### Video Transcript

In this video, weβre gonna look at
how to specify a sequence of numbers using recursive notation, and then weβll try
some typical questions.

A sequence is a list of numbers
thatβs generated using a consistent rule. For example, two, four, six, eight,
10, 12, and so on. This sequence starts with two, and
then we add two to each term to get the next term. This way of describing a sequence
by specifying the first term and setting out a term-to-term rule, telling you what
you need to do to one term to generate the next term in the sequence, is called
recursive. Now, rather than writing out the
description longhand like this, we can use some mathematical notation to specify the
rule. And in this case, we might put π
π plus one is equal to π π plus two, where π one is equal to two, and π is
greater than or equal to one, and π is an integer.

Now, this bit here tells you how
the sequence begins. The first term π one is equal to
two. And this bit here sets up our
numbering system for terms in the sequence, so weβre gonna have a first term, a
second term, a third term, and so on. So, π is gonna be a series of
integers starting at one. And this bit defines the
term-to-term rule. Itβs telling us that we need to
take the value of a term at position π, add two to that, and that will then
generate the next term in the sequence, the term at position π plus one in the
sequence. So, remember, π π means the term
thatβs at position π in the sequence. So, π one means the first term in
the sequence, π two means the second term in the sequence, π three means the third
term in the sequence, and so on.

So, just before we move on, letβs
make a point about this bit defining the values of π. Itβs important that we link the
values of π to the sequence formula. So, we get π one, π two, π
three, π four, etcetera, and not π zero or π half or anything like that. So, for example, if I just said
that π is greater than or equal to zero, then putting that in here wouldβve given
us an π zero term. And thatβs not a proper term in our
sequence. Likewise, if Iβve said π is
greater than or equal to two, then the smallest value that we couldβve got here is
π two. And we wouldnβt have defined the
first term in our sequence π one, so that formula wouldnβt work properly.

Now, another way of defining that
recursive sequence for the sequence we were looking at is π π is equal to π π
minus one plus two. And the first time is two. Now, letβs think very carefully
about what values of π we have to define if we put our sequence in this format. Well, if π took the values one,
two, three, four, five, and so on, then we would have a case where we had π one, π
is equal to one is equal to π π minus one is equal to π zero plus two. And we canβt have a zeroth term
because our terms need to be the first, second, third, fourth, and so on. So, we need to start counting at
two. So, weβve defined the first term as
being two, and then weβve got the second term is equal to the first term plus two.

Now, with this sort of notation, if
you know the value of any term in the sequence, you can easily work out the value of
the next term. For example, if I had told you the
102nd term was 204, then you could just add two to it to get the value of the 103rd
term. If the 102nd term is 204, that
means that π 102 is 204. And following this pattern, π π
plus one is equal to π π plus two, that means that π 103 is equal to π 102 plus
two. Well, weβve just said that the
102nd term is 204, so we can replace that in our formula. So, the 103rd term is 204 plus two,
which is 206.

But with a little bit of
rearranging of the formula, you could also work out the value of the 101st term. Looking at the formula, if I
subtract two from both sides of that equation, I get π π plus one minus two is
equal to π π. And now, Iβve got a formula that
tells me the current term if I know what the next term is. The next term take away two gives
me the current term. Or the 102nd term take away two
gives me the 101st term, which means that 204 take away two is the 101st term, which
means the 101st term is 202.

This means that using our recursive
formula, we can take any term in the sequence and work out what the next term was
and what the previous term was. But the big downside to this way of
doing things is if I want to know the millionth term, then youβve got to do a lot of
work to find out its value. You need to work out the value of
the second term then apply the rule again to find the third term then the fourth and
fifth and so on.

The other way of specifying a
sequence is using a position to term rule. So, thinking back to our original
sequence, two, four, five, eight, 10, 12, and so on, if we write the position of
each number in the sequence above the actual number, you might notice that the value
of each term is equal to twice its position in the sequence. So, two times one is two, two times
two is four, two times three is six, two times four is eight, and so on. This means that the value of a term
in our sequence is equal to twice, so two times, the position in that sequence. Or π π is equal to two π in this
case.

Now, itβs worth just taking a
moment to remind you that this π here is a suffix or a subscript, a little π just
below π, whereas this π here is a proper full-sized π right up against the two,
which means two times π. So, just watch out for that. When I say π π, I donβt mean π
times π in this case. I mean π suffix π. Now, the great thing about this
formula is it takes you straight to the value of the term if you know its
position. So, the millionth term π a million
is just equal to two times a million, which is two million.

