Sequences and Their Notations

Learning Objectives

Write the first few terms of a sequence (IA 12.1.1)
Find a formula for the general term (nth term) of a sequence (IA 12.1.2)

Objective 1: Write the first few terms of a sequence (IA 12.1.1).

A patient takes a 30 mg antibiotic capsule. At the end of that hour, the amount of antibiotic remaining in her body is only 90% of the amount in the beginning of that hour. The 30mg dose is taken at time t = 1 hour. How much of this dose remains at the end of 1 hour? 2hours? 3 hours? 4 hours?

.
Time t	Dose remaining after time t
1	0.90(30)=27mg
2	0.90(27)=24.3mg
3	0.90(24.3)=21.87mg
4	0.90(21.87)=19.68mg

This ordered list of numbers 27, 24.3, 21.87, 19.68, … is a sequence. Each number in the list is a term.

A sequence is a function whose domain is the counting numbers. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an infinite sequence.

Often when working with sequences we do not want to write out all the terms. We want a more compact way to show how each term is defined. When we worked with functions, we wrote $f (x) = 2 x$ and we said the expression 2x was the rule that defined values in the range.

While a sequence is a function, we do not use the usual function notation. Instead of writing the function as $f (x) = 2 x$ , we would write it as $a_{n} = 2 n$ . The $a_{n}$ is the nth term of the sequence, the term in the nth position where n is a value in the domain. The formula for writing the nth term of the sequence is called the general term or formula of the sequence.

General sequence terms are denoted as follows:
$\begin{matrix} a_{1} - first term \\ a_{2} - second term \\ a_{3} - third term \\ . \\ . \\ . \\ a_{n} - n^{t h} term \\ a_{n + 1} - (n + 1) term \\ . \\ . \\ . \end{matrix}$

Example 1

Write the first five terms of the sequence whose general term is $a_{n} = 2 n - 7$ .

Solution

$\begin{matrix} n & 1 & 2 & 3 & 4 & 5 \\ a_{n} & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} \\ 2 n - 7 & 2 (1) - 7 & 2 (2) - 7 & 2 (3) - 7 & 2 (4) - 7 & 2 (5) - 7 \\ - 5 & - 3 & - 1 & 1 & 3 \end{matrix}$

Practice Makes Perfect

Write the first few terms of a sequence.

Write the first five terms of the sequence whose general term is $a_{n} = 4 n + 2$ .

.
$n$	$1$	$2$	$3$	$4$	$5$
$a_{n}$	$a_{1}$	$a_{2}$	$a_{3}$	$a_{4}$	$a_{5}$
$4 n + 2$

Write the first five terms of the sequence whose general term is $a_{n} = 3^{n} - 1$ .

.
$n$	$1$	$2$	$3$	$4$	$5$
$a_{n}$	$a_{1}$	$a_{2}$	$a_{3}$	$a_{4}$	$a_{5}$
$3^{n} - 1$

Objective 2: Find a formula for the general term (nth term) of a sequence (IA 12.1.2)

Sometimes we have a few terms of a sequence and it would be helpful to know the general term or nth term. To find the general term, we look for patterns in the terms. Often the patterns involve multiples or powers. We also look for a pattern in the signs of the terms.

Example 2

Find a formula for the general term (nth term) of a sequence.

ⓐ Find a general term for the sequence whose first five terms are shown below:
4, 8, 12, 16, 20...
ⓑ Find a general term for the sequence whose first five terms are shown below:
$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}, \frac{1}{243}, . . .$

Solution

This figure shows five rows. The first row reads, “4”, “8”, “12”, “16”, “20”, and an “ellipsis”. The second row reads “n”, “1”, “2”, “3”, “4”, “5”, and an “ellipsis”. The third row reads “We look for a pattern in terms”, “Terms”, “4”, “8”, “12”, “16”, “20”, and an “ellipsis”. The four row reads, “The numbers are all multiples of 4”, “Pattern”, “4 times g times 1”, “4 times g times 3”, 4 times g times 4”, “4 times g times 5”, and an “ellipsis”, “4 times g times n”. The last row reads “The general term of the sequence is a nth term equals 4 times n”.
ⓐ

Look for a pattern in the terms
The numbers are all multiples of 4
The general term of the sequence:	$a_{n} = 4 n .$

This figure shows five rows. The first row reads, “one-third”, “one-ninth”, “one-twenty-seventh”, “1 divided by 81”, “1 divided by 243”, and an ellipsis. The second row reads, “n”, “1”, “2”, “3”, “4”, “5” and an ellipsis, “n”. The third row reads “We look for a pattern in the terms”, “Terms”, “one-third”, “one-ninth”, “one-twenty-seventh”, “1 divided by 81”, “1 divided by 243”, and an ellipsis. The fourth row reads, “The numerators are all at 1”, “Pattern”, “one-third to the power of 1 times 9, one-third to the power of 2 times 9, one-third to the power of 3 times 4 times 9, one-third to the power of 3 times 5 times 9, ellipsis, one-third to the power of n”. The fifth row reads, “The denominators are powers of 3. The general term of the sequence is a sub n equals one-third to the power of n”.
ⓑ	$A mathematical sequence showing fractions: 1/3, 1/9, 1/27, 1/81, 1/243, and so on, indicating a geometric progression where each term is (1/3)^n.$

