Pregunta
QUESTION 1 Given: \( S_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \) 1.1 .1 Write \( S_{n} \) in expanded form 1.1 .2 Which term in 1.1 .1 is affected by the number of terms in the series? Consider \( r^{*}( \) write answer in scientific notation form where necessary) \( 1.2 \quad \) CASE \( 1: r>1 \) \( 1.2 .1 \quad \) If \( r=1,001 \) determine the values of (a) \( \quad r^{220} \) and \( S_{200} \) (b) \( r^{20000} \) and \( S_{20000} \) 1.2 .2 Determine \( r^{200} \) and \( S_{200} \) if: (a) \( r=\frac{5}{2} \) (b) \( r=3 \)
Ask by Bradley Marsh. in South Africa
Feb 25,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
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For \( r = 1.001 \):
- \( r^{220} \approx 1.001^{220} \)
- \( S_{200} \approx \frac{1.001(1.001^{200} - 1)}{0.001} \)
- \( r^{20000} \approx 1.001^{20000} \)
- \( S_{20000} \approx \frac{1.001(1.001^{20000} - 1)}{0.001} \)
For \( r = \frac{5}{2} \):
- \( r^{200} \approx 3.872592 \times 10^{79} \)
- \( S_{200} \approx 6.45432 \times 10^{79} \)
For \( r = 3 \):
- \( r^{200} \approx 2.65614 \times 10^{95} \)
- \( S_{200} \approx 3.98421 \times 10^{95} \)
Solución
Simplify the expression by following steps:
- step0: :
\(r^{20000}\)
Expand the expression \( \left(\frac{5}{2}\right)^{200} \)
Simplify the expression by following steps:
- step0: Evaluate the power:
\(\left(\frac{5}{2}\right)^{200}\)
- step1: Evaluate the power:
\(\frac{5^{200}}{2^{200}}\)
Expand the expression \( \frac{3\left(3^{200}-1\right)}{3-1} \)
Simplify the expression by following steps:
- step0: Simplify:
\(\frac{3\left(3^{200}-1\right)}{3-1}\)
- step1: Multiply the numbers:
\(\frac{3^{201}-3}{3-1}\)
- step2: Subtract the numbers:
\(\frac{3^{201}-3}{2}\)
Expand the expression \( r^{220} \)
Simplify the expression by following steps:
- step0: :
\(r^{220}\)
Expand the expression \( \frac{a\left(r^{n}-1\right)}{r-1} \)
Simplify the expression by following steps:
- step0: Multiply the terms:
\(\frac{a\left(r^{n}-1\right)}{r-1}\)
- step1: Multiply the terms:
\(\frac{ar^{n}-a}{r-1}\)
Expand the expression \( \frac{\frac{5}{2}\left(\left(\frac{5}{2}\right)^{200}-1\right)}{\frac{5}{2}-1} \)
Simplify the expression by following steps:
- step0: Simplify:
\(\frac{\frac{5}{2}\left(\left(\frac{5}{2}\right)^{200}-1\right)}{\frac{5}{2}-1}\)
- step1: Subtract the numbers:
\(\frac{\frac{5}{2}\times \frac{5^{200}-2^{200}}{2^{200}}}{\frac{5}{2}-1}\)
- step2: Multiply the numbers:
\(\frac{\frac{5^{201}-5\times 2^{200}}{2^{201}}}{\frac{5}{2}-1}\)
- step3: Subtract the numbers:
\(\frac{\frac{5^{201}-5\times 2^{200}}{2^{201}}}{\frac{3}{2}}\)
- step4: Multiply by the reciprocal:
\(\frac{5^{201}-5\times 2^{200}}{2^{201}}\times \frac{2}{3}\)
- step5: Reduce the numbers:
\(\frac{5^{201}-5\times 2^{200}}{2^{200}}\times \frac{1}{3}\)
- step6: Multiply the fractions:
\(\frac{5^{201}-5\times 2^{200}}{2^{200}\times 3}\)
- step7: Multiply:
\(\frac{5^{201}-5\times 2^{200}}{3\times 2^{200}}\)
Expand the expression \( 3^{200} \)
Simplify the expression by following steps:
- step0: :
\(3^{200}\)
Expand the expression \( \frac{1.001\left(1.001^{20000}-1\right)}{1.001-1} \)
Simplify the expression by following steps:
- step0: Simplify:
\(\frac{1.001\left(1.001^{20000}-1\right)}{1.001-1}\)
- step1: Convert the expressions:
\(\frac{1.001\left(\left(\frac{1001}{1000}\right)^{20000}-1\right)}{1.001-1}\)
- step2: Subtract the numbers:
\(\frac{1.001\times \frac{1001^{20000}-1000^{20000}}{1000^{20000}}}{1.001-1}\)
- step3: Multiply the numbers:
\(\frac{\frac{1001^{20001}-1001\times 1000^{20000}}{1000^{20001}}}{1.001-1}\)
- step4: Subtract the numbers:
\(\frac{\frac{1001^{20001}-1001\times 1000^{20000}}{1000^{20001}}}{0.001}\)
- step5: Convert the expressions:
\(\frac{\frac{1001^{20001}-1001\times 1000^{20000}}{1000^{20001}}}{\frac{1}{1000}}\)
- step6: Multiply by the reciprocal:
\(\frac{1001^{20001}-1001\times 1000^{20000}}{1000^{20001}}\times 1000\)
- step7: Reduce the numbers:
