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 — remember, that’s 𝑛 — 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.
