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INVESTIGATING INFINITE GEOMETRIC SERIES RESOURCES: CALCULATOR, A4 PAPER. WHAT IS THE SUM OF AN INFINITE GEOMETRIC SERIES? QUESTION 1 \begin{tabular}{lll|}1.1 & 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 \( r^{n} \) (write answer in scientific notation form where necessary) \end{tabular}

Ask by Fleming Mcguire. in South Africa
Feb 25,2025

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Answer

The sum of an infinite geometric series is \( S = \frac{a}{1 - r} \) when \( |r| < 1 \). The term \( ar^{n} \) in the expanded form is affected by the number of terms \( n \).

Solution

To solve the problem regarding the infinite geometric series, let's break it down step by step. ### Step 1: Understanding the Formula The formula given is: \[ S_{n} = \frac{a(r^{n} - 1)}{r - 1} \] where: - \( S_{n} \) is the sum of the first \( n \) terms of a geometric series, - \( a \) is the first term, - \( r \) is the common ratio, - \( n \) is the number of terms. ### Step 1.1: Write \( S_{n} \) in Expanded Form To expand \( S_{n} \), we can distribute the numerator: \[ S_{n} = \frac{a(r^{n} - 1)}{r - 1} = \frac{ar^{n} - a}{r - 1} \] This can be interpreted as: \[ S_{n} = \frac{ar^{n}}{r - 1} - \frac{a}{r - 1} \] ### Step 1.2: Identify the Term Affected by the Number of Terms In the expanded form \( S_{n} = \frac{ar^{n}}{r - 1} - \frac{a}{r - 1} \), the term that is affected by the number of terms \( n \) is \( ar^{n} \). This term changes as \( n \) changes, since it directly depends on the exponent \( n \). ### Step 1.3: Consider \( r^{n} \) in Scientific Notation The term \( r^{n} \) can be expressed in scientific notation if \( r \) is a decimal (between 0 and 1) or a large number. If \( r < 1 \), as \( n \) approaches infinity, \( r^{n} \) approaches 0. Thus, in the context of an infinite geometric series, the sum converges to: \[ S = \frac{a}{1 - r} \quad \text{(for } |r| < 1\text{)} \] If \( r > 1 \), \( r^{n} \) grows without bound as \( n \) increases, leading to divergence. ### Summary of Answers 1.1.1: The expanded form of \( S_{n} \) is: \[ S_{n} = \frac{ar^{n}}{r - 1} - \frac{a}{r - 1} \] 1.1.2: The term affected by the number of terms in the series is \( ar^{n} \). 1.1.3: If \( r < 1 \), \( r^{n} \) approaches 0 as \( n \) increases, and can be expressed in scientific notation as \( r^{n} \approx 0 \) for large \( n \). If \( r > 1 \), \( r^{n} \) grows large, and can be expressed in scientific notation as \( r^{n} \) (e.g., \( 1.0 \times r^{n} \)).

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The Deep Dive

The sum of an infinite geometric series can be calculated using the formula \( S = \frac{a}{1 - r} \) where \( |r| < 1 \). This means that for an infinite number of terms, as long as the common ratio \( r \) is a fraction between -1 and 1, the series converges to a finite value. It's like taking a never-ending staircase that gets closer and closer to a ceiling without ever reaching it! When working with \( S_{n} = \frac{a(r^{n}-1)}{r-1} \), the term affected by the number of terms in the series is \( r^{n} \). As \( n \) increases, \( r^{n} \) shrinks towards zero if \( |r| < 1 \). So, when you're calculating \( r^{n} \), remember that the greater \( n \) becomes, the closer the term will approach \( 0 \). Keeping an eye on this limit can help visualize how the series behaves as you add more terms!

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