5.1: Sequences and Their Notations
In this section, you will:
- Write the terms of a sequence defined by an explicit formula.
- Write the terms of a sequence defined by a recursive formula.
- Use factorial notation.
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 \( \PageIndex{ 1 } \).
| Day | 1 | 2 | 3 | 4 | 5 | \( \ldots \) |
| Hits | 2 | 4 | 8 | 16 | 32 | \( \ldots \) |
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 natural numbers. The sequence established by the number of hits on the website is\[ \{ 2, 4, 8, 16, 32, \ldots \}. \nonumber \]The ellipsis ( \( \ldots \) ) 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 \( n^{\text{th}} \) term of the sequence, where \( n \) is any positive integer. 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 \( n^{\text{th}} \) term.
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 \( n^{\text{th}} \) term of the sequence can be found by raising \( 2 \) to the \( n^{\text{th}} \) 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\[ a_n = 2^n.\nonumber \]Now that we have a formula for the \( n^{\text{th}} \) 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 \( 31^{\text{st}} \) term of the sequence. We will substitute \( 31 \) for \( n \) in the formula.\[ a_{31} = 2^{31} = 2,147,483,648. \nonumber \]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 \( n^{\text{th}} \) term of the sequence are shown in Table \( \PageIndex{ 2 } \).
| 1 | 2 | 3 | 4 | 5 | \( n \) | |
| \( n^{\text{th}} \) 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 \( \PageIndex{ 1 } \) that the number of hits is rising at an exponential rate. This particular sequence forms an exponential function.
Figure \( \PageIndex{ 1 } \)
Lastly, we can write this particular sequence as\[ \{2, 4, 8, 16, 32, \ldots , 2n, \ldots \}. \nonumber \]A sequence that continues indefinitely is called an infinite sequence . The domain of an infinite sequence is the set of natural numbers. If we consider only the first \( 10 \) terms of the sequence, we could write\[ \{2, 4, 8, 16, 32, \ldots , 2n, \ldots , 1024 \}. \nonumber \]This sequence is called a finite sequence because it does not continue indefinitely.
A sequence is a function whose domain is the set of natural numbers. A finite sequence is a sequence whose domain consists of only the first \( n \) natural numbers. 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.\[ a_1, a_2, a_3, \ldots , a_n, \ldots. \nonumber \]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 \( n^{\text{th}} \) term of the sequence , or the general term of the sequence. An explicit formula defines the \( n^{\text{th}} \) term of a sequence using the position of the term. A sequence that continues indefinitely is an infinite sequence .
Does a sequence always have to begin with \( a_1 \)?
No. In certain problems, it may be useful to define the initial term as \( a_0 \) instead of \( a_1 \). In these problems, the domain of the function includes \( 0 \).
Write the first five terms of the sequence defined by the explicit formula \( a_n = −3n+8 \).
- Solution
-
Substitute \( n = 1 \) into the formula. Repeat with values \( 2 \) through \( 5 \) for \( n \).\[ \begin{array}{lcl} n = 1 & \quad & a_1 = -3(1) + 8 = 5 \\[6pt] n = 2 & \quad & a_2 = -3(2) + 8 = 2 \\[6pt] n = 3 & \quad & a_3 = -3(3) + 8 = -1 \\[6pt] n = 4 & \quad & a_4 = -3(4) + 8 = -4 \\[6pt] n = 5 & \quad & a_5 = -3(5) + 8 = -7 \\[6pt] \end{array} \nonumber \]The first five terms are \(\{5, 2, −1, −4, −7 \}\).
The sequence values in Example \( \PageIndex{ 1 } \) can be listed in a table. A table, such as Table \( \PageIndex{ 3 } \), is a convenient way to input the function into graphing technology like Desmos.
| \( n \) | 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 \( \PageIndex{ 2 } \), we can see that this sequence represents a linear function, but notice the graph is not continuous because the domain is over the natural numbers only.
Figure \( \PageIndex{ 2 } \)
Write the first five terms of the sequence defined by the explicit formula \(t_n = 5n−4\).
Investigating Alternating Sequences
Sometimes sequences have terms that alternate in sign. The steps to finding terms of such sequences are the same as if the signs did not alternate. However, the resulting terms will not show increasing or decreasing behvior as \( n \) increases. Let’s take a look at the following sequence.\[ \{2,−4,6,−8\} \nonumber \]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.
