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Magazine Chapter 13: Problem 53
Expand each sum. \(\sum_{k=0}^{n} \frac{1}{3^{k}}\)
Short Answer
Expert verified
1 + \(1/3\) + \(1/9\) + \(1/27\) + ... + \(1/3^{n}\)
Step by step solution
01
Understand the Summation
The summation notation \(\textstyle \sum_{k=0}^{n} \frac{1}{3^{k}}\) indicates that we sum the terms \(\frac{1}{3^{k}}\) starting from \(k = 0\) up to \(k = n\).
02
Identify Individual Terms
Write down the first few terms of the series to see the pattern: \(\frac{1}{3^{0}} + \frac{1}{3^{1}} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + ... + \frac{1}{3^{n}}\).
03
Expand the Series
Expand the series explicitly in terms of the general form: \(\frac{1}{3^{0}} + \frac{1}{3^{1}} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + ... + \frac{1}{3^{n}}\). Simplify the first few terms: \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^{n}}\).
04
Verify the Pattern
Verify that each term in the sequence follows the general form \(\frac{1}{3^{k}}\) for \(k = 0, 1, 2, ... n\). For \(k = 0\): \(\frac{1}{3^{0}} = 1\), for \(k = 1\): \(\frac{1}{3^{1}} = \frac{1}{3}\), and so on.
05
Summarize the Expanded Form
Put together the expanded form: \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^{n}}\). This is the final expanded form of the summation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Summation Notation
The summation notation is a way to represent the sum of a sequence of terms. It's denoted by the Greek letter sigma, \(\textstyle \sum\). For example, in the expression \(\textstyle \sum_{k=0}^{n} \frac{1}{3^{k}}\), we're summing the terms \(\frac{1}{3^{k}}\), starting from \(\text{k} = 0\) to \(\text{k} = n\). This is a concise way to denote the sum without writing out each term individually. The index of summation, k, starts at 0 and increments by 1 until it reaches n. So, \(\textstyle \sum_{k=0}^{n} \frac{1}{3^{k}}\) translates to \(\frac{1}{3^0} + \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + ... + \frac{1}{3^n}\).
Series Expansion
Series expansion involves expressing a summation in its detailed form by writing out all the individual terms. For example, if we have \(\textstyle \sum_{k=0}^{n} \frac{1}{3^{k}} \), we expand it to show each term explicitly: \(\frac{1}{3^{0}} + \frac{1}{3^{1}} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + ... + \frac{1}{3^{n}}\). Breaking it down:
\(\frac{1}{3^{0}} = 1\), since anything to the power of zero is 1
\(\frac{1}{3^{1}} = \frac{1}{3}\)
\(\frac{1}{3^{2}} = \frac{1}{9}\)
\(\frac{1}{3^{3}} = \frac{1}{27}\)
This expansion helps in understanding the pattern and nature of the series.
\(\frac{1}{3^{0}} = 1\), since anything to the power of zero is 1
\(\frac{1}{3^{1}} = \frac{1}{3}\)
\(\frac{1}{3^{2}} = \frac{1}{9}\)
\(\frac{1}{3^{3}} = \frac{1}{27}\)
This expansion helps in understanding the pattern and nature of the series.
Geometric Progression
A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant factor. This factor is called the common ratio. In the series \(\frac{1}{3^k}\), the common ratio is \(\frac{1}{3}\). Here's why:
Start with the first term: \(a = 1\)
Multiply by the common ratio: \( \frac{1}{3}\)
So, the sequence will be: \(1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, ...\)
Each subsequent term is found by multiplying the previous term by \(\frac{1}{3}\). This makes the series a geometric progression with a common ratio of \(\frac{1}{3}\).
Start with the first term: \(a = 1\)
Multiply by the common ratio: \( \frac{1}{3}\)
So, the sequence will be: \(1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, ...\)
Each subsequent term is found by multiplying the previous term by \(\frac{1}{3}\). This makes the series a geometric progression with a common ratio of \(\frac{1}{3}\).
General Term
The general term of a sequence or series gives us a formula that can be used to find any term in the sequence. For a geometric series, the general term can be expressed as:
\(\text{general term} = a \times r^k\)
where:
\a\ is the first term of the series
\r\ is the common ratio
\k\ is the position of the term in the sequence
For the series \(\textstyle \sum_{k=0}^{n} \frac{1}{3^{k}} \), the first term \a\ is 1, and the common ratio \r\ is \(\frac{1}{3}\). Therefore, the general term is:
\(\frac{1}{3^k}\)
This formula allows us to find any term in the series based on its position.
\(\text{general term} = a \times r^k\)
where:
\a\ is the first term of the series
\r\ is the common ratio
\k\ is the position of the term in the sequence
For the series \(\textstyle \sum_{k=0}^{n} \frac{1}{3^{k}} \), the first term \a\ is 1, and the common ratio \r\ is \(\frac{1}{3}\). Therefore, the general term is:
\(\frac{1}{3^k}\)
This formula allows us to find any term in the series based on its position.