Look for a pattern in the terms.	$The image displays the heading "Terms:" followed by a mathematical sequence of five fractions and an ellipsis, indicating that the sequence continues. The terms are 1/3, 1/9, 1/27, 1/81, and 1/243. Each term is a fraction with a numerator of 1, and the denominators are successive powers of 3 (3^1, 3^2, 3^3, 3^4, 3^5).$
The numerators are all 1 and the denominators are powers of 3	$A mathematical pattern displays a sequence of fractions: 1/3^1, 1/3^2, 1/3^3, 1/3^4, 1/3^5, ..., up to 1/3^n, indicating an increasing exponent in the denominator.$
The general term of the sequence:	$a_{n} = \frac{1}{3^{n}}$

Practice Makes Perfect

Find a general term for the sequence whose first five terms are shown:
8, 16, 24, 32, 40, ...

.

Look for a pattern in the terms	Terms: ________________
The general term of the sequence: ________________

Find a general term for the sequence whose first five terms are shown:

\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \frac{1}{256}, \frac{1}{1024}, . . .

.

Look for a pattern in the terms	Terms: ________________
The general term of the sequence: ________________

A video game company launches an exciting new advertising campaign. They predict the number of online visits to their website, or hits, will double each day. The model they are using shows 2 hits the first day, 4 hits the second day, 8 hits the third day, and so on. See Table 1.

Table 1
Day	1	2	3	4	5	…
Hits	2	4	8	16	32	…

If their model continues, how many hits will there be at the end of the month? To answer this question, we’ll first need to know how to determine a list of numbers written in a specific order. In this section, we will explore these kinds of ordered lists.

Writing the Terms of a Sequence Defined by an Explicit Formula

One way to describe an ordered list of numbers is as a sequence. A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is

{2, 4, 8, 16, 32, \dots} .

The ellipsis (…) indicates that the sequence continues indefinitely. Each number in the sequence is called a term. The first five terms of this sequence are 2, 4, 8, 16, and 32.

Listing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as 31 terms. A more efficient way to determine a specific term is by writing a formula to define the sequence.

One type of formula is an explicit formula, which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the nth term of the sequence, where $n$ is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the $n th$ term.

$Sequence of {2, 4, 8, 16, 32, ...} expressed in exponential form (i.e., {2^1, 2^2, 2^3, ..., 2^n, ...}$

The first term of the sequence is $2^{1} = 2,$ the second term is $2^{2} = 4,$ the third term is $2^{3} = 8,$ and so on. The $n th$ term of the sequence can be found by raising 2 to the $n th$ power. An explicit formula for a sequence is named by a lower case letter $a, b, c ...$ with the subscript $n .$ The explicit formula for this sequence is

a_{n} = 2^{n} .

Now that we have a formula for the $n th$ term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31^st term of the sequence. We will substitute 31 for $n$ in the formula.

\begin{array}{l} a_{31} = 2^{31} \\ = 2,147,483,648 \end{array}

If the doubling trend continues, the company will get $2,147,483,648$ hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions.

Another way to represent the sequence is by using a table. The first five terms of the sequence and the $n th$ term of the sequence are shown in Table 2.

Table 2
$n$	1	2	3	4	5	$n$
$n th$ term of the sequence, $a_{n}$	2	4	8	16	32	$2^{n}$

Graphing provides a visual representation of the sequence as a set of distinct points. We can see from the graph in Figure 1 that the number of hits is rising at an exponential rate. This particular sequence forms an exponential function.

Graph of a plotted exponential function, f(n) = 2^n, where the x-axis is labeled n and the y-axis is labeled a_n. — Figure 1

Lastly, we can write this particular sequence as

{2, 4, 8, 16, 32, \dots, 2^{n}, \dots} .

A sequence that continues indefinitely is called an infinite sequence. The domain of an infinite sequence is the set of counting numbers. If we consider only the first 10 terms of the sequence, we could write

{2, 4, 8, 16, 32, \dots, 2^{n}, \dots, 1024} .

This sequence is called a finite sequence because it does not continue indefinitely.

Sequence

A sequence is a function whose domain is the set of positive integers. A finite sequence is a sequence whose domain consists of only the first $n$ positive integers. The numbers in a sequence are called terms. The variable $a$ with a number subscript is used to represent the terms in a sequence and to indicate the position of the term in the sequence.

a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots

We call $a_{1}$ the first term of the sequence, $a_{2}$ the second term of the sequence, $a_{3}$ the third term of the sequence, and so on. The term $a_{n}$ is called the nth term of the sequence, or the general term of the sequence. An explicit formula defines the $n th$ term of a sequence using the position of the term. A sequence that continues indefinitely is an infinite sequence.