Now, just compare how long that
would have taken to work out the seventh term, the eighth, the ninth term, and so on
all the way up to the millionth term using the recursive formula. Anyway, back to the point of this
video, recursive sequences. Letβs try some questions.

Find the first three terms of the
sequence π π plus one is equal to two times π π plus five, where the first term
π one is 11, and π is greater than or equal one, and π is an integer.

Now, this formula tells that if we
take a term and we double it, and then we add five to that, we get the value of the
next term in the sequence. To calculate the second term, we
double the first term and add five. And the first term is 11. Thatβs what we weβre told in the
question. And that means the second term is
two times 11 plus five, which is 22 plus five, which is 27. And now, we can use that
information to work out the third term. The third term is equal to two
times the second term plus five. So, thatβs two times 27 plus five,
which is 54 plus five, which is 59. So, now, weβve got the first three
terms, 11, 27, and 59.

Next question then.

Find the first four terms in the
sequence π π plus one is equal to a half of π π minus four, where π two is 36,
and π is greater than or equal to one, and itβs an integer.

Now, this is a bit tricky because
weβve been told what the second term is, but we donβt know what the first term
is. So, weβve got to work that out. Now, the formula was telling us
that if we take a term and halve it and then subtract four, it gives us our next
term. So, for example, π two, the second
term, is equal to a half times π one, so a half of the first term, take away
four. But we were told that the value of
the second term was 36, so we know that 36 is equal to a half times π one minus
four. Now, I wanna work out what π one
is. So, if I add four to both sides of
that equation, I get 40 is equal to a half of π one, the first term. And then, if I double both sides of
the equation, I find that the first term must have been 80.

And we can apply our recursive
formula to work out what the third term is. Remember, to find the next term, we
have to halve this term and then subtract four. So, the third term is a half of the
second term take away four. And we know what the second term
is. Itβs 36. So, the third term is a half of 36
take away four. Well, half of 36 is 18. And 18 take away four is 14. And lastly, the fourth term is a
half of the third term take away four. And weβve just worked out that the
third term is 14, which means that the fourth term turns out to be three. So, to answer our question, the
first term is 80, the second is 36, the third term is 14, and the fourth term is three.

Letβs have a look at a slightly
more complicated recursive sequence then.

Find the first five terms in the
sequence π π is equal to two times π π minus one plus π π minus two, where π
one is three, and π two is five, and π is greater than or equal to three, and π
is an integer.

Now, this formula is telling us
that a term is some combination of the previous two terms. Itβs twice the immediately
preceding term, and then we have to add on the term before that. And to get this sort of formula
going, we need to provide two initial terms, the first and the second term, to
enable you to work out the third term. And now, π is greater than or
equal to three. If π was less than three, if π
started at two, then weβd have the second term is equal to the first term- two times
the first term plus the zeroth term. So, that wouldnβt work. So, this recursive sequence is only
set up for π is greater than or equal to three.

Well, the question asked us to work
out the first five terms. Weβve already been given the first
two, so letβs have a go at working out the third one. And as we said in the formula, to
work out a particular term, we double the previous term, π minus one, and then we
add on the term before that, π minus two. Letβs just plug in those values for
π one and π two then. So, π two is five and π one is
three. So, π three is two times five plus
three, which is 10 plus three, and thatβs 13.

Now, the fourth term is twice the
third term plus the second term. And thatβs two times 13 plus
five. Weβve just worked out the third
term. Well, two times 13 is 26. So, 26 plus five is 31. And then, finally, the fifth term
is two times the fourth term plus the third term. And plugging in the values of the
fourth and third term that weβve just worked out, we can see that the fifth term is
75. And we just need to write all those
out in full for our answer.

And thatβs a basic run through of
how to use recursive formulae. Now, some sequences are arithmetic
sequences. And that means that each
consecutive term has got a common difference. For example, three, seven, 11, 15,
19, and so on, I have to add four to each term to get the next term. That makes it an arithmetic
sequence. And 23, 21, 19, 17, 15, and so on,
Iβm having to subtract two from each term to get the next term. And thatβs also an arithmetic
sequence. And we can represent each of these
using a recursive formula. In the first case, one term is
equal to the previous term plus four. But donβt forget we have to tell it
where to start, and the π is greater than or equal to one in this case. And for the second sequence, one
term is equal to the previous term take away two. And again, we have to tell it where
to start and the relevant values of π.

