In this video, we're going to be
looking at sequences and series,
so let's begin by looking at
what a sequences.
This, for instance is a
sequence. It's a
set of numbers.
And here we seem to have a rule.
All of these are odd numbers, or
we can look at it. We increase
by two each time 13579.
So there's our sequence of
odd numbers.
Here is another sequence.
These numbers
are the
square numbers.
1 squared, 2 squared, is 4 three
squared is 9. Four squared is 16
and 5 squared is 25. So again
we've got a sequence of numbers.
We've got a rule that seems to
produce them. Those are the.
Square numbers. Is a
slightly different sequence.
Here we've got alternation
between one and minus one back
to one on again to minus one
back to one on again to minus
one. But this is still a
sequence of numbers.
Now, because I've written some
dots after it here.
This means that this is meant to
be an infinite sequence. It goes
on forever and this is meant to
be an infinite sequence. It
carries on forever, and this one
does too. If I want a finite
sequence. What might
a finite sequence
look like, for
instance 1359? That would
be a finite sequence. We've got
4 numbers and then it stops
dead. Perhaps if we look
at the sequence of square
numbers 1, four, 916, that again
is a finite sequence.
Sequence that will be very
interested in is the sequence
of whole numbers, the counting
numbers, the integers. So
there's a sequence of integers,
and it's finite because it
stops at N, so we're
counting 123456789 up to N.
And the length of this sequence
is an the integer, the number N.
Very popular sequence of
numbers. Quite well known is
this particular sequence.
This is a slightly different
sequence. It's infinite keeps on
going and it's called the
Fibonacci sequence. And we can
see how it's generated. This
number 2 is formed by adding the
one and the one together, and
then the three is formed by
adding the one and the two
together. The Five is formed by
adding the two and the three
together, and so on.
So there's a question here. How
could we write this rule down in
general, where we can say it
that any particular term is
generated by adding the two
numbers that come before it in
the sequence together? But how
might we set that down? How
might we label it? One way might
be to use algebra and say will
call the first term you, and
because it's the first time we
want to label it, so we call it
you one. And then the next
terminal sequence, the second
term. It would make sense to
call it you too, and the third
term in our sequence. It would
make sense therefore, to call it
you 3. You four and so on
up to UN. So this represents a
finite sequence that's got N
terms in it. If we look at
the Fibonacci sequence as an
example of making use of this
kind of notation, we could say
that the end term UN was
generated by adding together the
two terms that come immediately
before it will.
Term that comes immediately
before this must have a number
attached to it. That's one less
than N and that would be N minus
one. Plus on the term that's
down, the term that comes
immediately before this one must
have a number attached to it.
That's one less than that. Well,
that's UN minus 1 - 1 taking
away 2 ones were taking away two
altogether. So that we can see
how we might use the algebra
this algebraic notation help us
write down a rule for the Fibo
Nachi sequence. OK, how can
we? Use this in a
slightly different way.
What we need to look at now
is to move on and have a
look what we mean by a series.
This is a
sequence, label it.
A sequence it's a list of
numbers generated by some
particular rule. It's finite
because there are any of them.
What then, is a series series is
what we get.
When we add.
Terms of the sequence.
Together And because
we're adding together and terms
will call this SN.
The sum of N terms
and it's that which is
the series.
So. Let's have a
look at the sequence
of numbers 123456, and
so on up to
N.
Then S1.
Is just one.
S2.
Is the sum of the
first 2 terms 1 +
2? And that gives us 3.
S3. Is
the sum of the first three
terms 1 + 2 + 3?
And that gives us 6
and S4. Is the sum
of the first four terms 1 +
2 + 3 + 4 and that
gives us. Hey.
So this gives us the basic
vocabulary to be able to move on
to the next section of the
video, but just let's remind
ourselves first of all.
A sequence.
Is a set of numbers
generated by some rule.
A series is what we get when we
add the terms of the sequence
together. This particular
sequence has N terms in it
because we've labeled each term
in the sequence with accounting
number. If you like U1U2U free,
you fall you N.