Example 3

Writing the Terms of a Sequence Defined by an Explicit Formula

Write the first five terms of the sequence defined by the explicit formula $a_{n} = - 3 n + 8.$

Solution

Substitute $n = 1$ into the formula. Repeat with values 2 through 5 for $n .$

\begin{array}{l} n = 1 & a_{1} = - 3 (1) + 8 = 5 \\ n = 2 & a_{2} = - 3 (2) + 8 = 2 \\ n = 3 & a_{3} = - 3 (3) + 8 = - 1 \\ n = 4 & a_{4} = - 3 (4) + 8 = - 4 \\ n = 5 & a_{5} = - 3 (5) + 8 = - 7 \end{array}

The first five terms are ${5, 2, −1, −4, −7} .$

Analysis

The sequence values can be listed in a table. A table, such as Table 3, is a convenient way to input the function into a graphing utility.

Table 3
$n$	1	2	3	4	5
$a_{n}$	5	2	–1	–4	–7

A graph can be made from this table of values. From the graph in Figure 2, we can see that this sequence represents a linear function, but notice the graph is not continuous because the domain is over the positive integers only.

Graph of a scattered plot where the x-axis is labeled n and the y-axis is labeled a_n. — Figure 2

Investigating Alternating Sequences

Sometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. However, the resulting terms will not show increase or decrease as $n$ increases. Let’s take a look at the following sequence.

{2, −4, 6, −8}

Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence.

Example 4

Writing the Terms of an Alternating Sequence Defined by an Explicit Formula

Write the first five terms of the sequence.

a_{n} = \frac{{(- 1)}^{n} n^{2}}{n + 1}

Solution

Substitute $n = 1,$ $n = 2,$ and so on in the formula.

\begin{array}{l} n = 1 & \begin{matrix}  \end{matrix} & a_{1} = \frac{{(- 1)}^{1} 1^{2}}{1 + 1} = - \frac{1}{2} \\ n = 2 & \begin{matrix}  \end{matrix} & a_{2} = \frac{{(- 1)}^{2} 2^{2}}{2 + 1} = \frac{4}{3} \\ n = 3 & \begin{matrix}  \end{matrix} & a_{3} = \frac{{(- 1)}^{3} 3^{2}}{3 + 1} = - \frac{9}{4} \\ n = 4 & \begin{matrix}  \end{matrix} & a_{4} = \frac{{(- 1)}^{4} 4^{2}}{4 + 1} = \frac{16}{5} \\ n = 5 & a_{5} = \frac{{(- 1)}^{5} 5^{2}}{5 + 1} = - \frac{25}{6} \end{array}

The first five terms are ${- \frac{1}{2}, \frac{4}{3},- \frac{9}{4}, \frac{16}{5},- \frac{25}{6}} .$

Analysis

The graph of this function, shown in Figure 3, looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values.

Graph of a scattered plot with labeled points: (1, -1/2), (2, 4/3), (3, -9/4), (4, 16/5), and (5, -25/6). The x-axis is labeled n and the y-axis is labeled a_n. — Figure 3

Investigating Piecewise Explicit Formulas

We’ve learned that sequences are functions whose domain is over the positive integers. This is true for other types of functions, including some piecewise functions. Recall that a piecewise function is a function defined by multiple subsections. A different formula might represent each individual subsection.

Example 5

Writing the Terms of a Sequence Defined by a Piecewise Explicit Formula

Write the first six terms of the sequence.

a_{n} = {\begin{array}{l} n^{2} & if n is not divisible by 3 \\ \frac{n}{3} & if n is divisible by 3 \end{array}

Solution

Substitute $n = 1, n = 2,$ and so on in the appropriate formula. Use $n^{2}$ when $n$ is not a multiple of 3. Use $\frac{n}{3}$ when $n$ is a multiple of 3.

\begin{array}{l} a_{1} = 1^{2} = 1 \begin{matrix}  \end{matrix} & 1 is not a multiple of 3 .  Use n^{2} . \\ a_{2} = 2^{2} = 4 & 2 is not a multiple of 3 .  Use n^{2} . \\ a_{3} = \frac{3}{3} = 1 & 3 is a multiple of 3 .  Use \frac{n}{3} . \\ a_{4} = 4^{2} = 16 & 4 is not a multiple of 3 .  Use n^{2} . \\ a_{5} = 5^{2} = 25 & 5 is not a multiple of 3 .  Use n^{2} . \\ a_{6} = \frac{6}{3} = 2 & 6 is a multiple of 3 .  Use \frac{n}{3} . \end{array}

The first six terms are ${1, 4, 1, 16, 25, 2} .$

Analysis

Every third point on the graph shown in Figure 4 stands out from the two nearby points. This occurs because the sequence was defined by a piecewise function.

Finding an Explicit Formula

Thus far, we have been given the explicit formula and asked to find a number of terms of the sequence. Sometimes, the explicit formula for the $n th$ term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern in the terms. Keep in mind that the pattern may involve alternating terms, formulas for numerators, formulas for denominators, exponents, or bases.

