Lesson Video: Sequence Formulas Mathematics

Video Transcript

In this video, we will learn how to find general term or a recursive formula for a sequence and how to use these to work out terms in the sequence. We will also learn what an alternating sequence is and what increasing or decreasing sequences are.

Let’s begin with an example of a sequence. A sequence like this one — two, four, six, eight, and so on — can be described in terms of a position index. For example, the term with index one is two, the term with index two is four, and the term with index three is six, and so on. The term with index one can also be written as 𝑎 sub one. And we can read that as the first term. The second term can be referred to as 𝑎 sub two and the third term as 𝑎 sub three, and so on.

When it comes to writing the rules of sequences, we’re really trying to make life a little easier. For example, if we needed to find the 290th term, we wouldn’t want to list all the terms up to the 290th term. What we want is a relationship between the index and the term value that will allow us to very quickly write the term for any index.

The shorthand for any general term is the letter 𝑛. For a sequence, we want to find the 𝑛th term given an index 𝑛. You might have already worked out the relationship in this sequence between the index and the term. Every index is multiplied by two to give us the term. So for a term with index 𝑛, the 𝑛th term is two 𝑛. So the term that has the index of 290 would in fact be 580. And so for the sequence two, four, six, eight, and so on, the 𝑛th term, 𝑎 sub 𝑛, is equal to two 𝑛.

When we’re working with sequences, we can also be given the 𝑛th term and asked to work out the first few terms of a sequence. In the first example, we’ll see how we can do this. Let’s have a look.

Find the first five terms of the sequence whose general term is given by 𝑎 sub 𝑛 equals 𝑛 times 𝑛 minus 34, where 𝑛 is greater than or equal to one.

In this question, we’re given the general or the 𝑛th term of a sequence for an index 𝑛. To find any term in the sequence, we would substitute that value for 𝑛 into the general term. For example, if we wanted to find the 20th term, we would substitute 𝑛 equals 20 into the 𝑛th term. However, in this question, we need to find the first five terms. We’re told that the index 𝑛 is greater than or equal to one. So that means we’re going to start by substituting 𝑛 equals one then 𝑛 equals two, three, four, and five into the 𝑛th term to find the first five terms.

Let’s start by finding 𝑎 sub one, the first term, which occurs when 𝑛 is equal to one. This means that we’d have the first term 𝑎 sub one equals one times one minus 34. One minus 34 is negative 33. And when we multiply that by one, we get negative 33. And so the first term is negative 33. Now we can substitute 𝑛 equals two into the 𝑛th term. This time, we’ll have the second term, 𝑎 sub two, is equal to two times two minus 34. Simplifying this, we have two multiplied by negative 32, which gives us the second term as negative 64.

For the third term, we’ll follow the same process, only this time we’re substituting in 𝑛 equals three. This gives us that the third term is equal to three times three minus 34, which is negative 93. When 𝑛 is equal to four, the fourth term is negative 120. Finally, when 𝑛 is equal to five, the fifth term is equal to negative 145. And so we can give the answer for the first five terms of the sequence. And we’ve found that by substituting the five different values into the 𝑛th term formula.

We’ll now have a look at how we find a recursive formula for a sequence. Let’s have a look at this example sequence: one, four, seven, 10, and so on. We can compare the values of the index 𝑛 is greater than or equal to one to the values of three 𝑛. These values of three 𝑛 do not give us the same values as we have in the sequence. But if we subtracted two from each of the values of three 𝑛, we would get the terms of the sequence. In fact, we could write that the 𝑛th term of this sequence is three 𝑛 minus two. However, there’s also another way we could describe this sequence.

We might have noticed that the pattern between terms is to add three. For example, if we wanted to find the fifth term, we would take the fourth term and add three. So if we wanted to find any 𝑛th term, we would take the term before it and add three. Using the same form of notation, the term before the 𝑛th term — that’s the term with index 𝑛 — would be the term with index 𝑛 minus one. And so a different way to describe this sequence would be to say that the 𝑛th term 𝑎 sub 𝑛 is equal to 𝑎 sub 𝑛 minus one plus three.

When we’re given a formula for a sequence in this way, we also need to say what the first term is. We can write it as a list like this, so we have 𝑎 sub one is equal to one, and then we have the 𝑛th term. Notice that we’ve also given the values of the index as 𝑛 is greater than or equal to two. In this case, the index has to start with two. It can’t start with one, as we’ve been given the first term. And if we did substitute one into this part, we’d be trying to find the term with index zero. A rule that’s written in this way is called a recursive formula for a sequence.

