
A binomial expression is the
sum or difference of two

terms. So for example 2X
plus three Y.

Is an example of a binomial

expression. Because it's the sum
of the term 2X and the term 3 Y

is the sum of these two terms.

Some of the terms could be just
numbers, so for example X plus

one. Is the sum of the term X
and the term one, so that's two

is a binomial expression.

AB is the difference of
the two terms A&B, so

that too is a binomial

expression. Now in your previous
work you have seen many binomial

expressions and you have raised
them to different powers. So you

have squared them, cube them and

so on. You probably already be
very familiar with working with

the binomial expression like X
Plus One and squaring it.

And you have done that by
remembering that when we want to

square a bracket when
multiplying the bracket by

itself. So X Plus One squared is
X Plus One multiplied by X plus

one. And we remove the brackets
by multiplying all the terms in

the first bracket by all the
terms in the SEC bracket, so

they'll be an X multiplied by X.

Which is X squared.

X multiplied by one which
is just X.

1 multiplied by X, which is

another X. And 1 * 1, which
is just one.

So to tide you all that up X
Plus One squared is equal to X

squared. As an X plus another X
which is 2 X.

Plus the one at the end.

Note in particular that we
have two X here and that came

from this X here and another X
there. I'll come back to that

point later on and will see
why that's important.

Now suppose we want to raise a
binomial expression to our power

that's higher than two. So
suppose we want to cube it,

raise it to the power four or
five or even 32.

The process of removing the
brackets by multiplying term by

term over and over again is very
very cumbersome. I mean, if we

wanted to workout X plus one to
the Seven, you wouldn't really

want to multiply a pair of
brackets by itself several

times. So what we want is a
better way. Better way of doing

that. And one way of doing it
is by means of a triangle of

numbers, which is called
Pascals Triangle.

Pascal was a 17th century French
mathematician and he derived

this triangle of numbers that
will repeat for ourselves now,

and this is how we form the
triangle. We start by writing

down the number one.

Then we form a new row and on
this nuro we have a one.

And another one.

We're going to build up a
triangle like this and each nuro

that we write down will start

with a one. And will end with a

one. So my third row is
going to begin with the

one and end with a one.

And in a few minutes,
we'll write a number

in there in the gap.

The next row will begin with a
one and end with a one and will

write a number in there and a

number in there. And in this way
we can build a triangle of

numbers and we can build it as
big as we want to.

How do we find this number in

here? Well, the number that
goes in here we find by

looking on the row above.

And looking above to the left
and above to the right.

And adding what we find, there's
a one here. There's a one there.

We add them one and one gives 2
and we write the result in

there. So there's two.

On the 3rd row has come by
adding that one and that

one together.

Let's look at the next row down
the number that's going to go in

here. Is found by looking
on the previous row.

And we look above left which
gives us the one we look above

to the right, which gives us
two, and we add the numbers

that we find, so we're adding
a one and two which is 3 and

we write that in there.

What about the number here?

Well again previous row above
to the left is 2 above to the

right is one. We add what we
find 2 plus one is 3 and that

goes in there.

And we can carry on building
this triangle as big as we want

to. Let's just do one more row.

We start the row.

With a one and we finished with
a one and we put some numbers in

here and in here and in here.

The number that's going
to go in here.

Is found from the previous row
by adding the one and the three.

So 1 + 3.

Is 4 let me write that in there.

The number that's going to go in

here. Well, we look in the
previous row above left and

above right. 3 + 3 is 6 and we
write that in there.

And finally 3 Plus One is 4 when
we write that in there. So

that's another row.

And what you should do now is
practice generating additional

rows for yourself, and
altogether this triangle of

numbers is called pascals.

Pascal's triangle.

OK.
Now we're going to use this

triangle to expand binomial
expressions and will see that it

can make life very easy for us.

We'll start by
going back to

the expression A+B.

To the power 2.

So we have binomial binomial
expression here, a I'd be and

we're raising it to the power 2.

Let's do it the old way. First
of all by multiplying A&B by

itself. Because we're squaring

A&B. Let's write down
what will get.

A multiplied by a
gives us a squared.

A multiplied by B will give us
a Times B or just a B.

Be multiplied by a.
Gives us a BA.

And finally, be multiplied by B.
Give us a B squared.

And if we just tidy it, what we
found, there's a squared.

There's an AB.

And because BA is the same as a
bee, there's another a be here.

So altogether there's two lots
of a B.

And finally, AB squared
at the end.

Now that's the sort of
expansion. This sort of removing

brackets that you've seen many
times before. You were already

very familiar with, but what I
want to do is make some

observations about this result.