Example 6

Writing an Explicit Formula for the nth Term of a Sequence

Write an explicit formula for the $n th$ term of each sequence.

ⓐ ${- \frac{2}{11}, \frac{3}{13}, - \frac{4}{15}, \frac{5}{17}, - \frac{6}{19}, \dots}$
ⓑ ${- \frac{2}{25}, - \frac{2}{125}, - \frac{2}{625}, - \frac{2}{3, 125}, - \frac{2}{15, 625}, \dots}$
ⓒ ${e^{4}, e^{5}, e^{6}, e^{7}, e^{8}, \dots}$

Solution

Look for the pattern in each sequence.

ⓐThe terms alternate between positive and negative. We can use ${(- 1)}^{n}$ to make the terms alternate. The numerator can be represented by $n + 1.$ The denominator can be represented by $2 n + 9.$
$a_{n} = \frac{{(- 1)}^{n} (n + 1)}{2 n + 9}$
ⓑ
The terms are all negative.
$The image illustrates a sequence of fractions. The top line shows the sequence as {2/25, 2/125, 2/625, 2/3,125, 2/15,125, ...} and explicitly states that the numerator is 2. The bottom line reiterates the sequence with the denominators expressed as powers of 5: {2/5^2, 2/5^3, 2/5^4, 2/5^6, 2/5^7, ..., 2/5^n}, clarifying that the denominators are increasing powers of 5.$
So we know that the fraction is negative, the numerator is 2, and the denominator can be represented by $5^{n + 1} .$

$a_{n} = - \frac{2}{5^{n + 1}}$
ⓒ
The terms are powers of $e .$ For $n = 1,$ the first term is $e^{4}$ so the exponent must be $n + 3.$

$a_{n} = e^{n + 3}$

Writing the Terms of a Sequence Defined by a Recursive Formula

Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,…. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals.

Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a recursive formula, a formula that defines the terms of a sequence using previous terms.

A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining $a_{n}$ in terms of preceding terms. For example, suppose we know the following:

\begin{array}{l} a_{1} = 3 \\ a_{n} = 2 a_{n - 1} - 1 for n \geq 2 \end{array}

We can find the subsequent terms of the sequence using the first term.

\begin{array}{l} a_{1} = 3 \\ a_{2} = 2 a_{1} - 1 = 2 (3) - 1 = 5 \\ a_{3} = 2 a_{2} - 1 = 2 (5) - 1 = 9 \\ a_{4} = 2 a_{3} - 1 = 2 (9) - 1 = 17 \end{array}

So the first four terms of the sequence are ${3, 5, 9, 17}$ .

The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms.

\begin{array}{l} a_{1} = 1 \\ a_{2} = 1 \\ a_{n} = a_{n - 1} + a_{n - 2} for n \geq 3 \end{array}

To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so

a_{10} = a_{9} + a_{8} = 34 + 21 = 55

Example 7

Writing the Terms of a Sequence Defined by a Recursive Formula

Write the first five terms of the sequence defined by the recursive formula.

\begin{array}{l} \begin{array}{l} a_{1} = 9 \end{array} \\ a_{n} = 3 a_{n - 1} - 20, for n \geq 2 \end{array}

Solution

The first term is given in the formula. For each subsequent term, we replace $a_{n - 1}$ with the value of the preceding term.

\begin{array}{l} n = 1 \begin{array}{l}  \end{array} & a_{1} = 9 \\ n = 2 & a_{2} = 3 a_{1} - 20 = 3 (9) - 20 = 27 - 20 = 7 \\ n = 3 & a_{3} = 3 a_{2} - 20 = 3 (7) - 20 = 21 - 20 = 1 \\ n = 4 & a_{4} = 3 a_{3} - 20 = 3 (1) - 20 = 3 - 20 = - 17 \\ n = 5 & a_{5} = 3 a_{4} - 20 = 3 (- 17) - 20 = - 51 - 20 = - 71 \end{array}

The first five terms are ${9, 7, 1, - 17, - 71} .$ See Figure 5.

Graph of a scattered plot with labeled points: (1, 9), (2, 7), (3, 1), (4, -17), and (5, -71). The x-axis is labeled n and the y-axis is labeled a_n. — Figure 5

Example 8

Writing the Terms of a Sequence Defined by a Recursive Formula

Write the first six terms of the sequence defined by the recursive formula.

\begin{array}{l} a_{1} = 1 \\ a_{2} = 2 \\ a_{n} = 3 a_{n - 1} + 4 a_{n - 2}, for n \geq 3 \end{array}

Solution

The first two terms are given. For each subsequent term, we replace $a_{n - 1}$ and $a_{n - 2}$ with the values of the two preceding terms.

\begin{array}{l} n = 3 & a_{3} = 3 a_{2} + 4 a_{1} = 3 (2) + 4 (1) = 10 \\ n = 4 & a_{4} = 3 a_{3} + 4 a_{2} = 3 (10) + 4 (2) = 38 \\ n = 5 & a_{5} = 3 a_{4} + 4 a_{3} = 3 (38) + 4 (10) = 154 \\ n = 6 & a_{6} = 3 a_{5} + 4 a_{4} = 3 (154) + 4 (38) = 614 \end{array}

The first six terms are ${1,2,10,38,154,614} .$ See Figure 6.