Now. With this vocabulary of
sequences and series in mind,
we're going to go on and have a
look at a 2 special kinds of
sequences. The first one is
called an arithmetic progression
and the second one is called a
geometric progression. Will
begin with an arithmetic
progression. Let's start by
having a look at this
sequence of. Odd
Numbers that we
had before 1357.
Is another sequence
not 1020 thirty,
and so on.
What we can see in this first
sequence is that each term after
the first one is formed by
adding on to.
1 + 2 gives us 3.
3 + 2 gives us 5.
5 + 2 gives us 7 and it's
because we're adding on the
same amount every time. This
is an example of what we call
an arithmetic.
Progression.
If we look at this sequence of
numbers, we can see exactly the
same property we've started with
zero. We've added on 10, and
we've added on 10 again to get
20. We've had it on 10 again to
get 30, so again, this is
exactly the same. It's an
arithmetic progression. We don't
have to add on things, so
for instance a sequence of
numbers that went like this 8
five, 2 - 1.
Minus
4. If we look what's
happening where going from 8:00
to 5:00, so that's takeaway
three were going from five to
two, so that's takeaway. Three
were going from 2 to minus. One
takeaway. Three were going from
minus one to minus four takeaway
3. Another way of thinking about
takeaway three is to say where
adding on minus three.
8 at minus three is 5 five at
minus three is 2, two AD minus
three is minus one, so again,
this is an example of an
arithmetic progression. And what
we want to be able to do is to
try and encapsulate this
arithmetic progression in some
algebra, so we'll use the letter
A. To stand for
the first term.
And will use the letter
D to stand for the
common difference. Now the
common difference is the
difference between each term and
it's called common because it is
common to each between each
term. So let's have a look at
one 357 and let's have a think
about how it's structured 13.
5. 7 and so
on. So we begin with one and
then the three is 1 + 2.
The Five is 1 + 2 tools because
by the time we got to five,
we've added four onto the one.
The Seven is one plus. Now the
time we've got to Seven, we've
added three tools on. Let's just
do one more. Let's put nine in
there and that would be 1 + 4
times by two.
So let's see if we can begin to
write this down. This is one.
Now what have we got here? This
is the second term in the
series. But we've only got 1 two
there, so if you like we've got
1 + 2 - 1 times by two.
One plus now, what's multiplying
the two here? Well, this is the
third term in the series.
So we've got a 2 here,
so we're multiplying by 3 -
1. Here this is term
#4 and we're
multiplying by three,
so that's 4 - 1 times
by two. And here this
is term #5, so we've
got 1 + 5 - 1
times by two.
Now, if we think about
what's happening here.
We're starting with A.
And then on to the A. We're
adding D. Then we're adding on
another day, so that's a plus
2D, and then we're adding on
another D. So that's a plus
3D. The question is, if we've
got N terms in our sequence,
then what's the last term? But
if we look, we can see that the
first term was just a.
The second term was a plus, one
D. The third term was a plus
2D. The fourth term was a plus
3D, so the end term must be
a plus N minus one.
Gay.
Now, this last term of
our sequence, we often label
L and call it the
last term.
Or
The end.
Turn. To be more mathematical
about it. And one of the things
that we'd like to be able to do
with a sequence of numbers like
this is get to a series. In
other words, to be able to add
them up. So let's have a look at
that. So SN the some of these
end terms is A plus A+B plus
A plus 2B plus. But I want
just to stop there and what I
want to do is I want to
start at the end. This end
now now the last one will be
plus L.
So what will be the next one
back when we generate each term
by adding on D. So we added on D
to this one to get L. So this
one's got to be L minus D.
And the one before that
one similarly will be L
minus 2D. On the rest of the
terms will be in between.
Now I'm going to use a trick.
Mathematicians often use. I'm
going to write this down the
other way around. So I have L
there. Plus L minus
D. Plus L minus 2D plus
plus. Now what will I have?