A recursive formula is a formula in which the terms of a sequence are defined using one or more of the previous terms. In this case, our term with index 𝑛 is defined by the term before it. Before we finish with recursive formulas, there’s just one other point to note. In this case, we wrote the formula for 𝑎 sub 𝑛. But it could also have been given as a formula to find the term with index 𝑛 plus one. Notice, however, that we still have this relationship that it’s the term before it plus three. The first term will still be the same both times. But notice that the index is different. Because we’re given the formula 𝑎 sub 𝑛 plus one, we can start with a first value of 𝑛 as one to find the term with index two.

We’ll now see an example of how we can find a specific term in a sequence when we’re given a recursive formula.

If 𝑎 sub 𝑛 is a sequence defined as 𝑎 sub one equals 11 and 𝑎 sub 𝑛 plus one equals 𝑎 sub 𝑛 minus three, where 𝑛 is greater than or equal to one, then the fourth term equals what.

We’re given four answer options: two, four, five, or eight. In this question, we’re given a formula for a sequence. This type of formula is called a recursive formula. And that’s when the terms of a sequence are defined using one or more previous terms. If we wanted to describe this term in words, we would say that for any term with index 𝑛 plus one, we take the term before it — that’s the one with index 𝑛 — and we subtract three. And so if we wanted to find the fourth term — that’s the term with index four — that means that 𝑛 plus one must be equal to four, and so 𝑛 must be three. And so the fourth term must be equal to the third term minus three. But how do we find the third term?

Well, the third term — that’s the term with index three — must happen when 𝑛 plus one is three. And so 𝑛 must be equal to two. So the third term is equal to the second term minus three. Of course, we don’t know the second term either. But you’ve guessed it! It’s going to be the first term minus three. And this is also one of the disadvantages of recursive formulas because we need to work out every term up to the term that we need.

We do get a little bit of relief here because we’re actually given the first term. 𝑎 sub one is equal to 11. So now we can work forwards through the sequence. If 𝑎 sub one is equal to 11 and 𝑎 sub two is equal to 𝑎 sub one minus three, then 𝑎 sub two, the second term, is equal to 11 minus three. And that’s equal to eight. As the third term is equal to the second term minus three, then our third term must be equal to eight minus three, which is five. And finally then, the fourth term is the third term minus three. And so five minus three is equal to two. We can therefore give the answer that the fourth term of the sequence is that given in option (A). It’s the term two.

In this example, the terms of the sequence went from 11, eight, five, two, and so on. This type of sequence would be called a decreasing sequence. We’ll now define more formulae, what we mean by increasing, decreasing, or constant sequences along with the term monotonic.

A sequence of real numbers 𝑎 sub 𝑛 is said to be increasing if 𝑎 sub 𝑛 plus one is greater than 𝑎 sub 𝑛 for all values of 𝑛 in the natural numbers.

The terminology here really means that every term in the sequence must be greater than the term before it in order for the sequence to be increasing. For example, if we took the sequence of square numbers one, four, nine, 16, and so on, every value in that sequence is larger than the term before it. So the sequence of square numbers is an increasing sequence. Notice that this has to be true for all values of 𝑛. If we had another sequence that went one, two, three, one, and so on, this would not be increasing because although we have an increasing portion of the sequence, it’s not all increasing.

We can define a decreasing sequence in a similar way. This time, every term in the sequence must be less than the term before it. An example of a decreasing sequence could be the sequence one, one-half, one-third, one-quarter, and so on. When we have a sequence in which every term is equal to the term before it, then it’s called a constant sequence. An example of this type of sequence might be the sequence of all twos. If a sequence is one of these three types, that is, increasing, decreasing, or constant, then it’s called a monotonic sequence. In the next example, we’ll identify if a sequence is increasing, decreasing, or neither.

Is the sequence 𝑎 sub 𝑛 equals negative one to the power of 𝑛 over 11𝑛 minus 22 increasing, decreasing, or neither?

When we’re considering if a sequence is increasing or decreasing, we’re comparing any term to the term before it. If a sequence is increasing, then any term 𝑎 sub 𝑛 must be greater than 𝑎 sub 𝑛 minus one. That must be true for all values of 𝑛. Similarly, if a sequence is decreasing, then any term of index 𝑛 in a sequence must be less than the term before it. What we can do is to work out the first few terms of the sequence and see if the values are increasing, decreasing, or neither.