Now, letβs just rearrange those
formulae. For the first one, π π plus one
equals π π plus four. Iβm gonna subtract π π from both
sides, which gives me π π plus one minus π π is equal to four. Well, π π plus one minus π π,
thatβs the difference between two consecutive terms. And this formula is telling me that
thatβs always constant; itβs always four in this case. And if I do the same with the other
formula, I find that in this case the difference between two consecutive terms is
also a constant negative two in this case.

So, if youβve got a recursive
formula which you can rearrange to get the difference between two consecutive terms,
and that answer just comes out to be a constant, then you know that youβre looking
at an arithmetic sequence. And thatβs just an algebraic way of
saying that, you know, every time I go from one term to the other, Iβm always adding
the same amount. Lastly then, letβs take a look at a
couple of sequences and see if we can write down a recursive formula for each
one.

Write down a recursive formula for
the sequence five, seven, nine, 11, 13, and so on.

Well, a good place to start is to
work out whatβs the difference between each term. To get from five to seven, I need
to add two. From seven to nine, add two. And to get between the next two
terms, I add two. And the next two terms again, I add
two again. And itβs a good idea to try to
describe whatβs going on before you try to put it into a formula. So, to get from one term to the
next, Iβm adding two every time. And in our formula then, if I take
a term, letβs call it π π, and I add two to that term, itβs going to give me the
next term, π π plus one.

Well, actually, its sort of as easy
as that. Thatβs the basic formula. But we do need to tell it where to
start and we need to define the values of π that are gonna work in this
formula. Well, our first term was five, so
π one is five. And we want to use values of π
that generate terms π one, π two, π three, π four, π five, and so on. Now, in our formula, weβve got π
π, and then weβve got π π plus one. So, we can have π has all the
values from one upwards. So, thatβs our formula. π π plus one equals π π plus
two, where π one is five, and π is greater than or equal to one and an
integer.

Now, I couldβve formulated that
slightly differently. So, I couldβve said the πth term,
π π, is simply equal to the previous term π π minus one plus two. And again, the first term π one is
equal to five. But letβs think about the values of
π. If I put π equal to one, Iβd be
saying π one, the first term, is equal to π one minus one, π zero. Iβd be talking about the zeroth
term plus two. And I donβt have a zeroth term, so
Iβm gonna define π from two onwards. That gives me an alternative
recursive formula.

Right then, one last tricky
question.

Write down a recursive formula for
the sequence three, nine, 21, 45, 93, and so on.

Now, again, good starting point is
to actually say well whatβs the difference between each term and the next term. So, to get from three to nine, I
need to add six. But to get from nine to 21, I need
to add 12. To get from 21 to 45, I need to add
24. And to get from 45 to 93, I need to
add 48. So, this isnβt an arithmetic
sequence. This is a bit of a trickier
prospect.

Now, if you look at each term, and
then you look at the differences, we can see that in the first case we start off
with three but weβre adding six. Then we start, then we start off
with nine, and weβre adding 12. We start off with 21 and weβre
adding 24. We start off with 45, weβre adding
48. These differences are always three
more than the previous term. Now, if we label our terms π one,
π two, π three, π four, π five, and so on, the second term is equal to the first
term plus six. And the third term is equal to the
second term plus 12. But we said that the difference is
three bigger than the term- the previous term itself. So, this difference is three bigger
than this.

So, six is π one plus three. And for working out a third term,
that 12, that difference there, is the second term plus three again. So, the third term is equal to the
second term plus the second term plus three. And thatβs gonna be the case
generally. The π plus oneth term is equal to
the πth term plus the πth term plus three. And given that this is just three
values added together β I donβt need those parentheses β that gives me π π plus π
π plus three. Thatβs two lots of π π plus
three. So, π π plus one is equal to two
π π plus three.

And now, we need to think about the
starting conditions. Well, the first term π one is
equal to three. And to generate terms π one, π
two, π three, π four, and so on, I need to put π equal to one, two, three, and so
on. So, thatβs it; thereβs my
formula. Now, I could adjust this slightly
as well like I did the last time. And that will give me the πth term
is equal to two times the π minus oneth term plus three. And again, Iβd have to adjust the
starting point of π, π is greater than or equal two, so that I donβt end up with
the term π zero.