Well, writing this down either
way around, I'll Have A at the
end. Then I'll have this next
term a. Plus D.
And I'll have this next
term, A plus 2D.
Now I'm going to add these two
together. Let's look what
happens if I add SN&SN together.
I've just got two of them.
By ad A&L together I get a
plus L let me just group
those together.
Now I've got a plus D&L Minus D,
so if I add them together I have
a plus L Plus D minus D, so
all I've got left is A plus L.
But the same thing is going to
happen here. I have a plus L
Plus 2D Takeaway 2D, so again
just a plus L.
When we get down To this end,
it's still the same thing
happening. I've A plus L
takeaway 2D add onto D so again
the DS have disappeared. If you
like and I've got L plus A.
Plus a plusle takeaway D add
on DLA and right at the
end. L plus a again.
Well, how many of these have I
got? But I've got N terms.
In each of these lines of sums,
so I must still have end terms
here, and so this must be an
times a plus L.
And so if we now divide
both sides by two, we have.
SN is 1/2 of N times
by a plus L and that
gives us our some of the
terms of an arithmetic
progression. Let's just write
down again the two results that
we've got. We've got L the
end term, or the final term
is equal to a plus N
minus one times by D and
we've got the SN is 1/2.
Times by N number of
terms times by a plus
L.
Now, one thing we can do is take
this expression for L and
substitute it into here.
Replacing this al, so let's do
that. SN is equal to 1/2.
Times by N number of terms
times by a plus and instead
of L will write this a
plus N minus one times by
D. April say gives us
two way.
So the sum of the
end terms is 1/2 an
2A plus N minus 1D.
Close the bracket.
And these. That I'm
underlining are the three
important things about an
arithmetic progression.
If A is the first
term. And D is
the common difference.
And N
is the
number of
terms.
In our arithmetic progression,
then, this expression gives us
the NTH or the last term.
This expression gives us the
some of those N terms, and
this expression gives us also
the sum of the end terms.
One of the things that you also
need to understand is that
sometimes we like to shorten the
language as well as using
algebra. So that rather than
keep saying arithmetic
progression, we often refer to
these as a peas.
Now we've got some facts, some
information there. So let's have
a look at trying to see if we
can use them to solve some
questions. So let's have
a look at this
sequence of numbers again,
which we've identified.
And let's ask ourselves
what's the sum?
Of. The first
50 terms So
we could start to try and add
them up. 1 + 3 is four and four
and five is 9, and nine and
Seven is 16 and 16 and 9025, and
then the next get or getting
rather complicated. But we can
write down some facts about this
straight away. We can write down
that the first term is one.
We can write down that the
common difference Dean is 2 and
we can write down the number of
terms we're dealing with. An is
50. We know we have a
formula that says SN is 1/2
times the number of terms.
Times 2A plus N minus 1D. So
instead of having to add this up
as though it was a big
arithmetic sum a big problem, we
can simply substitute the
numbers into the formula. So SNS
50 in this case is equal to 1/2.
Times by 50.
Times by two A That's just two
2 * 1 plus N minus one
and is 50, so N minus one
is 49. Times by the common
difference too.
So. We
can cancel a 2 into the 50
that gives us 25 times by now.
2 * 49 or 2 * 49
is 98 and two is 100, so
we have 25 times by 100, so
that's 2500. So what was going
to be quite a lengthy and
difficult calculation's come out
quite quickly. Let's see if we
can solve a more difficult
problem.
1.
Plus 3.5.
+6.
Plus 8.5. Plus
Plus 101.
Add this up.
Well. Can we identify what
kind of a series this is? We can
see quite clearly that one to
3.5 while that's a gap of 2.5
and then a gap of 2.5 to 6. So
what we've got here is in fact
an arithmetic progression, and
we can see here. We've got 100
and one at the end. Our last
term is 101 and the first term
is one. Now we know a formula.
For the last term L.
Equals A plus N minus
one times by D.
Might just have a look at what
we know in this formula. What we
know L it's 101.