So we could take the 𝑛th term, and we’ll start by substituting 𝑛 is equal to one. So for the first term 𝑎 sub one, we have negative one to the power of one over 11 times one minus 22. When we simplify this, we get the fraction negative 243 over 11. Now that we’ve found the first term, we can find the second term by substituting in 𝑛 is equal to two. When we simplify negative one squared over 11 times two minus 22, we get the fraction negative 483 over 22. We can find the third term in the same way by substituting 𝑛 is equal to three. This means that we get a third term, 𝑎 sub three, equal to negative 727 over 33.

We have at this point got three terms in the sequence, but it’s a little difficult to see if they’re increasing or decreasing. So it might be helpful to find their decimal equivalents. The first term is approximately negative 22.09, the second term approximately negative 21.95, and the third term is approximately negative 22.03. We notice that the second term is greater than the first term. However, the third term is less than the second term. That means that we can’t say that for all values either 𝑎 sub 𝑛 is bigger than 𝑎 sub 𝑛 minus one or 𝑎 sub 𝑛 is less than 𝑎 sub 𝑛 minus one. And that means that the sequence isn’t increasing or decreasing, so it must be neither. The answer is that 𝑎 sub 𝑛 is neither increasing nor decreasing.

One other piece of terminology to introduce is that of alternating sequences. An alternating sequence is one where the terms of the sequence alternate between positive and negative. For example, the sequence negative two, three, negative four, five, negative six, and so on would be an alternating sequence. The values switch between positive and negative. We’ll now see an example of how we can find a general term of an alternating sequence.

The general term of the sequence three, negative six, nine, negative 12, 15 is 𝑎 sub 𝑛 equals what.

And we’re given four answer options. We might notice that the terms of this sequence alternate between positive and negative values. This type of sequence is defined as an alternating sequence. We can take the sequence and consider if we just had the absolute values of the sequence, then we would have the terms three, six, nine, 12, and 15. If we took the index in this case as 𝑛 is greater than or equal to one, then for any index 𝑛, the 𝑛th term of these absolute values would be 𝑎 sub 𝑛 is equal to three 𝑛. But as we don’t have just three, six, nine, 12, and so on — we have three, negative six, nine, negative 12, and so on — then the 𝑛th term of this sequence is not three 𝑛. Furthermore, we can also say that the 𝑛th term is not negative three 𝑛 either. In this case, the sequence would have the values negative three, negative six, negative nine, negative 12, and so on. However, we do have a sequence that does very closely match three 𝑛.

So one way to find a general term of a sequence that includes three 𝑛 but which alternates between positive and negative is to multiply three 𝑛 by a power of negative one. We notice that options (A) and (B) present two alternatives. Let’s have a look at the 𝑛th term option given in (A). In order to find the first term, we would substitute in 𝑛 is equal to one. Negative one to the power of one is negative one, and three times one is three. Multiplying these gives us the first term of negative three. However, if we look at the first term in the given sequence, it’s three and not negative three. Therefore, the 𝑛th term in option (A) is incorrect.

The 𝑛th term given in option (B) is different because the exponent of negative one is 𝑛 plus one. When we substitute in 𝑛 is equal to one to find the first term, we have negative one to the power of one plus one, which is two, and negative one squared gives us one, which when multiplied by three gives us three. This matches the given first term. Substituting in 𝑛 is equal to two, we get that the second term is equal to negative six. We can observe the pattern. When we have an even index, like we did here when 𝑛 is equal to two, then the exponent of negative one will be odd. Negative one with an odd power will give us the value of negative one. The result of this is that every even index gives us a term value which is negative.

If we continued by substituting an odd index of three, we would get an even value of nine. We can therefore give the answer that it’s option (B). 𝑎 sub 𝑛 is equal to negative one to the power of 𝑛 plus one times three 𝑛.

We can now summarize the key points of this video. Firstly, we saw that to find the terms of a sequence given a general term, we substitute values of 𝑛 is greater than or equal to one into the formula for the general term. We defined recursive formulas and saw that sometimes we might need to apply the formula several times in order to find the values of preceding terms. Finally, we defined increasing, decreasing, constant, and alternating sequences.