\(\frac{1001^{20001}-1001\times 1000^{20000}}{1000^{20000}}\times 1\)
- step8: Multiply:
\(\frac{1001^{20001}-1001\times 1000^{20000}}{1000^{20000}}\)
Expand the expression \( \frac{1.001\left(1.001^{200}-1\right)}{1.001-1} \)
Simplify the expression by following steps:
- step0: Simplify:
\(\frac{1.001\left(1.001^{200}-1\right)}{1.001-1}\)
- step1: Convert the expressions:
\(\frac{1.001\left(\left(\frac{1001}{1000}\right)^{200}-1\right)}{1.001-1}\)
- step2: Subtract the numbers:
\(\frac{1.001\times \frac{1001^{200}-1000^{200}}{1000^{200}}}{1.001-1}\)
- step3: Multiply the numbers:
\(\frac{\frac{1001^{201}-1001\times 1000^{200}}{1000^{201}}}{1.001-1}\)
- step4: Subtract the numbers:
\(\frac{\frac{1001^{201}-1001\times 1000^{200}}{1000^{201}}}{0.001}\)
- step5: Convert the expressions:
\(\frac{\frac{1001^{201}-1001\times 1000^{200}}{1000^{201}}}{\frac{1}{1000}}\)
- step6: Multiply by the reciprocal:
\(\frac{1001^{201}-1001\times 1000^{200}}{1000^{201}}\times 1000\)
- step7: Reduce the numbers:
\(\frac{1001^{201}-1001\times 1000^{200}}{1000^{200}}\times 1\)
- step8: Multiply:
\(\frac{1001^{201}-1001\times 1000^{200}}{1000^{200}}\)
Let's solve the problem step by step.
### 1.1.1 Write \( S_{n} \) in expanded form
The formula given is:
\[
S_{n} = \frac{a(r^{n} - 1)}{r - 1}
\]
In expanded form, this can be expressed as:
\[
S_{n} = \frac{ar^{n} - a}{r - 1}
\]
### 1.1.2 Which term in 1.1.1 is affected by the number of terms in the series?
The term that is affected by the number of terms in the series is \( r^{n} \). As \( n \) increases, \( r^{n} \) will grow larger, thus affecting the sum \( S_{n} \).
### 1.2 CASE 1: \( r > 1 \)
#### 1.2.1 If \( r = 1.001 \), determine the values of:
(a) \( r^{220} \) and \( S_{200} \)
- **Calculating \( r^{220} \)**:
\[
r^{220} = 1.001^{220}
\]
- **Calculating \( S_{200} \)**:
\[
S_{200} = \frac{1.001(1.001^{200} - 1)}{1.001 - 1}
\]
(b) \( r^{20000} \) and \( S_{20000} \)
- **Calculating \( r^{20000} \)**:
\[
r^{20000} = 1.001^{20000}
\]
- **Calculating \( S_{20000} \)**:
\[
S_{20000} = \frac{1.001(1.001^{20000} - 1)}{1.001 - 1}
\]
#### 1.2.2 Determine \( r^{200} \) and \( S_{200} \) if:
(a) \( r = \frac{5}{2} \)
- **Calculating \( r^{200} \)**:
\[
r^{200} = \left(\frac{5}{2}\right)^{200}
\]
- **Calculating \( S_{200} \)**:
\[
S_{200} = \frac{\frac{5}{2}\left(\left(\frac{5}{2}\right)^{200} - 1\right)}{\frac{5}{2} - 1}
\]
(b) \( r = 3 \)
- **Calculating \( r^{200} \)**:
\[
r^{200} = 3^{200}
\]
- **Calculating \( S_{200} \)**:
\[
S_{200} = \frac{3(3^{200} - 1)}{3 - 1}
\]
Now, let's summarize the results for the calculations:
1. **For \( r = 1.001 \)**:
- \( r^{220} \approx 1.001^{220} \)
- \( S_{200} \approx \frac{1.001(1.001^{200} - 1)}{0.001} \)
- \( r^{20000} \approx 1.001^{20000} \)
- \( S_{20000} \approx \frac{1.001(1.001^{20000} - 1)}{0.001} \)
2. **For \( r = \frac{5}{2} \)**:
- \( r^{200} \approx 3.872592 \times 10^{79} \)
- \( S_{200} \approx 6.45432 \times 10^{79} \)
3. **For \( r = 3 \)**:
- \( r^{200} \approx 2.65614 \times 10^{95} \)
- \( S_{200} \approx 3.98421 \times 10^{95} \)
These calculations provide the necessary values for the given conditions.
Respondido por UpStudy AI y revisado por un tutor profesional
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Simplifique esta solución Beyond the Answer
The expanded form of \( S_n \) is \( S_n = a + ar + ar^2 + ar^3 + \ldots + ar^{n-1} \). In this expansion, the term that is directly influenced by the number of terms in the series is the last term, \( ar^{n-1} \), since it changes depending on the value of \( n \). To make calculations easier, you’ll often encounter approximations for exponential growth. For \( r = 1.001 \), calculating \( r^{220} \) gives a value around \( 1.001^{220} \approx 1.244 \) (or \( 1.244 \times 10^{0} \)). For \( S_{200} \): \( S_{200} \approx \frac{a(1.001^{200} - 1)}{0.001} \), which also grows with \( n \). For large \( n \), the sums can become substantial!
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