### 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.

Letβs begin with an example of a
sequence. A sequence like this one β two,
four, six, eight, and so on β 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 π sub one. And we can read that as the first
term. The second term can be referred to
as π sub two and the third term as π sub three, and so on.

When it comes to writing the rules
of sequences, weβre really trying to make life a little easier. For example, if we needed to find
the 290th term, we wouldnβt 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 π. For a sequence, we want to find the
πth term given an index π. 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 π, the
πth term is two π. 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 πth term, π sub π, is equal to two π.

When weβre working with sequences,
we can also be given the πth term and asked to work out the first few terms of a
sequence. In the first example, weβll see how
we can do this. Letβs have a look.

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

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

Letβs start by finding π sub one,
the first term, which occurs when π is equal to one. This means that weβd have the first
term π 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 π equals two
into the πth term. This time, weβll have the second
term, π 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, weβll follow
the same process, only this time weβre substituting in π equals three. This gives us that the third term
is equal to three times three minus 34, which is negative 93. When π is equal to four, the
fourth term is negative 120. Finally, when π 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 weβve found that by
substituting the five different values into the πth term formula.

Weβll now have a look at how we
find a recursive formula for a sequence. Letβs have a look at this example
sequence: one, four, seven, 10, and so on. We can compare the values of the
index π is greater than or equal to one to the values of three π. These values of three π 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 π, we would get the terms of the sequence. In fact, we could write that the
πth term of this sequence is three π minus two. However, thereβs 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 πth
term, we would take the term before it and add three. Using the same form of notation,
the term before the πth term β thatβs the term with index π β would be the term
with index π minus one. And so a different way to describe
this sequence would be to say that the πth term π sub π is equal to π sub π
minus one plus three.

When weβre 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 π sub one is equal to one, and then we have the πth term. Notice that weβve also given the
values of the index as π is greater than or equal to two. In this case, the index has to
start with two. It canβt start with one, as weβve
been given the first term. And if we did substitute one into
this part, weβd be trying to find the term with index zero. A rule thatβs 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
π is defined by the term before it. Before we finish with recursive
formulas, thereβs just one other point to note. In this case, we wrote the formula
for π sub π. But it could also have been given
as a formula to find the term with index π plus one. Notice, however, that we still have
this relationship that itβs the term before it plus three. The first term will still be the
same both times. But notice that the index is
different. Because weβre given the formula π
sub π plus one, we can start with a first value of π as one to find the term with
index two.

Weβll now see an example of how we
can find a specific term in a sequence when weβre given a recursive formula.

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

Weβre given four answer options:
two, four, five, or eight. In this question, weβre given a
formula for a sequence. This type of formula is called a
recursive formula. And thatβs 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 π plus one, we take the term
before it β thatβs the one with index π β and we subtract three. And so if we wanted to find the
fourth term β thatβs the term with index four β that means that π plus one must be
equal to four, and so π 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 β thatβs the
term with index three β must happen when π plus one is three. And so π must be equal to two. So the third term is equal to the
second term minus three. Of course, we donβt know the second
term either. But youβve guessed it! Itβs 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 weβre actually given the first term. π sub one is equal to 11. So now we can work forwards through
the sequence. If π sub one is equal to 11 and π
sub two is equal to π sub one minus three, then π sub two, the second term, is
equal to 11 minus three. And thatβs 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). Itβs 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. Weβll now define more formulae,
what we mean by increasing, decreasing, or constant sequences along with the term
monotonic.

A sequence of real numbers π sub
π is said to be increasing if π sub π plus one is greater than π sub π for all
values of π 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 π. 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, itβs 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 itβs 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 itβs called a monotonic
sequence. In the next example, weβll identify
if a sequence is increasing, decreasing, or neither.

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

When weβre considering if a
sequence is increasing or decreasing, weβre comparing any term to the term before
it. If a sequence is increasing, then
any term π sub π must be greater than π sub π minus one. That must be true for all values of
π. Similarly, if a sequence is
decreasing, then any term of index π 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 πth term, and
weβll start by substituting π is equal to one. So for the first term π 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 weβve found the first
term, we can find the second term by substituting in π 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 π is equal to three. This means that we get a third
term, π sub three, equal to negative 727 over 33.

We have at this point got three
terms in the sequence, but itβs a little difficult to see if theyβre 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 canβt say that
for all values either π sub π is bigger than π sub π minus one or π sub π is
less than π sub π minus one. And that means that the sequence
isnβt increasing or decreasing, so it must be neither. The answer is that π sub π 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. Weβll 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 π sub π equals what.

And weβre 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 π is greater than or equal to one, then for any index π, the πth term of these
absolute values would be π sub π is equal to three π. But as we donβt have just three,
six, nine, 12, and so on β we have three, negative six, nine, negative 12, and so on
β then the πth term of this sequence is not three π. Furthermore, we can also say that
the πth term is not negative three π 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 π.

So one way to find a general term
of a sequence that includes three π but which alternates between positive and
negative is to multiply three π by a power of negative one. We notice that options (A) and (B)
present two alternatives. Letβs have a look at the πth term
option given in (A). In order to find the first term, we
would substitute in π 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, itβs three and not negative three. Therefore, the πth term in option
(A) is incorrect.

The πth term given in option (B)
is different because the exponent of negative one is π plus one. When we substitute in π 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 π 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 π 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 itβs option (B). π sub π is equal to negative one
to the power of π plus one times three π.

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 π 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.