This series can be rewritten as: \[ -8 \sum_{k=0}^{n-1} (0.5)^k = -\frac{63}{4} \] Calculating yields: \[ -8 \left( \frac{1 - (0.5)^n}{1 - 0.5} \right) = -\frac{63}{4} \implies -8(1 - (0.5)^n) \cdot 2 = -\frac{63}{4} \] Thus: \[ -16(1 - (0.5)^n) = -\frac{63}{4} \implies 16(1 - (0.5)^n) = \frac{63}{4} \] Now multiply by 4 to clear fractions: \[ 64(1 - (0.5)^n) = 63 \implies