Graph of a scattered plot with labeled points: (1, 1), (2, 2), (3, 10), (4, 38), (5, 154) and (6, 614). The x-axis is labeled n and the y-axis is labeled a_n. — Figure 6

Try It #8

Write the first 8 terms of the sequence defined by the recursive formula.

\begin{array}{l} \begin{array}{l} a_{1} = 0 \end{array} \\ a_{2} = 1 \\ a_{3} = 1 \\ a_{n} = \frac{a_{n - 1}}{a_{n - 2}} + a_{n - 3}, for n \geq 4 \end{array}

Solution

${0, 1, 1, 1, 2, 3, \frac{5}{2}, \frac{17}{6}} .$

Using Factorial Notation

The formulas for some sequences include products of consecutive positive integers. $n$ factorial, written as $n!,$ is the product of the positive integers from 1 to $n .$ For example,

\begin{array}{l} 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \end{array}

An example of formula containing a factorial is $a_{n} = (n + 1)! .$ The sixth term of the sequence can be found by substituting 6 for $n .$

a_{6} = (6 + 1)! = 7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040

The factorial of any whole number $n$ is $n (n - 1)!$ We can therefore also think of $5!$ as $5 \cdot 4! .$

Example 9

Writing the Terms of a Sequence Using Factorials

Write the first five terms of the sequence defined by the explicit formula $a_{n} = \frac{5 n}{(n + 2)!} .$

Solution

Substitute $n = 1, n = 2,$ and so on in the formula.

\begin{array}{l} n = 1 & a_{1} = \frac{5 (1)}{(1 + 2)!} = \frac{5}{3!} = \frac{5}{3 \cdot 2 \cdot 1} = \frac{5}{6} \\ n = 2 & a_{2} = \frac{5 (2)}{(2 + 2)!} = \frac{10}{4!} = \frac{10}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{5}{12} \\ n = 3 & a_{3} = \frac{5 (3)}{(3 + 2)!} = \frac{15}{5!} = \frac{15}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{1}{8} \\ n = 4 & a_{4} = \frac{5 (4)}{(4 + 2)!} = \frac{20}{6!} = \frac{20}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{1}{36} \\ n = 5 & a_{5} = \frac{5 (5)}{(5 + 2)!} = \frac{25}{7!} = \frac{25}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{5}{1, 008} \end{array}

The first five terms are ${\frac{5}{6}, \frac{5}{12}, \frac{1}{8}, \frac{1}{36}, \frac{5}{1,008}} .$

Analysis

Figure 7 shows the graph of the sequence. Notice that, since factorials grow very quickly, the presence of the factorial term in the denominator results in the denominator becoming much larger than the numerator as $n$ increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero.

Graph of a scattered plot with labeled points: (1, 5/6), (2, 5/12), (3, 1/8), (4, 1/36), and (5, 5/1008). The x-axis is labeled n and the y-axis is labeled a_n. — Figure 7

Key Equations

..
Formula for a factorial	$\begin{array}{l} 0! = 1 \\ 1! = 1 \\ n! = n (n - 1) (n - 2) \dots (2) (1), for n \geq 2 \end{array}$

Key Concepts

A sequence is a list of numbers, called terms, written in a specific order.
Explicit formulas define each term of a sequence using the position of the term. See Example 3, Example 4, and Example 5.
An explicit formula for the $n th$ term of a sequence can be written by analyzing the pattern of several terms. See Example 6.
Recursive formulas define each term of a sequence using previous terms.
Recursive formulas must state the initial term, or terms, of a sequence.
A set of terms can be written by using a recursive formula. See Example 7 and Example 8.
A factorial is a mathematical operation that can be defined recursively.
The factorial of $n$ is the product of all integers from 1 to $n$ See Example 9.

Section Exercises

Verbal

Exercise 1

Discuss the meaning of a sequence. If a finite sequence is defined by a formula, what is its domain? What about an infinite sequence?

Solution

A sequence is an ordered list of numbers that can be either finite or infinite in number. When a finite sequence is defined by a formula, its domain is a subset of the non-negative integers. When an infinite sequence is defined by a formula, its domain is all positive or all non-negative integers.

Exercise 2

Describe three ways that a sequence can be defined.

Exercise 3

Is the ordered set of even numbers an infinite sequence? What about the ordered set of odd numbers? Explain why or why not.

Solution

Yes, both sets go on indefinitely, so they are both infinite sequences.

Exercise 4

What happens to the terms $a_{n}$ of a sequence when there is a negative factor in the formula that is raised to a power that includes $n ?$ What is the term used to describe this phenomenon?

Exercise 5

What is a factorial, and how is it denoted? Use an example to illustrate how factorial notation can be beneficial.

