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Chapter 13: Problem 40

Write a formula for the general term of each infinite sequence. \(0,1,8,27,64, \dots\)

Short Answer

Expert verified
\(a_n = n^3\)

Step by step solution

01

Identify the Pattern

Observe the given sequence: 0, 1, 8, 27, 64, ... . Notice that each term can be expressed as the cube of non-negative integers. Specifically, 0 is the cube of 0, 1 is the cube of 1, 8 is the cube of 2, 27 is the cube of 3, and 64 is the cube of 4.
02

Generalize the Pattern

If each term in the sequence corresponds to the cube of an integer, then the general form for the nth term can be written as the cube of the integer n. The sequence follows the pattern: \(a_0 = 0^3, a_1 = 1^3, a_2 = 2^3, a_3 = 3^3, a_4 = 4^3, \text{and so on.}\)
03

Write the General Term

The general term for the sequence can, therefore, be formulated as: \(a_n = n^3\). Where n represents the position of the term in the sequence (starting from 0).

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Key Concepts

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

cube of integers
In mathematics, the cube of an integer refers to any integer raised to the power of three. This operation can be expressed as \(a^3\), where \(a\) is the integer. Cubing a number means multiplying the integer by itself three times. For example, \(2^3 = 2 \times 2 \times 2 = 8\). Cubes of integers are special because they grow very quickly as the integers increase. They are also used in various mathematical applications and problems.

Notice the pattern in the given sequence: 0, 1, 8, 27, 64, ... . Each term is simply the cube of non-negative integers:
  • 0 is \(0^3\)
  • 1 is \(1^3\)
  • 8 is \(2^3\)
  • 27 is \(3^3\)
  • 64 is \(4^3\)
infinite sequence
An infinite sequence is a sequence that continues indefinitely without ending. This means it has an unlimited number of terms. In this context, the given sequence 0, 1, 8, 27, 64, ... is an example of an infinite sequence, as indicated by the '...'.

Infinite sequences are critical in mathematics because they are used to explore limits, convergence, and other important concepts. The sequence of cubes we are examining will continue forever by cubing each successive integer:
  • First term: \(0^3 = 0\)
  • Second term: \(1^3 = 1\)
  • Third term: \(2^3 = 8\)
  • Fourth term: \(3^3 = 27\)
  • Fifth term: \(4^3 = 64\)
And so on, without an upper limit.
nth term formula
The nth term of a sequence is a formula that allows you to find any term in the sequence based on its position, \(n\). For this particular sequence of cubes, the nth term formula helps us to find the value of any term in the sequence quickly without having to cube each integer manually.

We have learned from the previous steps that each term in the sequence is the cube of its position number. Therefore, the general term formula for this sequence can be written as \(a_n = n^3\), where:
  • \(a_n\) represents the nth term in the sequence
  • \(n\) is the position of the term starting from 0.
This means:
  • For \(n = 0, a_n = 0^3 = 0\)
  • For \(n = 1, a_n = 1^3 = 1\)
  • For \(n = 2, a_n = 2^3 = 8\)
  • For \(n = 3, a_n = 3^3 = 27\)
  • For \(n = 4, a_n = 4^3 = 64\)
And so forth. This formula is essential because it provides a concise way to describe the sequence and enables us to find any term efficiently.

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Most popular questions from this chapter

Evaluate each expression. $$\frac{8 !}{5 ! 3 !}$$

Write out the first four terms in the expansion of each binomial. $$\left(\frac{a}{2}+\frac{b}{5}\right)^{8}$$

List all terms of each finite sequence. \(a_{n}=2 n\) for \(1 \leq n \leq 5\)

Write the first four terms of the infinite sequence whose nth term is given. \(a_{n}=(-1)^{2 n+1} 2^{n-1}\)

Write a formula for the general term of each infinite sequence. \(3,7,11,15, \dots\)
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