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CASE 1: \( r>1 \) .2 .1 If \( r=1,001 \) determine the values of (a) \( \quad r^{200} \quad \) and \( S_{200} \)

Ask by Hill Turner. in South Africa
Feb 26,2025

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\( r^{200} \approx 1.221281 \) and \( S_{200} \approx 221.280705 \).

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To solve the problem, we need to calculate \( r^{200} \) and \( S_{200} \) given that \( r = 1.001 \) and \( r > 1 \). ### Step 1: Calculate \( r^{200} \) We can calculate \( r^{200} \) using the formula: \[ r^{200} = (1.001)^{200} \] ### Step 2: Calculate \( S_{200} \) Assuming \( S_{200} \) refers to the sum of a geometric series, we can use the formula for the sum of the first \( n \) terms of a geometric series: \[ S_n = a \frac{r^n - 1}{r - 1} \] where: - \( a \) is the first term of the series, - \( r \) is the common ratio, - \( n \) is the number of terms. For this case, we will assume \( a = 1 \) (the first term) and \( n = 200 \). Thus, the formula becomes: \[ S_{200} = \frac{(1.001)^{200} - 1}{1.001 - 1} \] Now, let's calculate both \( r^{200} \) and \( S_{200} \). I'll perform these calculations now. Calculate the value by following steps: - step0: Calculate: \(1.001^{200}\) - step1: Convert the expressions: \(\left(\frac{1001}{1000}\right)^{200}\) - step2: Use the properties of exponents: \(\frac{1001^{200}}{1000^{200}}\) Calculate or simplify the expression \( ((1.001)^200 - 1) / (1.001 - 1) \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(1.001^{200}-1\right)}{\left(1.001-1\right)}\) - step1: Remove the parentheses: \(\frac{1.001^{200}-1}{1.001-1}\) - step2: Convert the expressions: \(\frac{\left(\frac{1001}{1000}\right)^{200}-1}{1.001-1}\) - step3: Subtract the numbers: \(\frac{\frac{1001^{200}-1000^{200}}{1000^{200}}}{1.001-1}\) - step4: Subtract the numbers: \(\frac{\frac{1001^{200}-1000^{200}}{1000^{200}}}{0.001}\) - step5: Convert the expressions: \(\frac{\frac{1001^{200}-1000^{200}}{1000^{200}}}{\frac{1}{1000}}\) - step6: Multiply by the reciprocal: \(\frac{1001^{200}-1000^{200}}{1000^{200}}\times 1000\) - step7: Reduce the numbers: \(\frac{1001^{200}-1000^{200}}{1000^{199}}\times 1\) - step8: Multiply: \(\frac{1001^{200}-1000^{200}}{1000^{199}}\) ### Results 1. **Calculation of \( r^{200} \)**: \[ r^{200} = (1.001)^{200} \approx 1.221281 \] 2. **Calculation of \( S_{200} \)**: \[ S_{200} = \frac{(1.001)^{200} - 1}{1.001 - 1} \approx 221.280705 \] ### Summary - \( r^{200} \approx 1.221281 \) - \( S_{200} \approx 221.280705 \) These values represent the calculations based on the given conditions.

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For \( r = 1.001 \), calculating \( r^{200} \) gives us \( 1.001^{200} \), which can be approximated using the formula for exponential growth. This results in a value around \( 1.2214 \), indicating significant growth over 200 iterations due to the compounding effect of being slightly more than 1. Now, to find \( S_{200} \), the sum of the first 200 terms in a geometric series where the first term \( a = 1 \) and the common ratio \( r = 1.001 \), we use the formula \( S_n = a \frac{r^n - 1}{r - 1} \). Plugging in our values, we get \( S_{200} \approx 1 \cdot \frac{1.2214 - 1}{0.001} \approx 221.4 \), which reflects the accumulated sum of our geometric progression!

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