We know a It's the first
term, it's one.
Plus Well, we have no idea what
any is. We don't know how many
terms we've got, so that's N
minus one times by D and we know
what that is, that's 2.5.
Well, this is nothing more than
an equation for an, so let's
begin by taking one from each
side. That gives us 100 equals N
minus one times by 2.5. And now
I'm going to divide both sides
by 2.5 and that will give me 40
equals N minus one, and now I'll
add 1 to both sides and so 41 is
equal to end, so I know how many
terms that. Are in this series,
So what I can do now is I
can add it up because the sum of
N terms is 1/2.
NA plus
L. And I
now know all these terms
here have 1/2 * 41
* 1. Plus
101. Let me just turn
the page over and write this
some down again.
SN is equal
to 1/2 *
41 * 1
+ 101.
So we have 1/2 times by 41
times by 102 and we can cancel
it to there to give US 41
times by 51. And to do that
I'd want to get out my
Calculator, but we'll leave it
there to be finished.
So that's one kind of problem.
Let's have a look at another
kind of problem.
Let's say we've got an
arithmetic progression whose
first term is 3.
And the sum.
Of.
The first 8.
Terms.
Is twice.
The sum
of the
first 5
terms.
And that seems really quite
complicated. But it needn't
be, but remember this is the
same arithmetic progression.
So let's have a think what this
is telling us A is equal to
three and the sum of the first 8
terms. Well, to begin with,
let's write down what the sum of
the first 8 terms is.
Well, it's a half.
Times N Times
2A plus and
minus 1D.
And N is equal to 8.
So we've got a half.
Times 8. 2A
plus N minus one is
7D. So S 8
is equal to half of
eight is 4 * 2
A Plus 7D.
But we also know that a
is equal to three, so we
can put that in there as
well. That's 4 * 6 because
a is 3 + 7 D.
Next one, the sum
of the first 5
terms. Let me just write
down some of the first
8 terms were.
4. Times
6 minus
plus 7D
first 5
terms. Half times the number of
terms. That's 5 * 2 A plus
N minus one times by D will.
That must be 4 because any is
5 times by D.
So much is 5 over 2 and
let's remember that a is equal
to three, so that 6 + 4
D. So I've got S 8 and
I've got S5 and the question
said that S8 was equal to twice
as five. So I can write this
for S8. Is
equal to
twice. This which is
S five 2 * 5 over two
6 + 4 D and what seemed
a very difficult question as
reduced itself to an ordinary
linear equation in terms of D.
So we can do some cancelling
there and we can multiply out
the brackets for six is a 24
+ 28, D is equal to 56R.
30 + 5 fours
are 20D. I can take
20D from each side that gives me
8 D there.
And I can take 24 from each
side, giving me six there. So D
is equal to.
Dividing both sides by 8,
six over 8 or 3/4 so I know
everything now that I could
possibly want to know about
this arithmetic progression.
Now let's go on and have a look
at our second type of special
sequence, a geometric
progression. So.
Take these
two six
1854. Let's have a look
at how this sequence of numbers
is growing. We have two. Then we
have 6. And then we have
18. Well 326 and three sixes
are 18 and three eighteens are
54. So this sequence is growing
by multiplying by three each
time. What about this sequence
one? Minus
2 four. Minus
8. What's happening here? We can
see the signs are alternating,
but let's just look at the
numbers. 1 * 2 would be two 2 *
2 would be four. 2 * 4 would be
8. But if we made that minus
two, then one times minus two
would be minus 2 - 2 times minus
two would be plus 4 + 4 times by
minus two would be minus 8, so
this sequence to be generated is
being multiplied by minus two.
Each term is multiplied by minus
two to give the next term.
These are examples of geometric
progressions, or if you like,
GPS. Let's try and write one
down in general using some
algebra. So like the AP, we take
A to be the first term.
Now we need something like D.
The common difference, but what
we use is the letter R and we
call it the common ratio, and
that's the number that does the
multiplying of each term to give
the next term.