Write the first five terms of the sequence.\[ a_n= \dfrac{(−1)^n n^2}{n+1} \nonumber \]
- Solution
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Substitute \( n = 1 \), \( n = 2 \), and so on into the formula.\[ \begin{array}{lcl} n = 1 & \quad & a_1 = dfrac{(−1)^1 1^2}{1+1} = -\dfrac{1}{2} \\[6pt] n = 2 & \quad & a_2 = dfrac{(−1)^2 2^2}{2+1} = \dfrac{4}{3} \\[6pt] n = 3 & \quad & a_3 = dfrac{(−1)^3 3^2}{3+1} = -\dfrac{9}{4} \\[6pt] n = 4 & \quad & a_4 = dfrac{(−1)^4 4^2}{4+1} = \dfrac{16}{5} \\[6pt] n = 5 & \quad & a_5 = dfrac{(−1)^5 5^2}{5+1} = -\dfrac{25}{6} \\[6pt] \end{array} \nonumber \]The first five terms are \(\left\{ - \frac{1}{2}, \frac{4}{3}, −\frac{9}{4}, \frac{16}{5}, −\frac{25}{6} \right\}\).
The graph of this function, shown in Figure \( \PageIndex{ 3 } \), looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values.
Figure \( \PageIndex{ 3 } \)
In Example \( \PageIndex{ 2 } \), does the \((–1)\) to the power of \( n \) account for the oscillations of signs?
Yes, the power might be \( n \), \( n+1 \), \( n−1 \), and so on, but any odd powers will result in a negative term, and any even power will result in a positive term.
Write the first five terms of the sequence.\[ a_n = \dfrac{4n}{(−2)^n} \nonumber \]
Investigating Piecewise Explicit Formulas
We’ve learned that sequences are functions whose domain is over the natural numbers. 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.
Write the first six terms of the sequence.\[ a_n = \begin{cases} n^2, & \text{ if } n \text{ is not divisible by }3 \\[6pt] \frac{n}{3}, & \text{ if }n \text{ is divisible by }3 \\[6pt] \end{cases} \nonumber \]
- Solution
-
Substitute \( n = 1 \), \( n = 2 \), and so on into the appropriate formula. Use \( n^2 \) when \( n \) is not a multiple of \( 3 \). Use \( \frac{n}{3} \) when \( n \) is a multiple of \( 3 \).\[ \begin{array}{lclcl} n = 1 & \quad & a_1 = 1^2 = 1 & \quad & \left( 1\text{ is not a multiple of }3 \implies \text{ Use }n^2 \right) \\[6pt] n = 2 & \quad & a_2 = 2^2 = 4 & \quad & \left( 2\text{ is not a multiple of }3 \implies \text{ Use }n^2 \right) \\[6pt] n = 3 & \quad & a_3 = \dfrac{3}{3} = 1 & \quad & \left( 3\text{ is a multiple of }3 \implies \text{ Use }\dfrac{n}{3} \right) \\[6pt] n = 4 & \quad & a_4 = 4^2 = 16 & \quad & \left( 4\text{ is not a multiple of }3 \implies \text{ Use }n^2 \right) \\[6pt] n = 5 & \quad & a_5 = 5^2 = 25 & \quad & \left( 5\text{ is not a multiple of }3 \implies \text{ Use }n^2 \right) \\[6pt] n = 6 & \quad & a_6 = \dfrac{6}{3} = 2 & \quad & \left( 6\text{ is a multiple of }3 \implies \text{ Use }\dfrac{n}{3} \right) \\[6pt] \end{array} \nonumber \]The first six terms are \(\left\{ 1, 4, 1, 16, 25, 2 \right\}\).
Every third point on the graph shown in Figure \( \PageIndex{ 4 } \) stands out from the two nearby points. This occurs because the sequence was defined by a piecewise function.
Figure \( \PageIndex{ 4 } \)
Write the first six terms of the sequence.\[ a_n = \begin{cases} 2n^3, & \text{ if } n \text{ is odd} \\[6pt] \frac{5n}{2}, & \text{ if }n \text{ is even} \\[6pt] \end{cases} \nonumber \]
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 \( n^{\text{th}} \) 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.
Write an explicit formula for the \( n^{\text{th}} \) term of each sequence.
