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.

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.

Sequences 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

[latex]\left\{2,4,8,16,32,\dots \right\}[/latex].

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 [latex]n\text{th}[/latex] term of the sequence, where [latex]n[/latex] 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 [latex]n\text{th}[/latex] term.

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

Now that we have a formula for the [latex]n\text{th}[/latex] 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 [latex]n[/latex] in the formula.

[latex]\begin{array}{l}{a}_{31}={2}^{31}\hfill \\ \text{ }=\text{2,147,483,648}\hfill \end{array}[/latex]

If the doubling trend continues, the company will get [latex]\text{2,147,483,648}[/latex] 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 [latex]n\text{th}[/latex] term of the sequence are shown in the table.

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

Lastly, we can write this particular sequence as

[latex]\left\{2,4,8,16,32,\dots ,{2}^{n},\dots \right\}[/latex].

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

[latex]\left\{2,4,8,16,32,\dots ,{2}^{n},\dots ,1024\right\}[/latex].

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

Investigating 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.

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 [latex]n\text{th}[/latex] 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.

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 [latex]n[/latex] increases. Let’s take a look at the following sequence.

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.

Sequences 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 [latex]{a}_{n}[/latex] in terms of preceding terms. For example, suppose we know the following:

[latex]\begin{array}{l}{a}_{1}=3\hfill \\ {a}_{n}=2{a}_{n - 1}-1, \text{for} n\ge 2\hfill \end{array}[/latex]

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

[latex]\begin{array}{l}{a}_{1}=3\\ {a}_{2}=2{a}_{1}-1=2\left(3\right)-1=5\\ {a}_{3}=2{a}_{2}-1=2\left(5\right)-1=9\\ {a}_{4}=2{a}_{3}-1=2\left(9\right)-1=17\end{array}[/latex]

So the first four terms of the sequence are [latex]\left\{3,\text{ }5,\text{ }9,\text{ }17\right\}[/latex] .

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.

[latex]\begin{array}{l}{a}_{1}=1\hfill \\ {a}_{2}=1\hfill \\ {a}_{n}={a}_{n - 1}+{a}_{n - 2}, \text{for} n\ge 3\hfill \end{array}[/latex]

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

[latex]{a}_{10}={a}_{9}+{a}_{8}=34+21=55[/latex]

Using Factorial Notation

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

An example of formula containing a factorial is [latex]{a}_{n}=\left(n+1\right)![/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[/latex].

The factorial of any whole number [latex]n[/latex] is [latex]n\left(n - 1\right)![/latex] We can therefore also think of [latex]5![/latex] as [latex]5\cdot 4!\text{.}[/latex]

Analysis of the Solution

Figure 8 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 [latex]n[/latex] increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero.

Figure 8