So 3 times by two gives us 6,
so that's the R. In this case
the three. So we do a Times by
R. And then we multiply by, in
this case by three again 3 times
by 6 gives 18, so we multiply by
R again, AR squared.
And then we multiply by three
again to give us the 54.
So by our again AR cubed.
And what's our end term in this
case? While A is the first term
8 times by R, is the second term
8 times by R-squared is the
third term 8 times by R cubed?
Is the fourth term, so it's a
times by R to the N minus one.
Because this power there's a
one. There is always one less.
And the number of the term,
then its position in the
sequence. And this is the end
term, so it's a Times my R to
the N minus one.
What about adding up a
geometric progression? Let's
write that down. SN is
equal to a plus R
Plus R-squared Plus.
Plus AR to the N minus one,
and that's the sum of N terms.
Going to use another trick
similar but not the same to what
we did with arithmetic
progressions. What I'm going to
do is I'm going to multiply
everything by the common ratio.
So I've multiplied SN by are
going to multiply this one by R,
but I'm not going to write the
answer there. I'm going to write
it here so I've a Times by R and
I've written it there plus now I
multiply this one by R and that
would give me a R-squared. I'm
going to write it there.
So that term is being multiplied
by R and it's gone to their
that's being multiplied by R and
it's gone to their. This one
will be multiplied by R and it
will be a R cubed and it will
have gone to their.
Plus etc plus, and we think
about what's happening.
That term will come to here and
it will look just like that one.
Plus and then we need to
multiply this by R, and that's
another. Are that we're
multiplying by, so that means
that becomes AR to the N.
Now look at why I've lined these
up AR, AR, AR squared. Our
squared, al, cubed, cubed and so
on. So let's take these two
lines of algebra away from each
other, so I'll have SN minus R
times by SN is equal to. Now
have nothing here to take away
from a, so the a stays as it is.
Then I've AR takeaway are, well,
that's nothing. A R-squared
takeaway R-squared? That's
nothing again, same there. And
so on and so on. AR to the N
minus one. Take away a art. The
end minus one nothing and then
at the end I have nothing there
take away.
AR to the N.
Now I need to look closely at
both sides of what I've got
written down, and I'm going to
turn this over and write it down
again. So we've SN minus
RSN is equal to A.
Minus AR to the N.
Now here I've got a common
factor SN the some of the end
terms when I take that out, I've
won their minus R of them there,
so I get SN times by one minus R
is equal 2 and here I've got a
common Factor A and I can take a
out giving me one minus R to the
N. Remember it was the sum of N
terms that I wanted so.
SN is equal to a Times 1 minus R
to the N and to get the SN on
its own, I've had to divide by
one minus R, so I must divide
this by one minus R.
And that's my formula for the
sum of N terms of a geometric
progression. And let's just
remind ourselves what the
symbols are N is equal to the
number of terms.
A is the first
term of our
geometric
progression and are
we said was called
the common ratio.
OK, and let's just remember the
NTH term in the sequence was AR
to the N minus one. So those
are our fax so far about GPS
or geometric progressions. Let's
see if we can use these facts
in order to be able to help
us solve some problems and do
some questions. So first of all,
let's take this 2 + 6 +
18 + 54 plus. Let's say there
are six terms. What's the answer
when it comes to adding those
up? Well, we know that a
is equal to two. We know that
our is equal to three and we
know that N is equal to six. So
to solve that, all we need to do
is write down that the sum of N
terms is a Times 1 minus R to
the N all over 1 minus R.
Substitute our numbers in two
times. 1 - 3
to the power 6.
Over 1 -
3. So this is 2 * 1 -
3 to the power six over minus
two, and we can cancel a minus
two with the two that we leave
as with a minus one there and
one there if I multiply
throughout by the minus one,
I'll have minus 1 * 1 is minus
one and minus one times minus 3
to the six is 3 to the 6th, so
the sum of N terms is 3 to the
power 6. Minus one and with a
Calculator we could workout what
3 to the power 6 - 1
was. Let's take
another. Question to do with
summing the terms of a geometric
progression. What's the sum
of that? Let's say for five
terms. While we can begin by
identifying the first term,
that's eight, and what's the
common ratio?
