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 number of terms in ti Consider \( r^{n} \) (write answer in scientific notation form where necessary 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) \( r=3 \) 1.2 .3 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \)

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: \[ S_n = \frac{a \cdot r^n - a}{r - 1} \] This is the expanded form of \( S_n \). #### 1.1.2 Which term in 1.1.1 is affected by the number of terms in \( n \)? The term that is affected by the number of terms \( n \) is \( r^n \). As \( n \) increases, \( r^n \) grows exponentially, 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 \( \mathrm{S}_{200} \) 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 \( \mathrm{S}_{20000} \) 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 \( \mathrm{S}_{200} \) if: ##### (a) \( r=\frac{5}{2} \) Calculating \( r^{200} \): \[ r^{200} = \left(\frac{5}{2}\right)^{200} \] Calculating \( S_{200} \): \[ S_{200} = \frac{a \left( \left(\frac{5}{2}\right)^{200} - 1 \right)}{\frac{5}{2} - 1} \] ##### (b) \( r=3 \) Calculating \( r^{200} \): \[ r^{200} = 3^{200} \] Calculating \( S_{200} \): \[ S_{200} = \frac{a \left( 3^{200} - 1 \right)}{3 - 1} \] #### 1.2.3 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \) As \( n \) approaches infinity, if \( r > 1 \), both \( r^n \) and \( S_n \) will grow without bound. Specifically, \( S_n \) will approach \( \frac{a \cdot r^n}{r - 1} \) as the term \( -1 \) becomes negligible compared to \( r^n \). Thus, we can conjecture that: \[ \lim_{n \to \infty} S_n \approx \frac{a \cdot r^n}{r - 1} \] Now, let's calculate the specific values for \( r^{200} \), \( S_{200} \), \( r^{20000} \), and \( S_{20000} \) for \( r = 1.001 \). I'll perform these calculations now. Simplify the expression by following steps: - step0: : \(3^{200}\) 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 \( (1.001)^{20000} \) 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 \( (\frac{5}{2})^{200} \) 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 \( \mathrm{S}_{200} \) - \( r^{200} = (1.001)^{200} \approx 1.221386025 \) - To calculate \( S_{200} \), we need to substitute \( r^{200} \) into the formula: \[ S_{200} = \frac{a \left( (1.001)^{200} - 1 \right)}{1.001 - 1} = \frac{a \left( 1.221386025 - 1 \right)}{0.001} \approx \frac{a \cdot 0.221386025}{0.001} = 221.386025a \] ##### (b) \( r^{20000} \) and \( \mathrm{S}_{20000} \) - \( r^{20000} = (1.001)^{20000} \approx 4.803409 \times 10^{8} \) - For \( S_{20000} \): \[ S_{20000} = \frac{a \left( (1.001)^{20000} - 1 \right)}{1.001 - 1} \approx \frac{a \left( 4.803409 \times 10^{8} - 1 \right)}{0.001} \approx 4.803409 \times 10^{11} a \] ### 1.2.2 Determine \( r^{200} \) and \( \mathrm{S}_{200} \) if: ##### (a) \( r=\frac{5}{2} \) - \( r^{200} = \left(\frac{5}{2}\right)^{200} \approx 3.872592 \times 10^{79} \) - For \( S_{200} \): \[ S_{200} = \frac{a \left( \left(\frac{5}{2}\right)^{200} - 1 \right)}{\frac{5}{2} - 1} = \frac{a \left( 3.872592 \times 10^{79} - 1 \right)}{\frac{3}{2}} \approx \frac{a \cdot 3.872592 \times 10^{79}}{\frac{3}{2}} \approx 2.582194 \times 10^{79} a \] ##### (b) \( r=3 \) - \( r^{200} = 3^{200} \approx 2.65614 \times 10^{95} \) - For \( S_{200} \): \[ S_{200} = \frac{a \left( 3^{200} - 1 \right)}{3 - 1} = \frac{a \left( 2.65614 \times 10^{95} - 1 \right)}{2} \approx 1.32807 \times 10^{95} a \] ### 1.2.3 Conjecture As \( n \rightarrow \infty \): - \( r^n \) and \( S_n \) both grow without bound. - Specifically, \( S_n \) approaches \( \frac{a \cdot r^n}{r - 1} \) as the term \( -1 \) becomes negligible. This concludes the detailed solution to the problem. If you have any further questions or need additional calculations, feel free to ask!