Solution

A factorial is the product of a positive integer and all the positive integers below it. An exclamation point is used to indicate the operation. Answers may vary. An example of the benefit of using factorial notation is when indicating the product It is much easier to write than it is to write out $13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 .$

Algebraic

For the following exercises, write the first four terms of the sequence.

Exercise 6

$a_{n} = 2^{n} - 2$

Exercise 7

$a_{n} = - \frac{16}{n + 1}$

Solution

First four terms: $- 8, - \frac{16}{3}, - 4, - \frac{16}{5}$

Exercise 8

$a_{n} = - {(- 5)}^{n - 1}$

Exercise 9

$a_{n} = \frac{2^{n}}{n^{3}}$

Solution

First four terms: $2, \frac{1}{2}, \frac{8}{27}, \frac{1}{4}$ .

Exercise 10

$a_{n} = \frac{2 n + 1}{n^{3}}$

Exercise 11

$a_{n} = 1.25 \cdot {(- 4)}^{n - 1}$

Solution

First four terms: $1.25, - 5, 20, - 80$ .

Exercise 12

$a_{n} = - 4 \cdot {(- 6)}^{n - 1}$

Exercise 13

$a_{n} = \frac{n^{2}}{2 n + 1}$

Solution

First four terms: $\frac{1}{3}, \frac{4}{5}, \frac{9}{7}, \frac{16}{9}$ .

Exercise 14

$a_{n} = {(- 10)}^{n} + 1$

Exercise 15

$a_{n} = - (\frac{4 \cdot {(- 5)}^{n - 1}}{5})$

Solution

First four terms: $- \frac{4}{5}, 4, - 20, 100$

For the following exercises, write the first eight terms of the piecewise sequence.

Exercise 16

$a_{n} = {\begin{array}{l} {(- 2)}^{n} - 2 & if n is even \\ {(3)}^{n - 1} & if n is odd \end{array}$

Exercise 17

$a_{n} = {\begin{array}{l} \frac{n^{2}}{2 n + 1} & if n \leq 5 \\ n^{2} - 5 & if n >5 \end{array}$

Solution

$\frac{1}{3}, \frac{4}{5}, \frac{9}{7}, \frac{16}{9}, \frac{25}{11}, 31, 44, 59$

Exercise 18

$a_{n} = {\begin{array}{l} {(2 n + 1)}^{2} & if n is divisible by 4 \\ \frac{2}{n} & if n is not divisible by 4 \end{array}$

Exercise 19

$a_{n} = {\begin{array}{l} - 0.6 \cdot 5^{n - 1} & if n is prime or 1 \\ 2.5 \cdot {(- 2)}^{n - 1} & if n is composite \end{array}$

Solution

$- 0.6, - 3, - 15, - 20, - 375, - 80, - 9375, - 320$

Exercise 20

$a_{n} = {\begin{array}{l} 4 (n^{2} - 2) & if n \leq 3 or n > 6 \\ \frac{n^{2} - 2}{4} & if 3 < n \leq 6 \end{array}$

For the following exercises, write an explicit formula for each sequence.

Exercise 21

$4, 7, 12, 19, 28, \dots$

Solution

$a_{n} = n^{2} + 3$

Exercise 22

$- 4, 2, - 10, 14, - 34, \dots$

Exercise 23

$1, 1, \frac{4}{3}, 2, \frac{16}{5}, \dots$

Solution

$a_{n} = \frac{2^{n}}{2 n} or \frac{2^{n - 1}}{n}$

Exercise 24

$0, \frac{1 - e^{1}}{1 + e^{2}}, \frac{1 - e^{2}}{1 + e^{3}}, \frac{1 - e^{3}}{1 + e^{4}}, \frac{1 - e^{4}}{1 + e^{5}}, \dots$

Exercise 25

$1, - \frac{1}{2}, \frac{1}{4}, - \frac{1}{8}, \frac{1}{16}, \dots$

Solution

$a_{n} = {(- \frac{1}{2})}^{n - 1}$

For the following exercises, write the first five terms of the sequence.

Exercise 26

$a_{1} = 9, a_{n} = a_{n - 1} + n$

Exercise 27

$a_{1} = 3, a_{n} = (- 3) a_{n - 1}$

Solution

First five terms: $3, - 9, 27, - 81, 243$

Exercise 28

$a_{1} = - 4, a_{n} = \frac{a_{n - 1} + 2 n}{a_{n - 1} - 1}$

Exercise 29

$a_{1} = - 1, a_{n} = \frac{{(- 3)}^{n - 1}}{a_{n - 1} - 2}$

Solution

First five terms: $- 1, 1, - 9, \frac{27}{11}, \frac{891}{5}$

Exercise 30

$a_{1} = - 30, a_{n} = (2 + a_{n - 1}) {(\frac{1}{2})}^{n}$

For the following exercises, write the first eight terms of the sequence.