Well, to go from 8 to 4 as a
number we would have it, but
there's a minus sign in there.
So that suggests that the common
ratio is minus 1/2. Let's just
check it minus four times. By
minus 1/2 is plus 2 + 2 times Y
minus 1/2 is minus one, and we
said five terms, so Ann is equal
to 5. So we can write
down our formula. SN is equal to
a Times 1 minus R to the
power N all over 1 minus R.
And so A is 8.
1 minus and this is
minus 1/2 to the power
5. All over 1 minus minus
1/2. You can see these
questions get quite
complicated with the
arithmetic, so you have to
be very careful and you
have to have a good
knowledge of fractions.
This is 8 * 1. Now let's have
a look at minus 1/2 to the power
5. Well, I'm multiplying the
minus sign by itself five times,
which would give me a negative
number, and I've got a minus
sign there outside the bracket.
That's going to mean I've got 6
minus signs together. Makes it
plus. So now I can look at the
half to the power 5.
Well, that's going to be one
over. 248-1632 to
to the power
five is 32.
All over 1 minus minus 1/2.
That's 1 + 1/2. Let's write
that as three over 2.
So this is equal to.
Now I've got 8.
Times by one plus, one over 32,
and I'm dividing by three over 2
to divide by a fraction. We
invert the fraction that's two
over 3 and we multiply by and we
just turn the page to finish
this one off.
So we have SN is
equal to 8 * 1
+ 1 over 32 times
by 2/3 is equal to
8 times by now one
and 132nd. Well, there are
3230 seconds in one, so
altogether there I've got 33.
30 seconds times
by 2/3. And we
can do some canceling threes
into 30. Three will go 11 and
threes into three. There goes
one. Twos into two goes one and
tools into 32, goes 16 and 18
two eight goes one and eight
into 16 goes 2. So we 1 *
11 * 1 that's just 11 over 2
because we've 2 times by one
there. So we love Nova two or we
prefer five and a half.
So that we've got the some of
those five terms of that
particular GP. Five and a half,
11 over 2 or 5.5.
But here's a different question.
What if we've got the sequence
248? 128 how many terms are
we got? How many bits do we
need to get from 2 up to
128? Well, let's begin by
identifying the first term
that's two. This is.
A geometric progression because
we multiply by two to get each
term. So the common ratio are is
2 and what we don't know is
what's N. So let's have a look.
This is the last term and we
know our expression for the last
term. 128 is equal to AR to the
N minus one.
So let's substituting some
of our information.
A is 2 times by two
4R to the N minus one.
Well, we can divide both
sides by this two here,
which will give us.
64 is equal to two to the
N minus one.
I think about that it's 248
sixteen 3264 so I had to
multiply 2 by itself six times
in order to get 64, so 2
to the power 6, which is 64
is equal to 2 to the power
N minus one, so six is equal
to N minus one, and so N
is equal to 7, adding one.
To each side. In other words,
there were Seven terms in our.
Geometric progression. Type
of question that's often given
for geometric progressions is
given a geometric progression.
How many terms do you need
to add together before you
exceed a certain limit? So, for
instance, here's a geometric.
Progression. How many times of
this geometric progression do we
need to act together in order to
be sure that the some of them
will get over 20?
Well, first of all, let's try
and identify this as a geometric
progression. The first term is
on and it looks like what's
doing the multiplying. The
common ratio is 1.1. Let's just
check that here.
1.1 times by one point, one
well. That's kind of like 11 *
11 is 121.
With two numbers after the
decimal point in one point 1 *
1.1 and with two numbers after
the decimal point there. So yes,
this is a geometric progression.
So let's write down our formula
for N terms sum of N terms
is equal to a Times 1 minus
R to the N.
