Question
QUESTION 1 1.1 Given: \( \mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \) 1.1 .1 Write \( S_{n} \) in expanded form 1.1 .2 Which term in 1.1 .1 is affected by the Consider \( r^{n} \) (write answer in scientific notation f 1.2 CASE \( 1: r>1 \) 1.2 .1 If \( r=1,001 \) determine the values of (a) \( r^{200} \) and \( \mathrm{S}_{200} \) (b) \( r^{20000} \) and \( \mathrm{S}_{20000} \) 1.2 .2 Determine \( r^{200} \) and \( \mathrm{S}_{200} \) if: (a) \( r=\frac{5}{2} \) (b) \( r=3 \)\( ~ ل ~\left(\begin{array}{l}\text { (b) }\end{array}\right. \)
Ask by Harris Phillips. in South Africa
Feb 25,2025
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Tutor-Verified Answer
Answer
For \( r = 1.001 \):
- \( r^{200} \approx 1.221 \times 10^{0} \)
- \( S_{200} \approx 221.386a \)
- \( r^{20000} \approx 4.803409 \times 10^{8} \)
- \( S_{20000} \approx 4.803409 \times 10^{11} a \)
For \( r = \frac{5}{2} \):
- \( r^{200} \approx 3.872592 \times 10^{79} \)
- \( S_{200} \approx 2.582 \times 10^{79} a \)
For \( r = 3 \):
- \( r^{200} \approx 2.65614 \times 10^{95} \)
- \( S_{200} \approx 1.32807 \times 10^{95} a \)
Solution
Let's solve the problem step by step.
### 1.1 Given:
\[
\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}
\]
#### 1.1.1 Write \( S_{n} \) in expanded form
To expand \( S_n \), we can distribute the terms in the formula. The expression can be rewritten as:
\[
S_n = \frac{a \cdot r^n - a}{r - 1}
\]
This shows that \( S_n \) consists of two main components: \( a \cdot r^n \) and \( -a \).
#### 1.1.2 Which term in 1.1.1 is affected by \( r^{n} \)?
The term that is affected by \( r^{n} \) is \( a \cdot r^n \). This term grows exponentially as \( n \) increases, especially when \( r > 1 \).
### 1.2 CASE \( 1: r > 1 \)
#### 1.2.1 If \( r = 1.001 \), determine the values of:
##### (a) \( r^{200} \) and \( S_{200} \)
First, we calculate \( r^{200} \):
\[
r^{200} = (1.001)^{200}
\]
Next, we calculate \( S_{200} \):
\[
S_{200} = \frac{a \left( (1.001)^{200} - 1 \right)}{1.001 - 1}
\]
##### (b) \( r^{20000} \) and \( S_{20000} \)
Now, we calculate \( r^{20000} \):
\[
r^{20000} = (1.001)^{20000}
\]
Next, we calculate \( S_{20000} \):
\[
S_{20000} = \frac{a \left( (1.001)^{20000} - 1 \right)}{1.001 - 1}
\]
#### 1.2.2 Determine \( r^{200} \) and \( S_{200} \) if:
##### (a) \( r = \frac{5}{2} \)
First, we calculate \( r^{200} \):
\[
r^{200} = \left( \frac{5}{2} \right)^{200}
\]
Next, we calculate \( S_{200} \):
\[
S_{200} = \frac{a \left( \left( \frac{5}{2} \right)^{200} - 1 \right)}{\frac{5}{2} - 1}
\]
##### (b) \( r = 3 \)
First, we calculate \( r^{200} \):
\[
r^{200} = 3^{200}
\]
Next, we calculate \( S_{200} \):
\[
S_{200} = \frac{a \left( 3^{200} - 1 \right)}{3 - 1}
\]
Now, let's perform the calculations for \( r^{200} \), \( S_{200} \), \( r^{20000} \), and \( S_{20000} \) using the specified values of \( r \).
I'll calculate these values now.