- \( \left\{−\frac{2}{11}, \frac{3}{13}, −\frac{4}{15}, \frac{5}{17}, −\frac{6}{19}, \ldots \right\} \)
- \( \left\{−\frac{2}{25}, −\frac{2}{125}, −\frac{2}{625}, −\frac{2}{3,125}, −\frac{2}{15,625}, \ldots \right\} \)
- \( \left\{e^4, e^5, e^6, e^7, e^8, \ldots \right\}\)
- Solutions
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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 \( 2n+9 \). Therefore,\[ a_n = \dfrac{(-1)^n (n + 1)}{2n + 9} \nonumber \]
- The terms are all negative.\[ \left\{−\frac{\textcolor{red}{2}}{25}, −\frac{\textcolor{red}{2}}{125}, −\frac{\textcolor{red}{2}}{625}, −\frac{\textcolor{red}{2}}{3,125}, −\frac{\textcolor{red}{2}}{15,625}, \ldots \right\} \nonumber \]The numerator is always \( 2 \).\[ \left\{−\frac{2}{\textcolor{blue}{5^2}}, −\frac{2}{\textcolor{blue}{5^3}}, −\frac{2}{\textcolor{blue}{5^4}}, −\frac{2}{\textcolor{blue}{5^5}}, −\frac{2}{\textcolor{blue}{5^6}}, \ldots \right\} \nonumber \]The denominators are increasing powers of \( 5 \), starting with the second power when \( n = 1 \). So we know that the denominator can be represented by \( 5^{n + 1} \). Therefore,\[ a_n = -\dfrac{2}{5^{n + 1}}. \nonumber \]
- The terms are powers of \( e \). For \( n = 1 \), we get \( e^4 \). Therefore, the exponent must be \( n + 3 \). That is,\[ a_n = e^{n + 3}. \nonumber \]
Write an explicit formula for the \( n^{\text{th}} \) term of each sequence.
- \( \left\{ 9, −81, 729, −6561, 59049, \ldots \right\}\)
- \(\left\{−\frac{3}{4}, −\frac{9}{8}, −\frac{27}{12}, −\frac{81}{16}, −\frac{243}{20}, \ldots \right\} \)
- \( \left\{ \frac{1}{e^2}, \frac{1}{e}, 1, e, e^2, \ldots \right\}\)
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 Fibonacci sequence are \( 1 \), \( 1 \), \( 2 \), \( 3 \), \( 5 \), \( 8 \), \( 13 \), \( 21 \), \( 34 \), \(\ldots\). 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:\[ \begin{array}{rclcl} a_1 & = & 3 & & \\[6pt] a_n & = & 2a_{n - 1} - 1 & \quad & \text{for }n \geq 2 \\[6pt] \end{array} \nonumber \]We can find the subsequent terms of the sequence using the first term.\[ \begin{array}{rcl} a_1 & = & 3 \\[6pt] a_2 & = & 2a_{1} - 1 = 2(3) - 1 = 5 \\[6pt] a_3 & = & 2a_{2} - 1 = 2(5) - 1 = 9 \\[6pt] a_4 & = & 2a_{3} - 1 = 2(9) - 1 = 17 \\[6pt] \end{array} \nonumber \]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.\[ \begin{array}{rclcl}
a_1 & = & 1 & & \\[6pt]
a_2 & = & 1 & & \\[6pt]
a_n & = & a_{n - 1} + a_{n - 2} & \quad & \text{for }n \geq 3 \\[6pt]
\end{array} \nonumber \]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\[ a_{10} = a_9 + a_8 = 34 + 21 = 55. \nonumber \]
A
recursive formula
is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence.
Must the first two terms always be given in a recursive formula?
No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined.
Write the first five terms of the sequence defined by the recursive formula.\[ \begin{array}{rclcl}
a_1 & = & 9 & & \\[6pt]
a_n & = & 3a_{n - 1} - 20 & \quad & \text{for }n \geq 2 \\[6pt]
\end{array} \nonumber \]
The first term is given in the formula. For each subsequent term, we replace \( a_{n−1} \) with the value of the preceding term.\[ \begin{array}{lcl}
n = 1 & \quad & a_1 = 9 \\[6pt]
n = 2 & \quad & a_2 = 3a_{1} - 20 = 3(9) - 20 = 27 - 20 = 7 \\[6pt]
n = 3 & \quad & a_3 = 3a_{2} - 20 = 3(7) - 20 = 21 - 20 = 1 \\[6pt]
n = 4 & \quad & a_4 = 3a_{3} - 20 = 3(1) - 20 = 3 - 20 = -17 \\[6pt]
n = 5 & \quad & a_5 = 3a_{4} - 20 = 3(-17) - 20 = -51 - 20 = -71 \\[6pt]
\end{array} \nonumber \]The first five terms are \(\left\{ 9, 7, 1, -17, -71 \right\}\). See Figure \( \PageIndex{ 5 } \)
Write the first five terms of the sequence defined by the recursive formula.\[ \begin{array}{rclcl}
a_1 & = & 2 & & \\[6pt]
a_n & = & 2a_{n - 1} + 1 & \quad & \text{for }n \geq 2 \\[6pt]
\end{array} \nonumber \]
Write the first six terms of the sequence defined by the recursive formula.\[ \begin{array}{rclcl}
a_1 & = & 1 & & \\[6pt]
a_2 & = & 2 & & \\[6pt]
a_n & = & 3a_{n - 1} + 4 a_{n - 2} & \quad & \text{for }n \geq 3 \\[6pt]
\end{array} \nonumber \]
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.\[ \begin{array}{lcl}
n = 3 & \quad & a_3 = 3a_2 + 4a_1 = 3(2) + 4(1) = 10 \\[6pt]
n = 4 & \quad & a_4 = 3a_3 + 4a_2 = 3(10) + 4(2) = 38 \\[6pt]
n = 5 & \quad & a_5 = 3a_4 + 4a_3 = 3(38) + 4(10) = 154 \\[6pt]
n = 6 & \quad & a_6 = 3a_5 + 4a_4 = 3(154) + 4(38) = 614 \\[6pt]
\end{array} \nonumber \]The first six terms are \(\left\{ 1, 2, 10, 38, 154, 614 \right\}\). See Figure \( \PageIndex{ 6 } \).
