Chapter 6: Problem 20
In \(15-22 :\) a. Write each sum as a series. b. Find the sum of each series. $$ 1+\sum_{n=1}^{5}\left(\frac{2}{3}\right)^{n} $$
Short Answer
Expert verified
The sum of the series is approximately 0.131.
Step by step solution
01
Understanding the Series
The given expression is \(1+\sum_{n=1}^{5}\left(\frac{2}{3}\right)^{n}\). This means we have a sum of 1 and a geometric series starting from \(n=1\) to \(n=5\) with the term \(\left(\frac{2}{3}\right)^{n}\).
02
Writing the Series
To write out the series, list each term in the sum: \(1 + \left(\frac{2}{3}\right)^{1} + \left(\frac{2}{3}\right)^{2} + \left(\frac{2}{3}\right)^{3} + \left(\frac{2}{3}\right)^{4} + \left(\frac{2}{3}\right)^{5}\).
03
Calculating Each Term
Calculate each term of the series:- \(\left(\frac{2}{3}\right)^{1} = \frac{2}{3}\)- \(\left(\frac{2}{3}\right)^{2} = \frac{4}{9}\)- \(\left(\frac{2}{3}\right)^{3} = \frac{8}{27}\)- \(\left(\frac{2}{3}\right)^{4} = \frac{16}{81}\)- \(\left(\frac{2}{3}\right)^{5} = \frac{32}{243}\)
04
Summing the Series
Add up all the terms of the series:\[1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \frac{16}{81} + \frac{32}{243} = \text{Total Sum}\]
05
Find the Sum of Geometric Series
Use the formula for the sum of a finite geometric series:\[ S_n = a \frac{r^n-1}{r-1} \]Where \(a = \frac{2}{3}\) (first term), \(r = \frac{2}{3}\) (common ratio), and \(n = 5\) (number of terms):\[ S_5 = \frac{2}{3} \frac{(\frac{2}{3})^5 - 1}{\frac{2}{3} - 1} = \frac{2}{3} \left(\frac{\frac{32}{243} - 1}{\frac{-1}{3}}\right) = \frac{2}{3} \left(\frac{-211}{243} \right) \]Now add \(1\) to the sum of the series approximately to find total sum.
06
Calculate the Total Sum Approximatively
Approximating the sum:1. Calculate the sum \( \frac{2}{3} \cdot \left(-\frac{211}{243} \right) \approx -0.869\)2. Add 1 to the sum of the geometric series: \[1 - 0.869 \approx 0.131\]
07
Final Answer
The sum of the series is approximately 0.131.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sum of Series
When dealing with a series, our goal is often to find a single number that represents the addition of its terms. This single number is called the "sum of the series." In mathematical terms, a series is the sum of the terms of a sequence. In our exercise, we specifically deal with a geometric series that combines multiple exponential terms.
- The sum could be finite or infinite, but in this case, we're dealing with a finite series bound by a specific number of terms.
- The series uses terms that follow a specific rule or pattern - here, each term is a power of the common ratio.
Finite Geometric Series
A finite geometric series is a series with a limited number of terms, as opposed to an infinite geometric series, which continues indefinitely. The distinctive feature of a geometric series, whether finite or infinite, is that each term is obtained by multiplying the previous term by a constant factor called the common ratio.
- To find the sum of a finite geometric series, one can use the formula: \( S_n = a \frac{r^n - 1}{r - 1} \).
- In this formula, \(a\) represents the first term of the series, \(r\) is the common ratio, and \(n\) is the number of terms.
Geometric Sequence
A geometric sequence is a list of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The backbone of solving any geometric series problem is understanding the sequence involved.
- To break it down, if our first term is \(a\), then the second term becomes \(ar\), the third is \(ar^2\), and so on.
- In our series \( 1+ \sum_{n=1}^{5}\left(\frac{2}{3}\right)^n \), the geometric sequence part is the list \(\left(\frac{2}{3}\right)^n\) for \(n = 1\) to \(5\).
Common Ratio
The common ratio in a geometric sequence or series is the factor by which we multiply a term to get the next term. It holds everything together in a geometric sequence.
- This multiplier is crucial because it dictates the exponential growth (or decay) of the series.
- In our problem, the common ratio is consistently \(\frac{2}{3}\), impacting how each subsequent term is calculated as \(\left(\frac{2}{3}\right)^{n}\).