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Chapter 13: Problem 23

List the first five terms of each sequence. \(\left\\{t_{n}\right\\}=\left\\{\frac{(-1)^{n}}{(n+1)(n+2)}\right\\}\)

Short Answer

Expert verified
[\frac{1}{2}, \frac{-1}{6}, \frac{1}{12}, \frac{-1}{20}, \frac{1}{30}]

Step by step solution

01

Identify the sequence formula

The given sequence formula is \( t_{n} = \frac{(-1)^{n}}{(n+1)(n+2)} \).
02

Calculate the first term \( t_{0} \)

Substitute \( n = 0 \) into the sequence formula: \( t_{0} = \frac{(-1)^{0}}{(0+1)(0+2)} = \frac{1}{1 \times 2} = \frac{1}{2} \).
03

Calculate the second term \( t_{1} \)

Substitute \( n = 1 \) into the sequence formula: \( t_{1} = \frac{(-1)^{1}}{(1+1)(1+2)} = \frac{-1}{2 \times 3} = \frac{-1}{6} \).
04

Calculate the third term \( t_{2} \)

Substitute \( n = 2 \) into the sequence formula: \( t_{2} = \frac{(-1)^{2}}{(2+1)(2+2)} = \frac{1}{3 \times 4} = \frac{1}{12} \).
05

Calculate the fourth term \( t_{3} \)

Substitute \( n = 3 \) into the sequence formula: \( t_{3} = \frac{(-1)^{3}}{(3+1)(3+2)} = \frac{-1}{4 \times 5} = \frac{-1}{20} \).
06

Calculate the fifth term \( t_{4} \)

Substitute \( n = 4 \) into the sequence formula: \( t_{4} = \frac{(-1)^{4}}{(4+1)(4+2)} = \frac{1}{5 \times 6} = \frac{1}{30} \).

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Key Concepts

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

Sequence Formula
A sequence is a list of numbers in a specific order, defined by a particular formula. In this case, the sequence formula given is:
\( t_{n} = \frac{(-1)^{n}}{(n+1)(n+2)} \).
This formula defines how each term in the sequence is calculated based on its position.
You can see that the formula includes an alternating sign, produced by \( (-1)^{n} \), and a fraction term that depends on n.
This formula allows us to find each term by substituting different values for 'n'.
Term Calculation
To calculate a term in a sequence, you need to replace 'n' in the formula with the specific term number starting from 0.
Below are the steps and calculations for the first five terms:
  • First Term (\( t_{0} \)):

    Substitute n = 0 into the formula: \( t_{0} = \frac{(-1)^{0}}{(0+1)(0+2)} = \frac{1}{1 \times 2} = \frac{1}{2} \).

  • Second Term (\( t_{1} \)):

    Substitute n = 1 into the formula: \( t_{1} = \frac{(-1)^{1}}{(1+1)(1+2)} = \frac{-1}{2 \times 3} = \frac{-1}{6} \).

  • Third Term (\( t_{2} \)):

    Substitute n = 2 into the formula: \( t_{2} = \frac{(-1)^{2}}{(2+1)(2+2)} = \frac{1}{3 \times 4} = \frac{1}{12} \).

  • Fourth Term (\( t_{3} \)):

    Substitute n = 3 into the formula: \( t_{3} = \frac{(-1)^{3}}{(3+1)(3+2)} = \frac{-1}{4 \times 5} = \frac{-1}{20} \).

  • Fifth Term (\( t_{4} \)):

    Substitute n = 4 into the formula: \( t_{4} = \frac{(-1)^{4}}{(4+1)(4+2)} = \frac{1}{5 \times 6} = \frac{1}{30} \).

Each term involves simple substitution and arithmetic operations.
Alternating Series
An alternating series is a sequence where the terms change sign in a regular pattern.
This is achieved using \( (-1)^{n} \) in the sequence formula.
When 'n' is even (e.g., 0, 2, 4), \( (-1)^{n} \) becomes 1, making the term positive.
When 'n' is odd (e.g., 1, 3), \( (-1)^{n} \) becomes -1, making the term negative.
The given sequence: \( t_{n} = \frac{(-1)^{n}}{(n+1)(n+2)} \), is an example of an alternating series.
This results in a pattern of signs like positive, negative, positive, and so forth.
Such series are useful in various mathematical contexts, including solving problems in calculus and understanding behavior in certain functions.

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Most popular questions from this chapter

If \(y=\frac{5}{3} x^{3}+2 x+C\) and \(y=5\) when \(x=3,\) find the value of \(C\).

Suppose that, throughout the U.S. economy, individuals spend \(90 \%\) of every additional dollar that they earn. Economists would say that an individual's marginal propensity to consume is \(0.90 .\) For example, if Jane earns an additional dollar, she will spend \(0.9(1)=\$ 0.90\) of it. The individual who earns \(\$ 0.90\) (from Jane) will spend \(90 \%\) of it, or \(\$ 0.81 .\) This process of spending continues and results in an infinite geometric series as follows: $$1,0.90,0.90^{2}, 0.90^{3}, 0.90^{4}, \ldots$$ The sum of this infinite geometric series is called the multiplier. What is the multiplier if individuals spend \(90 \%\) of every additional dollar that they earn?

Determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference; if it is geometric, find the common ratio. If the sequence is arithmetic or geometric, find the sum of the first 50 terms. $$ \left\\{\left(\frac{5}{4}\right)^{n}\right\\} $$

Find the horizontal asymptote, if one exists, of \(f(x)=\frac{9 x}{3 x^{2}-2 x-1}\)

Bode's Law In \(1772,\) Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: $$ a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2} $$ where \(n \geq 2\) is the number of the planet from the sun. (a) Determine the first eight terms of the sequence. (b) At the time of Bode's publication, the known planets were Mercury \((0.39 \mathrm{AU}),\) Venus \((0.72 \mathrm{AU}),\) Earth \((1 \mathrm{AU})\) Mars \((1.52 \mathrm{AU}),\) Jupiter \((5.20 \mathrm{AU}),\) and Saturn \((9.54 \mathrm{AU})\) How do the actual distances compare to the terms of the sequence? (c) The planet Uranus was discovered in \(1781,\) and the asteroid Ceres was discovered in \(1801 .\) The mean orbital distances from the sun to Uranus and Ceres " are \(19.2 \mathrm{AU}\) and \(2.77 \mathrm{AU},\) respectively. How well do these values fit within the sequence? (d) Determine the ninth and tenth terms of Bode's sequence. (e) The planets Neptune and Pluto" were discovered in 1846 and \(1930,\) respectively. Their mean orbital distances from the sun are \(30.07 \mathrm{AU}\) and \(39.44 \mathrm{AU},\) respectively. How do these actual distances compare to the terms of the sequence? (f) On July \(29,2005,\) NASA announced the discovery of a dwarf planet \((n=11),\) which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.
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