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Magazine Chapter 0: Problem 9
$$ \text { If } a, r \in \mathbb{R} \text { and } r \neq 1, \text { show that } a+a r+a r^{2}+\ldots+a r^{n}=a\left(1-r^{n+1}\right) /(1-r) \text { for } n \geq 1 $$
Short Answer
Expert verified
The sum of the series equals \( \frac{a(1-r^{n+1})}{1-r} \).
Step by step solution
01
Understand the Problem
We are asked to prove the formula for the sum of a geometric series. The series is given as \( a + ar + ar^2 + \ldots + ar^n \) and must be shown to equal \( \frac{a(1-r^{n+1})}{1-r} \). This is applicable for \( n \geq 1 \) provided \( r eq 1 \) to avoid division by zero.
02
Define the Series
Recognize that the series \( a + ar + ar^2 + \ldots + ar^n \) is a geometric series with first term \( a \) and common ratio \( r \). It can be denoted as \( S_n \).
03
Write the Expression for the Sum
The sum of the first \( n+1 \) terms of a geometric series is given by the formula \( S_n = a \frac{1-r^{n+1}}{1-r} \) when \( r eq 1 \). This is the standard formula for the sum of a geometric series.
04
Derive the Formula
Start the derivation from the series: \( S_n = a + ar + ar^2 + \ldots + ar^n \). Multiply through by \( r \): \( rS_n = ar + ar^2 + \ldots + ar^{n+1} \).
05
Subtract the Two Equations
Subtracting the equation \( rS_n \) from \( S_n \), we have: \( S_n - rS_n = a - ar^{n+1} \), which simplifies to \( S_n(1-r) = a(1 - r^{n+1}) \).
06
Solve for \( S_n \)
Isolate \( S_n \) in the equation from Step 5: \( S_n = \frac{a(1-r^{n+1})}{1-r} \). This confirms the original statement.
07
Conclusion
Thus, it is proven that \( a + ar + ar^2 + \ldots + ar^n = \frac{a(1-r^{n+1})}{1-r} \) for \( n \geq 1 \) and \( r eq 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sum of Geometric Series Formula
The sum of a geometric series is an important formula in mathematics that helps us find the total when you add together multiple terms. These terms each increase by a consistent factor, known as the common ratio. The formula for the sum of the first \( n+1 \) terms of a geometric series is \( S_n = \frac{a(1-r^{n+1})}{1-r} \). Here, \( a \) represents the first term of the series, \( r \) is the common ratio, and \( n \) is the number of terms minus one.
- The formula is valid when the absolute value of \( r \) is not equal to 1 (i.e., \( r eq 1 \)). This is to prevent division by zero.
- The denominator, \((1-r)\), essentially determines how the sum grows with each additional term.
Mathematical Proof
A mathematical proof is a logical argument that establishes the truth of a given statement. For the sum of a geometric series, we prove \( S_n = \frac{a(1-r^{n+1})}{1-r} \) by logically manipulating equations.
- Start with the series \( S_n = a + ar + ar^2 + \ldots + ar^n \).
- Multiply the entire series by \( r \) to shift each term: \( rS_n = ar + ar^2 + \ldots + ar^{n+1} \).
- Subtract this new series from the original \( S_n \), which cleverly cancels most terms leaving us \( S_n - rS_n = a - ar^{n+1} \).
Series Derivation
Series derivation involves breaking down a series into understandable pieces to derive its sum formula. Let鈥檚 explore the derivation of the sum of a geometric series.
- We begin with the series expression: \( S_n = a + ar + ar^2 + \ldots + ar^n \).
- The trick is recognizing that multiplying by \( r \) shifts terms: \( rS_n = ar + ar^2 + \ldots + ar^{n+1} \).
- Subtract \( rS_n \) from \( S_n \) to find simplified terms that allow us to unravel the series naturally: \( S_n(1-r) = a(1 - r^{n+1}) \).
Common Ratio
In the context of a geometric series, the common ratio, \( r \), is a constant that each term is multiplied by to get the next term in the sequence. It鈥檚 crucial in determining the behavior and total sum of the series.
- The series \( a, ar, ar^2, \ldots \) consistently multiplies by \( r \), illustrating growth or decay.
- If \( |r| < 1 \), terms decrease, converging to a limit, while \( |r| > 1 \) causes terms to grow larger.
- \( r eq 1 \) is essential because a common ratio of one yields no progression, invalidating the geometric nature.