Simplify the expression by following steps:
- step0: Evaluate the power:
\(1.001^{20000}\)
- step1: Convert the expressions:
\(\left(\frac{1001}{1000}\right)^{20000}\)
- step2: Evaluate the power:
\(\frac{1001^{20000}}{1000^{20000}}\)
Expand the expression \( (1.001)^{200} \)
Simplify the expression by following steps:
- step0: Evaluate the power:
\(1.001^{200}\)
- step1: Convert the expressions:
\(\left(\frac{1001}{1000}\right)^{200}\)
- step2: Evaluate the power:
\(\frac{1001^{200}}{1000^{200}}\)
Expand the expression \( 3^{200} \)
Simplify the expression by following steps:
- step0: :
\(3^{200}\)
Expand the expression \( (\frac{5}{2})^{200} \)
Simplify the expression by following steps:
- step0: Evaluate the power:
\(\left(\frac{5}{2}\right)^{200}\)
- step1: Evaluate the power:
\(\frac{5^{200}}{2^{200}}\)
Here are the calculated values for the specified cases:
### 1.2.1 If \( r = 1.001 \):
##### (a) \( r^{200} \) and \( S_{200} \)
- \( r^{200} = (1.001)^{200} \approx 1.221386 \) (in scientific notation: \( \approx 1.221 \times 10^{0} \))
- To calculate \( S_{200} \), we need to substitute \( r^{200} \) into the formula:
\[
S_{200} = \frac{a \left( (1.001)^{200} - 1 \right)}{1.001 - 1} = \frac{a \left( 1.221386 - 1 \right)}{0.001} = \frac{a \cdot 0.221386}{0.001} \approx 221.386a
\]
##### (b) \( r^{20000} \) and \( S_{20000} \)
- \( r^{20000} = (1.001)^{20000} \approx 4.803409 \times 10^{8} \)
- To calculate \( S_{20000} \):
\[
S_{20000} = \frac{a \left( (1.001)^{20000} - 1 \right)}{1.001 - 1} = \frac{a \left( 4.803409 \times 10^{8} - 1 \right)}{0.001} \approx 4.803409 \times 10^{11} a
\]
### 1.2.2 Determine \( r^{200} \) and \( S_{200} \) if:
##### (a) \( r = \frac{5}{2} \)
- \( r^{200} = \left( \frac{5}{2} \right)^{200} \approx 3.872592 \times 10^{79} \)
- To calculate \( S_{200} \):
\[
S_{200} = \frac{a \left( \left( \frac{5}{2} \right)^{200} - 1 \right)}{\frac{5}{2} - 1} = \frac{a \left( 3.872592 \times 10^{79} - 1 \right)}{\frac{3}{2}} \approx \frac{2a \cdot (3.872592 \times 10^{79} - 1)}{3} \approx \frac{2a \cdot 3.872592 \times 10^{79}}{3} \approx 2.582 \times 10^{79} a
\]
##### (b) \( r = 3 \)
- \( r^{200} = 3^{200} \approx 2.65614 \times 10^{95} \)
- To calculate \( S_{200} \):
\[
S_{200} = \frac{a \left( 3^{200} - 1 \right)}{3 - 1} = \frac{a \left( 2.65614 \times 10^{95} - 1 \right)}{2} \approx \frac{a \cdot 2.65614 \times 10^{95}}{2} \approx 1.32807 \times 10^{95} a
\]
### Summary of Results:
- For \( r = 1.001 \):
- \( r^{200} \approx 1.221 \times 10^{0} \)
- \( S_{200} \approx 221.386a \)
- \( r^{20000} \approx 4.803409 \times 10^{8} \)
- \( S_{20000} \approx 4.803409 \times 10^{11} a \)
- For \( r = \frac{5}{2} \):
- \( r^{200} \approx 3.872592 \times 10^{79} \)
- \( S_{200} \approx 2.582 \times 10^{79} a \)
- For \( r = 3 \):
- \( r^{200} \approx 2.65614 \times 10^{95} \)
- \( S_{200} \approx 1.32807 \times 10^{95} a \)
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Simplify this solution Beyond the Answer
The expanded form of \( S_n \) can be written as \( S_n = a + ar + ar^2 + ar^3 + \ldots + ar^{n-1} \). This series sums up the first \( n \) terms of a geometric sequence where each term is the product of the initial term \( a \) and the common ratio raised to the power of the term's position in the sequence. Therefore, if \( r=1.001 \), for \( n=200 \), \( r^{200} \approx 10^{0.877} \) and \( S_{200} \approx 1000.746 \). When considering large values of \( n \) like \( r^{20000} \) with \( r=1.001 \), you can see a rapid increase. In scientific notation, \( r^{20000} \) will yield an incredibly large number \( \approx 10^{8.73} \), signifying exponential growth. Similarly, \( S_{20000} \) includes the previous terms influenced by \( r^{20000} \), yielding even larger results, making the power of the initial terms significant in determining \( S_n \) values.
