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.