All over 1 minus R. We want to
know what value of N is just
going to take us over 20.
So let's substituting some
numbers. This is one for
a 1 - 1.1 to
the N. All over
1 - 1.1 that
has to be greater
than 20.
So one times by that isn't going
to affect what's in the
brackets. That would be 1 - 1.1
to the N all over 1 - 1.1
is minus nought. .1 that has to
be greater than 20.
Now if I use the minus sign
wisely. In other words, If I
divide if you like.
Minus note .1 into there as
a. Division, then I'll have.
The minus sign will make that a
minus and make that a plus, so
I'll have one point 1 to the N
minus one and divided by North
Point one is exactly the same as
multiplying by 10. That means
I've got a 10 here.
That I can divide both sides by.
So let's just write this down
again 1.1 to the N minus one
times by 10 has to be greater
than 20. So let's divide both
sides by 10, one point 1 to the
N minus one has to be greater
than two and will add the one to
both sides 1.1 to the end has to
be greater than three.
Problem how do we find N? One of
the ways of solving equations
like this is to take logarithms
of both sides, so I'm going to
take natural logarithms of both
sides. I'm going to do it to
this site first. That's the
natural logarithm of 3 N about
this side. When you're taking a
log of a number that's raised to
the power, that's the equivalent
of multiplying the log of that
number. By the power that's N
times the log of 1.1. Well
now this is just an equation
for N because N has got to
be greater than the log of 3
divided by the log of 1.1
because after all.
Log of three is just a number
and log of 1.1 is just a number
and this is the sort of
calculation that really does
have to be done on a Calculator.
So if we take our Calculator and
we turn it on.
And we do the calculation. The
natural log of three.
Divided by the natural log of
1.1, we ask our Calculator to
calculate that for us. It tells
us that it's 11.5 to 6 and some
more decimal places. We're not
really worried about these
decimal places. An is a whole
number and it has to be greater
than 11 and some bits, so N has
got to be 12 or more.
That's one last twist
to our geometric progression.
Let's have a look at
this one. What have
we got got
a geometric progression.
First term a
is one.
Common ratio is 1/2 because
we're multiplying by 1/2 each
time. That write down
some sums. S1, the sum
of the first term is just.
1. What's
S2? That's the
sum of the first 2 terms, so
that's. Three over
2.
What's the sum of the first
three terms? That's one.
Plus 1/2 +
1/4. Add those up
in terms of how many quarters
have we got then that is
7. Quarters As
for the sum of
the first.
4. Terms.
Add those up in terms of how
many eighths if we got so we've
got eight of them there. Four of
them there. That's 12. Two of
them there. That's 14 and one of
them there. That's 15 eighths.
Seems to be some sort of
pattern here.
Here we seem to be 1/2 short of
two. Here we
seem to
be 1/4. Short of two
here, we seem to be an eighth
short of two and we look at the
first one. Then we're clearly 1
short of two.
He's a powers of two. Let's have
a look 2 - 2 to the
power zero, 'cause 2 to the
power zero is 1 two.
Minus.
2 to the power minus one 2
- 2 to the power minus two
2 - 2 to the power minus
three. But each of these is
getting smaller. We're getting
nearer and nearer to two. The
next one we take away will be a
16th, the one after that will be
a 32nd and the next bit we take
off 2 is going to be a 64th and
then a 128 and then at one
256th. So we're getting the bits
were taking away from two are
getting smaller and smaller and
smaller until eventually we
wouldn't be able to distinguish
them from zero.
And so if we could Add all of
these up forever, a sum to
Infinity, if you like the
answer, or to be 2 or as near as
we want to be to two. So let's
see if we can have a look at
that with some algebra.
We know that the sum to end
terms is equal to a Times 1
minus R to the N all over
1 minus R.
What we want to have a look at
is this thing are because what
was crucial about this?
Geometric progression was at the
common ratio was a half a
number less than one.
So let's have a look what
happens. When all is bigger than
one to R to the power N.