When we expanded A+B to the
power two, what we find is that

as we successively move through
these terms that we've written

down the power of a decreases,
it starts off here with an A

squared. The highest power being
two corresponding to the power

in the original binomial
expression, and then every

subsequent term that power
drops. So it was 8 to the power

2. There's A to the power one
in here, although we don't

normally right the one in and
then know as at all, so the

powers of a decrease as we
move from left to right.

What about bees?

There's no bees in here.
There's a beta. The one in

there, although we just
normally right B&AB to the

power two there. So as we move
from left to right, the powers

of B increase until we reach
the highest power B squared,

and the squared corresponds to
the two in the original

problem.

What else can we observe if we
look at the coefficients of

these terms now the coefficients
are the numbers in front of each

of these terms. Well, there's a
one in here, although we

wouldn't normally write it in,
there's a two there, and there's

a one inference of the B
squared, although we wouldn't

normally write it in.

So the coefficients
are 1, two and one.

Now let me remind you again
about pascals triangle. Have a

copy of the triangle here so we
can refer to it. If we look at

pascals triangle here will see
that one 2 one is the numbers

that's in the 3rd row of pascals
triangle. 121 other numbers that

occur in the expansion of A+B to
the power 2.

There's something else I want to
point out that this 2A. B in

here. Came from a term here 1A B
on one BA in there and together

the one plus the one gave the
two in exactly the same way as

the two in pascals triangle came
from adding the one and the one

in the previous row.

So Pascal's triangle will give
us an easy way of evaluating a

binomial expression when we want
to raise it to an even higher

power. Let me look at what
happens if we want a plus B to

the power three and will see
that we can do this almost

straight away. What we note is
that the highest power now is 3.

So we start with an
A to the power 3.

Each successive term that power
of a will reduce, so they'll be

a term in a squared.

That'll be a term in A.

And then they'll be a term
without any Asian at all.

So as we move from left to
right, the powers of a decrease.

Similarly, as we move from left
to right, we want the powers of

be to increase just as they did
here. There will be no bees in

the first term. ABB to the power
one or just B.

In the second term.

B to the power two in the next

term. And then finally there
will be a B to the power

three and we stop it be to
the power three that highest

power corresponding to the
power in the original

binomial expression.

We need some coefficients.
That's the numbers in front of

each of these terms.

And the numbers come from the
relevant row in pascals

triangle, and we want the row
that begins 1, three and the

reason why we want the row
beginning 1. Three is because

three is the power in the
original expression. So I go

back to my pascals triangle and
I look for the robe beginning 1

three, which is 1331.

So these numbers are the
coefficients that I need.

In this expansion I want one.

331
And just to tidy that up a

little bit 1A Cube would
normally just be written as a

cubed. 3A squared

B. 3A B

squared. And finally 1B cubed
which would normally write as

just be cubed.

Now, I hope you'll agree that
using pascals triangle to expand

A+B to the power 3.

Is much simpler than multiplying
A+B Times A+B times A+B?

What I want to do for just
before we go on is just actually

go and do it the long way, just
to point something out.

Let's go back to
a plus B.

To the power three and work
it out the long way by noting

that we can work this out as
a plus B multiplied by a plus

B or squared.

We've already expanded A+B to
the power two, so let's write

that down. Well remember A+B to
the power two we've already seen

is A squared.

2AB And
B squared.

Now to expand this,
everything in the first

bracket must multiply
everything in the SEC

bracket, so we've been a
multiplied by a squared

which is a cubed.

A multiplied by two AB.

Which is 2.

A squared B.

A multiplied by

B squared. Which
is a B squared.

We be multiplied by a squared.

She's BA squared.

We be multiplied by two AB which
is 2A B squared.

And finally, be multiplied by AB
squared is AB cubed.

To tidy this up as a
cubed and then notice there's a

squared B terms in here.

And there's also an A squared B
turn there, one of them, so

we've into there and a one there
too, and the one gives you three

lots of A squared fee.

There's an AB squared.

Here, and there's more AB
squared's there. There's one

there, two of them there so
altogether will have three lots

of AB squared.

And finally, the last term at
the end B cubed.

That's working out the expansion
the long way. Why have I done

that? Well, I've only done that
just to point out something to

you and I want to point out that
the three in here in the three A

squared B came from adding a 2

here. And a one in there 2 plus
the one gave you the three.

Similarly, this three here came
from a one lot of AB squared

there and two lots of AB squared
there. So the one plus the two

gave you the three, and that
mirrors exactly what we had when

we generated the triangle,
because the three here came back

from adding the one in the two
in the row above and the three

here came from adding two and
one in the row above.

