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Chapter 12: Problem 35

Find the first four partial sums and the \(n\)th partial sum of the sequence \(a_{n} .\) \(a_{n}=\frac{2}{3^{n}}\)

Short Answer

Expert verified
The first four partial sums are \(\frac{2}{3}, \frac{8}{9}, \frac{26}{27}, \frac{80}{81}\), and the nth partial sum is \(2\left(1 - \left(\frac{1}{3}\right)^n\right)\).

Step by step solution

01

Write Down the First Term

To find the first partial sum, first calculate the first term of the sequence. Evaluate \(a_1 = \frac{2}{3^1} = \frac{2}{3}\). The first partial sum, \(S_1\), is just the first term: \(S_1 = a_1 = \frac{2}{3}\).
02

Calculate the Second Partial Sum

Find the second term: \(a_2 = \frac{2}{3^2} = \frac{2}{9}\). The second partial sum is the sum of the first two terms: \[S_2 = a_1 + a_2 = \frac{2}{3} + \frac{2}{9}.\]To add these fractions, find a common denominator: \(\frac{2}{3} = \frac{6}{9}\), so \(S_2 = \frac{6}{9} + \frac{2}{9} = \frac{8}{9}\).
03

Calculate the Third Partial Sum

Find the third term: \(a_3 = \frac{2}{3^3} = \frac{2}{27}\). The third partial sum is: \[S_3 = S_2 + a_3 = \frac{8}{9} + \frac{2}{27}.\]Convert \(\frac{8}{9}\) to a fraction over 27: \(\frac{8}{9} = \frac{24}{27}\). Now add: \(S_3 = \frac{24}{27} + \frac{2}{27} = \frac{26}{27}\).
04

Calculate the Fourth Partial Sum

Find the fourth term: \(a_4 = \frac{2}{3^4} = \frac{2}{81}\). The fourth partial sum is: \[S_4 = S_3 + a_4 = \frac{26}{27} + \frac{2}{81}.\]Convert \(\frac{26}{27}\) to a fraction over 81: \(\frac{26}{27} = \frac{78}{81}\). Add: \(S_4 = \frac{78}{81} + \frac{2}{81} = \frac{80}{81}\).
05

General Formula for the nth Partial Sum

The nth partial sum is the sum of the first n terms: \(S_n = a_1 + a_2 + a_3 + \ldots + a_n\).This series is geometric with first term \(a = \frac{2}{3}\) and common ratio \(r = \frac{1}{3}\). The sum of the first n terms of a geometric series is: \[S_n = a \left(\frac{1-r^n}{1-r}\right).\]Substitute the values: \[S_n = \frac{2}{3} \left(\frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}}\right) = 2\left(1 - \left(\frac{1}{3}\right)^n\right).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a type of sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Let's unravel this with the sequence given in the exercise: \(a_n = \frac{2}{3^n}\).
Here, each term is derived from multiplying the previous term by the common ratio, \(r = \frac{1}{3}\). This defines a geometric sequence.
  • The first term \(a_1 = \frac{2}{3}\) is the foundation of our series.
  • Recognizing it is geometric helps us use specific formulas for finding sums.
In geometric series, the behavior of the sequence is predictable, which assists in finding the sum of the first \(n\) terms, called the partial sum. Each partial sum builds from the one before it, incorporating additional terms continuously.
Common Ratio
The common ratio is a crucial part of a geometric series. It determines how the sequence develops by constantly scaling each term to obtain the next. In the sequence \(a_n = \frac{2}{3^n}\), the common ratio \(r\) is \(\frac{1}{3}\).
  • To find this ratio, divide any term by the previous term.
  • It remains constant throughout: \(a_2/a_1 = \left(\frac{2}{9}\right) / \left(\frac{2}{3}\right) = \frac{1}{3}\).
Understanding the common ratio clarifies how rapidly or slowly the series grows or converges, significantly impacting the calculation of sums and the general properties of the series. In this exercise, knowing \(r = \frac{1}{3}\) allowed us to use the sum formula for geometric series efficiently.
Nth Term
In sequences and series, particularly geometric ones, the nth term \(a_n\) is a formula that stands for any term in the sequence. For our series \(a_n = \frac{2}{3^n}\), it allows calculation of any specific term directly.
  • The nth term is derived from the first term times the common ratio raised to a power: \(a_n = a \cdot r^{n-1}\).
  • For this series, \(a = \frac{2}{3}\), so the nth term formula becomes: \(a_n = \frac{2}{3} \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{2}{3^n}\).
Calculating specific terms without computing previous ones is a great advantage offered by the nth term formula and highlights the efficiency of working with geometric series.
Sequence
A sequence is an ordered list of numbers following a particular pattern. In our exercise, \(a_n = \frac{2}{3^n}\) represents a sequence where the pattern is guided by the common ratio.
  • The sequence starts with the first term \(\frac{2}{3}\), and each subsequent term is generated by multiplying by \(\frac{1}{3}\).
  • Recognizing the sequence involves identifying this consistent pattern of changes between terms.
Sequences like this geometric one are prevalent in mathematical contexts because they embody regular, predictable behaviors. This allows for detailed analysis and easy computation of sums and specific terms, demonstrating why they appear so frequently in both theoretical and applied mathematics.

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Most popular questions from this chapter

Express the repeating decimal as a fraction. $$ 0.030303 \ldots $$

\(43-44\) Simplify using the Binomial Theorem. $$ \frac{(x+h)^{3}-x^{3}}{h} $$

Sums of Binomial Coefficients Add each of the first five rows of Pascal's triangle, as indicated. Do you see a pattern? $$ \begin{array}{c}{1+1=?} \\ {1+2+1=?} \\ {1+3+3+1=?} \\ {1+4+6+4+1=?} \\\ {1+5+10+10+5+1=?}\end{array} $$ Based on the pattern you have found, find the sum of the nth row: $$ \left(\begin{array}{l}{n} \\ {0}\end{array}\right)+\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)+\left(\begin{array}{l}{n} \\\ {2}\end{array}\right)+\cdots+\left(\begin{array}{l}{n} \\\ {n}\end{array}\right) $$ Prove your result by expanding \((1+1)^{n}\) using the Binomial Theorem.

Logarithms of a Geometric Sequence If \(a_{1}, a_{2}, a_{3}, \ldots\) is a geometric sequence with a common ratio \(r>0\) and \(a_{1}>0,\) show that the sequence $$\log a_{1}, \log a_{2}, \log a_{3}, \ldots$$ is an arithmetic sequence, and find the common difference.

Find the middle term in the expansion of \(\left(x^{2}+1\right)^{18}\)
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