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Chapter 1: Problem 3

List the first five terms of the sequence. $$a_{n}=\frac{(-1)^{n-1}}{5^{n}}$$

Short Answer

Expert verified
First five terms: \(\frac{1}{5}, \frac{-1}{25}, \frac{1}{125}, \frac{-1}{625}, \frac{1}{3125}\).

Step by step solution

01

Understand the sequence formula

The given sequence formula is \(a_{n} = \frac{(-1)^{n-1}}{5^{n}}\). This formula defines the general term of the sequence, where \(n\) is the position of the term in the sequence starting from \(n=1\).
02

Calculate the first term \(a_{1}\)

Substitute \(n=1\) into the formula: \[ a_{1} = \frac{(-1)^{1-1}}{5^{1}} = \frac{(-1)^{0}}{5} = \frac{1}{5}. \] So, the first term \(a_{1}\) is \(\frac{1}{5}\).
03

Calculate the second term \(a_{2}\)

Substitute \(n=2\) into the formula:\[ a_{2} = \frac{(-1)^{2-1}}{5^{2}} = \frac{(-1)^{1}}{25} = \frac{-1}{25}. \] So, the second term \(a_{2}\) is \(\frac{-1}{25}\).
04

Calculate the third term \(a_{3}\)

Substitute \(n=3\) into the formula:\[ a_{3} = \frac{(-1)^{3-1}}{5^{3}} = \frac{(-1)^{2}}{125} = \frac{1}{125}. \] So, the third term \(a_{3}\) is \(\frac{1}{125}\).
05

Calculate the fourth term \(a_{4}\)

Substitute \(n=4\) into the formula:\[ a_{4} = \frac{(-1)^{4-1}}{5^{4}} = \frac{(-1)^{3}}{625} = \frac{-1}{625}. \] So, the fourth term \(a_{4}\) is \(\frac{-1}{625}\).
06

Calculate the fifth term \(a_{5}\)

Substitute \(n=5\) into the formula:\[ a_{5} = \frac{(-1)^{5-1}}{5^{5}} = \frac{(-1)^{4}}{3125} = \frac{1}{3125}. \] So, the fifth term \(a_{5}\) is \(\frac{1}{3125}\).
07

Summarize the first five terms

The first five terms of the sequence are \(\frac{1}{5}, \frac{-1}{25}, \frac{1}{125}, \frac{-1}{625}, \frac{1}{3125}\).

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Key Concepts

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

General Term
A general term of a sequence is an expression that allows us to find any term in the sequence by substituting the term's position number. In the sequence formula given, the general term is \(a_{n} = \frac{(-1)^{n-1}}{5^{n}}\), where \(n\) is the term number.
  • "General term" reduces the need to compute the entire sequence to find a specific term.
  • It helps in predicting or calculating any term of the sequence easily by substituting values into the formula.
By understanding and using the general term, it's easier to determine other characteristics of the sequence, such as its behavior as terms progress.
Alternating Sign
Alternating signs in a sequence mean that the terms change signs in a regular pattern. In our sequence, the numerator \((-1)^{n-1}\) ensures alternation of signs.
  • For odd \(n\), the term \((-1)^{n-1}\) results in 1, giving the term a positive sign.
  • For even \(n\), the term \((-1)^{n-1}\) results in -1, giving the term a negative sign.
This alternating sign pattern creates a sequence oscillating between positive and negative values, impacting the sequence's progression as a whole.
In practical terms, this property can imply alternating gains and losses in applications like finance, or other phenomena where positive and negative variations are essential.
Sequence Formula
The sequence formula serves as a blueprint for calculating terms of a sequence. Our specific sequence formula, \(a_{n} = \frac{(-1)^{n-1}}{5^{n}}\), uses a combination of alternating signs and exponential growth.
  • The denominator, \(5^{n}\), determines how quickly the term values decrease as \(n\) increases.
  • The numerator, \((-1)^{n-1}\), implements the alternating sign.
Understanding this formula gives insight into how the sequence behaves as we progress through its terms.
This understanding is vital in recognizing patterns or predicting future terms without direct computation, offering insight into the sequence's theoretical and practical implications.
Term Calculation
Calculating the specific terms of a sequence involves substituting consecutive integers into the sequence's general term. This allows us to determine each distinct term step by step.
For the sequence formula \(a_{n} = \frac{(-1)^{n-1}}{5^{n}}\):
  • The calculation of \(a_{1}\), \(a_{2}\), \(a_{3}\), etc., is done by substituting \(n=1\), \(n=2\), \(n=3\), respectively, and simplifying the expression.
  • This step-by-step method ensures accurate computation and expresses the specific properties such as magnitude and sign for each term.
Through practice of this calculation, the process becomes intuitive, enabling faster and more accurate understanding of sequences in various contexts.

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Most popular questions from this chapter

Express the function in the form \(f \circ g\) \(F(x)=\left(2 x+x^{2}\right)^{4}\)

\(\begin{array}{ll}{\text { Use the table to evaluate each expression. }} \\\ {\text { (a) } f(g(1))} & {\text { (b) } g(f(1))} & {\text { (c) } f(f(1))} \\\ {\text { (d) } g(g(1))} & {\text { (e) }(g \circ f)(3)} & {\text { (f) }(f \circ g)(6)}\end{array}\) \(\begin{array}{|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {3} & {1} & {4} & {2} & {2} & {5} \\ \hline g(x) & {6} & {3} & {2} & {1} & {2} & {3} \\ \hline\end{array}\)

38\. Methyl groups in DNA DNA sometimes has chemical groups attached, called methyl groups, that affect gene expression. Suppose that, during each hour, first a fraction \(m\) of unmethylated locations on the DNA become methyl- ated, and then a fraction \(u\) of methylated locations become unmethylated. Find a recursion for the fraction \(f\) of the DNA molecule that is methylated.

Data points \((x, y)\) are given. (a) Draw a scatter plot of the data points. (b) Make semilog and log-log plots of the data. (c) Is a linear, power, or exponential function appropriate for modeling these data? (d) Find an appropriate model for the data and then graph the model together with a scatter plot of the data. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {2} & {4} & {6} & {8} & {10} & {12} \\\ \hline y & {0.08} & {0.12} & {0.18} & {0.26} & {0.35} & {0.53} \\\ \hline\end{array}$$ \(\begin{array}{|c|c|c|c|c|c|}\hline x & {5} & {10} & {15} & {20} & {25} & {30} \\\ \hline y & {0.013} & {0.046} & {0.208} & {0.930} & {4.131} & {18.002} \\\ \hline\end{array}\)

Find a formula for the inverse of the function. \(y=\frac{e^{x}}{1+2 e^{x}}\)
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