Video: Identifying the 𝑛th Term of a Quadratic Sequence

Which of the following expressions can be used to find the 𝑛th term of the sequence {βˆ’2, 1, 6, ...}? [A] 𝑛 βˆ’ 2 [B] 𝑛 βˆ’ 3 [C] 𝑛² [D] 𝑛² βˆ’ 3 [E] 𝑛² + 3

05:23

Video Transcript

Which of the following expressions can be used to find the 𝑛th term of the sequence containing the terms negative two, one, six, and so on? Is it (A) 𝑛 minus two, (B) 𝑛 minus three, (C) 𝑛 squared, (D) 𝑛 squared minus three, or (E) 𝑛 squared plus three?

We’re looking to find the 𝑛th term of the sequence whose first three terms are given. The 𝑛th term is an algebraic expression that will allow us to find any term in the sequence given the term number. And there are two ways we can answer this question. We’re given five expressions which represent 𝑛th term rules of sequences. So, one method that we have is to find the first three terms given these 𝑛th term rules.

Let’s look at this method first. We’re going to generate the first three terms of each sequence given its 𝑛th term. Let’s begin with the sequence given by 𝑛 minus two. We can do this in table form. We know that to find the first three terms given an 𝑛th term rule, we begin by letting 𝑛 be equal to one. Then, for the second term, we let it be equal to two. And for the third term, we let 𝑛 be equal to three.

So, for the rule 𝑛 minus two, when 𝑛 is one, we get one minus two, which is negative one. Then, when 𝑛 is two, this gives us the second term. We get two minus two, which is equal to zero. Finally, to calculate the third time in this sequence, we do three minus two, which gives us one. And so, the first three terms here are negative one, zero, and one.

In a similar way, if we calculated the first three terms for the sequence given by 𝑛th term 𝑛 minus three, we get negative two, negative one, zero. What about the sequence 𝑛 squared? For this one, we take the term number β€” remember, that’s 𝑛 β€” and we square it. So, the first term is given by one squared, which is equal to one. The second and third terms are two squared and three squared. That’s four and nine, respectively, giving us the first three terms as one, four, and nine.

Now that we have the sequence given by 𝑛 squared, we can quite easily work out the sequence 𝑛 squared minus three. We subtract three from one, four, and nine. That gives us the terms negative two, one, and six. Similarly, we can find the first three terms of our sequence given by 𝑛th term 𝑛 squared plus three by adding three to one, four, and nine. That gives us the terms four, seven, and 12.

We’ll now compare each of these with our sequence. Our sequence in the question is negative two, one, six, and so on. That corresponds to the sequence created with (D), with the expression 𝑛 squared minus three. And so, we can see the correct answer is (D). The 𝑛th term is 𝑛 squared minus three.

But we said there was two methods. Now, the second method is to find the 𝑛th term given the sequence. We’ve done it the other way round so far. We’ve generated the terms given the 𝑛th term. And actually, we’ll see whilst we’re finding the 𝑛th term of our sequence that we would have been able to disregard two of the 𝑛th terms pretty quickly.

We’ll begin by finding the first difference between each of our terms. The difference between negative two and one is three, and the difference between one and six is five. We might then deduce that the difference between the next two terms will be seven, giving us a fourth term of 13, although this isn’t entirely necessary. Next, we’ll find the second difference. That’s the difference between the first differences. Well, the difference between both three and five and five and seven is two.

When we have a common second difference like in this example, we know we have a quadratic sequence. If we’d had a common first difference, that would have given a linear sequence. (A) and (B) are both examples of 𝑛th terms of linear sequences, so we could have quite quickly disregarded (A) and (B).

So, what do we do next? Well, we take the second difference and we halve it. And that tells us the coefficient of 𝑛 squared. Two divided by two is one. So, we know the coefficient of the 𝑛 squared part of our 𝑛th term rule to be one, but we can just write that as 𝑛 squared.

Our next job is to list the first three terms at least for our 𝑛 squared sequence. We’ll list these above the original terms in our sequence. When 𝑛 is one, 𝑛 squared is one. When 𝑛 is two, 𝑛 squared is four. And when 𝑛 is three, 𝑛 squared is nine.

We now compare the 𝑛 squared sequence with our sequence. What do we do to get from one to negative two or four to one or nine to six each time? Well, in fact, we see we do the same thing each time. We take our 𝑛 squared sequence, and we subtract three. This means the 𝑛th term of our sequence negative two, one, six, and so on is given by 𝑛 squared minus three. Once again, we see that this matches answer (D). The 𝑛th term of our sequence is 𝑛 squared minus three.

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