Write the first 8 terms of the sequence defined by the recursive formula.\[ \begin{array}{rclcl}
a_1 & = & 0 & & \\[6pt]
a_2 & = & 1 & & \\[6pt]
a_3 & = & 1 & & \\[6pt]
a_n & = & \dfrac{a_{n - 1}}{a_{n - 2}} + a_{n - 3} & \quad & \text{for }n \geq 4 \\[6pt]
\end{array} \nonumber \]
Figure \( \PageIndex{ 5 } \)
Figure \( \PageIndex{ 6 } \)
Using Factorial Notation
The formulas for some sequences include products of consecutive natural numbers. \( n \) factorial , written as \(n!\), is the product of the natural numbers from \( 1 \) to \( n \). For example,\[ \begin{array}{rcl} 4! & = & 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\[6pt] 5! & = & 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \\[6pt] \end{array} \nonumber \]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 \( n \).\[ a_6 = (6+1)! = 7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040 \nonumber \]The factorial of any natural number \( n \) is \( n(n−1)!\) We can therefore also think of \(5!\) as \(5 \cdot 4!\).
\(n\) factorial is a mathematical operation that can be defined using a recursive formula. The factorial of \(n\), denoted \( n! \), is defined for a natural number \( n \) as:\[ \begin{array}{rclcl} 0! & = & 1 & & \\[6pt] 1! & = & 1 & & \\[6pt] n! & = & n(n-1)(n-2)\cdots (2)(1) & \quad & \text{for }n \geq 2 \\[6pt] \end{array} \nonumber \]The special case \( 0! \) is defined as \( 0!=1 \).
Can factorials always be found using a calculator?
No. Factorials get large very quickly - faster than exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial.
Write the first five terms of the sequence defined by the explicit formula \( a_n = \frac{5n}{(n+2)!}\).
- Solution
-
Substitute \( n = 1 \), \( n = 2 \), and so on in the formula.\[ \begin{array}{rcrcl} n = 1 & \quad & a_1 & = & \dfrac{5(1)}{(1 + 2)!} = \dfrac{5}{3!} = \dfrac{5}{3 \cdot 2 \cdot 1} = \dfrac{5}{6} \\[6pt] n = 2 & \quad & a_2 & = & \dfrac{5(2)}{(2 + 2)!} = \dfrac{10}{4!} = \dfrac{10}{4 \cdot 3 \cdot 2 \cdot 1} = \dfrac{5}{12} \\[6pt] n = 3 & \quad & a_3 & = & \dfrac{5(3)}{(3 + 2)!} = \dfrac{15}{5!} = \dfrac{15}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \dfrac{1}{8} \\[6pt] n = 4 & \quad & a_4 & = & \dfrac{5(4)}{(4 + 2)!} = \dfrac{20}{6!} = \dfrac{20}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \dfrac{1}{36} \\[6pt] n = 5 & \quad & a_5 & = & \dfrac{5(5)}{(5 + 2)!} = \dfrac{25}{7!} = \dfrac{25}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \dfrac{5}{1008} \\[6pt] \end{array} \nonumber \]The first five terms are \( \left\{ \frac{5}{6}, \frac{5}{12}, \frac{1}{8}, \frac{1}{36}, \frac{5}{1008} \right\} \).
Figure \( \PageIndex{ 7 } \) 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 nn increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero.
Figure \( \PageIndex{ 7 } \)
Write the first five terms of the sequence defined by the explicit formula \(a_n = \frac{(n+1)!}{2n}\).
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