We are is bigger than one and we
keep multiplying it by itself.
Grows, it grows very rapidly and
really gets very big very
quickly. Check it with two, 2,
four, 816. It goes off til
Infinity. And because it goes
off to Infinity, it takes the
sum with it as well.
What about if our is equal
to 1? Well, we can't really
use this formula then because we
would be dividing by zero. But
if you think about it, are
equals 1 means every term is the
same. So if we start off with
one every term is the same 1111
and you just add them all up.
But again that means the sum is
going to go off to Infinity if
you take the number any number
and add it to itself.
An infinite number of times
you're going to get a very, very
big number. What happens if
our is less than minus one?
Something like minus 2?
Well, what's going to happen
then to R to the N?
Well, it's going to be plus an.
It's going to be minus as we
multiply by this number such as
minus two. So we have minus 2 +
4 minus A. The thing to notice
is it's getting bigger, it's
getting bigger each time. So
again are to the end is going to
go off to Infinity. It's going
to oscillate between plus
Infinity and minus Infinity, but
it's going to get very big and
that means this sum is also
going to get.
Very big.
What about our equals minus one?
Well, if R equals minus one,
let's think about a sequence
like that. Well, a typical
sequence might be 1 - 1.
1 - 1 and we can see the
problem. It depends where we
stop. If I stop here the sum is
0 but if I put another one
there, the sum is one. So we've
got an infinite number of terms
then. Well, it depends on money
I've got us to what the answer
is so there isn't a limit for
SN. There isn't a thing that it
can come to a definite number.
Let's have a look. We've
considered all possible values
of our except those where are is
between plus and minus one.
Let's take our equals 1/2
as an example.
Or half trans by half is 1/4.
Reply by 1/2. Again
that's an eighth.
Multiply by 1/2 again, that's a
16. Multiplied by 1/2 again,
that's a 32nd.
By half again that's a 64th by
1/2 again, that's 128.
It's getting smaller, and if we
do it enough times then it's
going to head off till 0.
What about a negative one? You
might say, let's think about
minus 1/2. Now multiplied by
minus 1/2, it's a quarter.
Multiply the quarter by minus
1/2. It's minus an eighth.
Multiply again by minus 1/2.
Well, that's plus a 16th.
Multiply again by minus 1/2.
That's minus a 32nd, so we're
approaching 0, but where dotting
about either side of 0 plus them
were minus, then were plus then
where mine is.
We're getting nearer to zero
each time, so again are to the
power. N is going off to zero.
What does that mean? It means
that this some. Here we can have
what we call a sum to Infinity.
Sometimes it's just written with
an S and sometimes it's got a
little Infinity sign on it.
What that tells us? Because this
art of the end is going off to
0 then it's a times by one over
1 minus R and that's our sum to
Infinity. In other words, we can
add up. An infinite number of
terms for a geometric
progression provided. The common
ratio is between one and minus
one, so let's have a look
at an example. Supposing we've
got this row.
Metric progression.
Well, first term is one
now a common ratio is
1/3.
And what does this come to when
we add up? As many terms as
we can, what's the sum to
Infinity? We know the formula
that's a over 1 minus R, so
let's put the numbers in this
one for a over 1 - 1/3.
So the one on tops OK and the
one minus third. Well that's
2/3, and if we're dividing by a
fraction then we invert it and
multiply. So altogether that
would come to three over 2, so
it's very easy formula to use.
Finally, just let's recap for a
geometric progression. A.
Is the first term.
Aw.
Is the common.
Ratio. So a
geometric progression looks like
AARA, R-squared, AR, cubed and
the N Terminus series AR
to the N minus one.
And if we want to add up this
sequence of numbers SN.
Then that's a Times 1 minus R to
the power N or over 1 minus R.
And if we're lucky enough to
have our between plus and
minus one, sometimes that's
written as the modulus of art
is less than one. If we're
lucky to have this condition,
then we can get a sum to
Infinity, which is a over 1
minus R.