Exercise 31

$a_{1} = \frac{1}{24}, a_{2} = 1, a_{n} = (2 a_{n - 2}) (3 a_{n - 1})$

Solution

$\frac{1}{24}, 1, \frac{1}{4}, \frac{3}{2}, \frac{9}{4}, \frac{81}{4}, \frac{2187}{8}, \frac{531, 441}{16}$

Exercise 32

$a_{1} = - 1, a_{2} = 5, a_{n} = a_{n - 2} (3 - a_{n - 1})$

Exercise 33

$a_{1} = 2, a_{2} = 10, a_{n} = \frac{2 (a_{n - 1} + 2)}{a_{n - 2}}$

Solution

$2, 10, 12, \frac{14}{5}, \frac{4}{5}, 2, 10, 12$

For the following exercises, write a recursive formula for each sequence.

Exercise 34

$- 2.5, - 5, - 10, - 20, - 40, \dots$

Exercise 35

$- 8, - 6, - 3, 1, 6, \dots$

Solution

$a_{1} = - 8, a_{n} = a_{n - 1} + n$

Exercise 36

$2, 4, 12, 48, 240, \dots$

Exercise 37

$35, 38, 41, 44, 47, \dots$

Solution

$a_{1} = 35, a_{n} = a_{n - 1} + 3$

Exercise 38

$15, 3, \frac{3}{5}, \frac{3}{25}, \frac{3}{125}, \dots$

For the following exercises, evaluate the factorial.

Exercise 39

$6!$

Solution

$720$

Exercise 40

$(\frac{12}{6})!$

Exercise 41

$\frac{12!}{6!}$

Solution

$665, 280$

Exercise 42

$\frac{100!}{99!}$

For the following exercises, write the first four terms of the sequence.

Exercise 43

$a_{n} = \frac{n!}{n^{2}}$

Solution

First four terms: $1, \frac{1}{2}, \frac{2}{3}, \frac{3}{2}$

Exercise 44

$a_{n} = \frac{3 \cdot n!}{4 \cdot n!}$

Exercise 45

$a_{n} = \frac{n!}{n^{2} - n - 1}$

Solution

First four terms: $- 1, 2, \frac{6}{5}, \frac{24}{11}$

Exercise 46

$a_{n} = \frac{100 \cdot n}{n (n - 1)!}$

Graphical

For the following exercises, graph the first five terms of the indicated sequence

Exercise 47

$a_{n} = \frac{{(- 1)}^{n}}{n} + n$

Solution

Graph of a scattered plot with points at (1, 0), (2, 5/2), (3, 8/3), (4, 17/4), and (5, 24/5). The x-axis is labeled n and the y-axis is labeled a_n.

Exercise 48

$a_{n} = {\begin{array}{l} \frac{4 + n}{2 n} & if n is even \\ 3 + n & if n is odd \end{array}$

Exercise 49

$a_{1} = 2, a_{n} = {(- a_{n - 1} + 1)}^{2}$

Solution

Graph of a scattered plot with points at (1, 2), (2, 1), (3, 0), (4, 1), and (5, 0). The x-axis is labeled n and the y-axis is labeled a_n.

Exercise 50

$a_{1} = 1, a_{n} = a_{n - 1} + 8$

Exercise 51

$a_{n} = \frac{(n + 1)!}{(n - 1)!}$

Solution

Graph of a scattered plot with labeled points: (1, 2), (2, 6), (3, 12), (4, 20), and (5, 30). The x-axis is labeled n and the y-axis is labeled a_n.

For the following exercises, write an explicit formula for the sequence using the first five points shown on the graph.

Exercise 52

Graph of a scattered plot with labeled points: (1, 5), (2, 7), (3, 9), (4, 11), and (5, 13). The x-axis is labeled n and the y-axis is labeled a_n.

Exercise 53

Graph of a scattered plot with labeled points: (1, 0.5), (2, 1), (3, 2), (4, 4), and (5, 8). The x-axis is labeled n and the y-axis is labeled a_n.

Solution

$a_{n} = 2^{n - 2}$

Exercise 54

Graph of a scattered plot with labeled points: (1, 12), (2, 9), (3, 6), (4, 3), and (5, 0). The x-axis is labeled n and the y-axis is labeled a_n.

For the following exercises, write a recursive formula for the sequence using the first five points shown on the graph.

Exercise 55

Graph of a scattered plot with labeled points: (1, 6), (2, 7), (3, 9), (4, 13), and (5, 21). The x-axis is labeled n and the y-axis is labeled a_n.

Solution

$a_{1} = 6, a_{n} = 2 a_{n - 1} - 5$

Exercise 56

Graph of a scattered plot with labeled points: (1, 16), (2, 8), (3, 4), (4, 2), and (5, 1). The x-axis is labeled n and the y-axis is labeled a_n.

Technology

Follow these steps to evaluate a sequence defined recursively using a graphing calculator:

On the home screen, key in the value for the initial term $a_{1}$ and press [ENTER].
Enter the recursive formula by keying in all numerical values given in the formula, along with the key strokes [2ND] ANS for the previous term $a_{n - 1} .$ Press [ENTER].
Continue pressing [ENTER] to calculate the values for each successive term.

For the following exercises, use the steps above to find the indicated term or terms for the sequence.

