Representing Sequences with Recursive and Explicit Formulas

Recursive formulas and explicit formulas are two ways of expressing arithmetic and geometric sequences of numbers. In this video, we are going to discuss what each of them are and how to write sequences of numbers using each of these two techniques.

Let’s consider the sequence of numbers 3, 8, 13, 18, 23, and so on.

Can you detect a pattern for this sequence? To get from 3 to 8, we add 5. We also add 5 to get from 8 to 13. This pattern continues with the following terms as well. For this reason, we can express the terms in this way:

This is a way of saying that the \(n\)th term of the sequence is equal to the previous, \(n-1\), term, plus five. The lowercase a’s denote that we are talking about terms in a sequence of numbers. The lowercase \(n\)’s represent the number of the term we are looking at. For example, if we wanted the 7th term of the sequence, we would be looking for \(a_{7}\) and we would need to know the previous term, \(a_{n-1}\), or \(a_{7-1}=a_{6}\). This formula for the sequence is called a recursive formula, because it is written in terms of the previous term of the sequence. If we know that the fifth term in the sequence is 23, we can find the sixth term by using the recursive formula \(a_{6}=a_{5}+5\), or \(a_{6}=28\).

When we write recursive formulas, it is important to include a couple extra details. We need to specify which values of \(n\) the formula works for, and we need to include the value for the first term in the sequence. In this case, the sequence started with 3, so we would say that the first term is \(a_{1}=3\). After the first term, the recursive nature of the formula kicks in, so we would say that the formula applies for \(n\geq 2\).

One of the problems that arises when working with recursive formulas is when we need to find the value of a very large term. For example, if we were asked to find the 100th term in the sequence, we would have to know the 99th term in order to use the recursive formula. Working our way up to the 99th term would be very time-consuming. Because of this problem, explicit formulas tend to be used more often.

When working with explicit formulas, we express this sequence in terms of \(n\), rather than in terms of the previous term. That is, we can write another type of formula which will give us the value of any given term based on what position it is in. Because we already said that each term is five more than the previous, we can start by writing \(a_{n}= 5\) times \(n\). Every time \(n\) increases by one, the term of the sequence increases by five. But we also know that the first term is 3, which is not 5, but 2 less than 5. That is, \(a_{1}=3\). To make this work, we can just subtract 2 from \(5n\).

Let’s check this formula to be sure that it works for the sequence. The first term, \(a_{1}\), is \(5(1)-2\), or 3. That’s correct. The second term, \(a_{2}\), is \(5(2)-2\), or 8. Also correct! The third term, \(a_{3}\), is \(5(3)-2\), or 13. It looks like we’ve got this formula written correctly. This type of formula for a sequence, where each term is written in terms of \(n\), is called an explicit formula. Like with recursive formulas, though, we need to include the values of \(n\) for which the formula applies. In this case, our formula works for values of \(n\) greater than or equal to one.

While recursive and explicit formulas are both useful in math, it is more common to use explicit formulas because you can find any term you want without having to first find all the terms before it.

Let’s now discuss how to convert an explicit formula to a recursive one, and vice versa.

How can we change the explicit formula \(a_{n}=10n+1\), where \(n\geq 1\), into a recursive formula? Recall that one of the things a recursive formula needs defined is its first term, \(a_{1}\). Let’s start there.

Start with \(a_{1}\)

We can find the first term in this sequence by using \(n=1\), since that is the smallest allowable value of \(n\). This gives us \(a_{1}=10(1)+1=11\). Next, we need to figure out how one term relates to the next so we can write an in terms of \(a_{n-1}\).

Find the relationship between each term and the previous term

In the last example, we said that the sequence increasing by five with each term corresponded to the explicit formula containing 5 times \(n\). In this explicit formula, we see 10 times \(n\), which implies that for each time \(n\) increases by one, the term increases by 10. In other words, each term should be ten greater than the previous. This gives us \(a_{n}=a_{n-1}+10\). Now, which values of \(n\) does this formula work for?

Define which values of \(n\) apply

We said that the first term is \(a_{1}=11\), where \(n=1\). So, the recursive nature of the formula will kick in at the next term, and work for \(n\geq 2\).

Let’s check that the first few terms are equivalent to the explicit formula we started with.

Check for equivalence

For \(n=1\), we know that \(10(1)+1=11\), and we said that \(a_{1}=11\) for the recursive formula. For \(n=2\), \(10(2)+1=21\), and \(11+10\) is also 21. Similarly, for \(n=3\), \(10(3)+1=31\), and \(21+10 = 31\). This recursive formula works perfectly!

Let’s now try to form an explicit formula from a recursive one. Consider the formula \(a_{n}=2a_{n-1}\) where \(a_{1}=2\) and \(n\geq 2\). To convert this to an explicit formula, let’s try to understand the pattern. Let’s use the given formula to find the first four terms. The first term in the sequence is 2. Then \(a_{2}=2(2)=4\). Next, \(a_{3}=2(4)=8\), and \(a_{4}=2(8)=16\). Another way of looking at this, though, is that \(a_{3}=2\times 2\times 2\), or 23, and \(a_{4}=2\times 2\times 2\times 2\), or \(2^{4}\).

Because each term is twice the previous one, the \(n\)th term should be equal to \(2^{n}\). To add the finishing touch to this explicit formula, we need to define which values of \(n\) it works for. We know that the first term in the sequence is \(a_{1}=2\), and because \(2^{1}=2\), \(n=1\) is valid as the first value of \(n\). Therefore, we say that \(a_{n}=2^{n}\) for \(n\geq 1\).

When it comes to sequences of numbers, explicit formulas are usually regarded as more convenient to use. We can typically express sequences as either recursive or explicit formulas. Recursive formulas present an in terms of the previous term, \(a_{n-1}\), while explicit formulas write an in terms of \(n\). Both types of formulas require defining which values of \(n\) they apply to, but recursive formulas also require you to define what the first term is. To rewrite an explicit formula as a recursive one, and vice versa, identify the first term, determine the relationship between each term and either \(n\) or \(a_{n-1}\), and define which values of \(n\) the formula allows. Finally, it is always a good idea to check your work and be sure that the two formulas are equivalent by testing the values in the given sequence.

I hope this video was helpful. Thanks for watching, and happy studying!