Exercise 57

Find the first five terms of the sequence $a_{1} = \frac{87}{111}$ , $a_{n} = \frac{4}{3} a_{n - 1} + \frac{12}{37} .$ Use the >Frac feature to give fractional results.

Solution

First five terms: $\frac{29}{37}$ , $\frac{152}{111}$ , $\frac{716}{333}$ , $\frac{3188}{999}$ , $\frac{13724}{2997}$

Exercise 58

Find the 15^th term of the sequence $a_{1} = 625$ , $a_{n} = 0.8 a_{n - 1} + 18.$

Exercise 59

Find the first five terms of the sequence $a_{1} = 2$ , $a_{n} = 2^{[(a_{n} - 1) - 1]} + 1.$

Solution

First five terms: 2, 3, 5, 17, 65537

Exercise 60

Find the first ten terms of the sequence $a_{1} = 8$ , $a_{n} = \frac{(a_{n - 1} + 1)!}{a_{n - 1}!} .$

Exercise 61

Find the tenth term of the sequence $a_{1} = 2$ , $a_{n} = n a_{n - 1}$

Solution

$a_{10} = 7, 257, 600$

Follow these steps to evaluate a finite sequence defined by an explicit formula. Using a TI-84, do the following.

In the home screen, press [2ND] LIST.
Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER].
In the line headed “Expr:” type in the explicit formula, using the $[X,T, θ, n]$ button for $n$
In the line headed “Variable:” type in the variable used on the previous step.
In the line headed “start:” key in the value of $n$ that begins the sequence.
In the line headed “end:” key in the value of $n$ that ends the sequence.
Press [ENTER] 3 times to return to the home screen. You will see the sequence syntax on the screen. Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms.

Using a TI-83, do the following.

In the home screen, press [2ND] LIST.
Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER].
Enter the items in the order “Expr”, “Variable”, “start”, “end” separated by commas. See the instructions above for the description of each item.
Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms.

For the following exercises, use the steps above to find the indicated terms for the sequence. Round to the nearest thousandth when necessary.

Exercise 62

List the first five terms of the sequence $a_{n} = - \frac{28}{9} n + \frac{5}{3} .$

Exercise 63

List the first six terms of the sequence $a_{n} = \frac{n^{3} - 3.5 n^{2} + 4.1 n - 1.5}{2.4 n} .$

Solution

First six terms: 0.042, 0.146, 0.875, 2.385, 4.708

Exercise 64

List the first five terms of the sequence $a_{n} = \frac{15 n \cdot {(- 2)}^{n - 1}}{47}$

Exercise 65

List the first four terms of the sequence $a_{n} = {5.7}^{n} + 0.275 (n - 1)!$

Solution

First four terms: 5.975, 2.765, 185.743, 1057.25, 6023.521

Exercise 66

List the first six terms of the sequence $a_{n} = \frac{n!}{n} .$

Extensions

Exercise 67

Consider the sequence defined by $a_{n} = - 6 - 8 n .$ Is $a_{n} = - 421$ a term in the sequence? Verify the result.

Solution

If $a_{n} = - 421$ is a term in the sequence, then solving the equation $- 421 = - 6 - 8 n$ for $n$ will yield a non-negative integer. However, if $- 421 = - 6 - 8 n,$ then $n = 51.875$ so $a_{n} = - 421$ is not a term in the sequence.

Exercise 68

What term in the sequence $a_{n} = \frac{n^{2} + 4 n + 4}{2 (n + 2)}$ has the value $41 ?$ Verify the result.

Exercise 69

Find a recursive formula for the sequence 1, 0, −1, −1, 0, 1, 1, 0, −1, −1, 0, 1, 1, .... (Hint: find a pattern for $a_{n}$ based on the first two terms.)

Solution

$a_{1} = 1, a_{2} = 0, a_{n} = a_{n - 1} - a_{n - 2}$

Exercise 70

Calculate the first eight terms of the sequences $a_{n} = \frac{(n + 2)!}{(n - 1)!}$ and $b_{n} = n^{3} + 3 n^{2} + 2 n,$ and then make a conjecture about the relationship between these two sequences.

Exercise 71

Prove the conjecture made in the preceding exercise.

Solution

$\frac{(n + 2)!}{(n - 1)!} = \frac{(n + 2) \cdot (n + 1) \cdot (n) \cdot (n - 1) \cdot ... \cdot 3 \cdot 2 \cdot 1}{(n - 1) \cdot ... \cdot 3 \cdot 2 \cdot 1} = n (n + 1) (n + 2) = n^{3} + 3 n^{2} + 2 n$

explicit formula

a formula that defines each term of a sequence in terms of its position in the sequence

finite sequence

a function whose domain consists of a finite subset of the positive integers

{1, 2, \dots n}

for some positive integer

n

infinite sequence

a function whose domain is the set of positive integers

n factorial

the product of all the positive integers from 1 to

n

nth term of a sequence

a formula for the general term of a sequence

recursive formula

a formula that defines each term of a sequence using previous term(s)

sequence

a function whose domain is a subset of the positive integers

term

a number in a sequence