Let's have a look at another
example and see if we can just

write the answer down
straightaway. Suppose we want to

expand A+B or raised to the

power 4. Well, this is
straightforward to do. We know

that when we expand this, our
highest power of a will be 4

because that's the power in
the original expression.

And thereafter every subsequent
term will have a power reduced

by one each time. So there will
be an A cubed.

And a squared and A and then, no
worries at all.

As we move from left to
right, the powers of B will

increase. There will be none
at all in the first term.

And they'll be a big to the one.

Or just be. A bit of the two.

Beta three will be cubed and
finally the last term will be to

the four and again the highest
power corresponding to the power

four in the original expression.

And all we need now are the
coefficients. The coefficients

come from the appropriate role
in the triangle and this time

because we're looking at power
four, we want to look at the Roo

beginning 14. The row beginning
1 four is 14641.

Those are the coefficients that

will need.
14641

And just to tidy it up, we
wouldn't normally right the one

in there and the one in there so
A&B to the four is 8 to 4

four A cubed B that's that.

6A squared, B squared.

4A B

cubed. And finally, be
to the power 4.

OK, so I hope you'll agree that
using pascals triangle to get

this expansion was much simpler
than multiplying this bracket

over and over by itself. Lots
and lots of times that way is

also prone to error, so if you
can get used to using pascals

triangle. We can use the same
technique even when we have

slightly more complicated
expressions. Let's do another

example. Suppose we want to
expand 2X plus Y all to

the power 3.

So it's more complicated this
time because I just haven't got

a single term here, but I've
actually got a 2X in there.

The principle is
exactly the same.

What will do is will write this
term down first. The whole of

2X. And just like before, it
will be raised to the highest

possible power which is 3 and
that corresponds to the three in

the original problem.

Every subsequent term will have
a 2X in it, but as we go from

left to right, the power of 2X
will decrease, so the next term

will have a 2X or squared.

The next term will have a 2X to
the power one or just 2X, and

then there won't be any at all
in the last term.

Powers of Y will increase as we
move from the left to the right,

so there won't be any in the

first term. Then they'll be Y.

Then they'll be Y squared and
finally Y cubed.

And then we remember the
coefficients. Where do we get

the coefficients from?

Well, because we're looking at
power three, we go to pascals

triangle and we look for the row

beginning 13. You might even
remember those numbers now.

We've seen it so many times. The
numbers are 1331. Those are the

coefficients we require, 1331.

So I want one of those
three of those three of

those, one of those.

And there's just a bit
more tidying up to do to

finish it off.

Here we've got 2 to the Power 3,
two cubed that's eight.

X cubed
And the one just is, one could

just stay there 1. Multiply by
all that is not going to do

anything else, just 8X cubed.

What about this term? There's a
2 squared, which is 4, and it's

got to be multiplied by three.
So 4 threes are 12, so we have

12. What about powers of X?
Well, there be an X squared.

Why?

In this term, we've just got
2X to the power one. That's

just 2X, so this is just
three times 2X, which is 6X,

and there's a Y squared.

And finally, there's just the Y
cubed at the end. One Y cubed is

just Y cubed. So there we've
expanded the binomial expression

2X plus Y to the power three in
just a couple of lines using

pascals triangle. Let's look at
another one. Suppose this time

we want one plus P different
letter just for a change one

plus P or raised to the power 4.

In lots of ways, this is going
to be a bit simpler.

Because as we move through the
terms from left to right, we

want powers of the first term,

which is one. It won't want to
the Power 4 one to the Power 3,

one to the power two and so on,
but want to any power is still

one that's going to make life

easier for ourselves. So 1 to
the power four is just one.

And then thereafter they'll be

just one. All the way through.

We want the powers of P to
increase. We don't want any

peace in the first term.

We want to be there.

P squared there the next time
will have a P cubed in and the

last term will have a Peter. The
four in these ones.

Are the powers of the first term
one, so 1 to the 4th, one to

three, 1 to the two, 1 to the
one which is just one?

And no ones there at all.

And finally, we want some
coefficients and the

coefficients come from pascals
triangle. This time the row

beginning 1, four. Because of
this powerful here.

So the numbers we
want our 14641.

1.

4.
6.

4. One, let's
just tidy it

up as one.

4 * 1 is just four P.

6 * 1 is 66

P squared. 4 * 1
is 4 P cubed.

And last of all, one times Peter
the four is just Peter the four.

Again, another example of a
binomial expression raised to a

power, and we can almost write
the answer straight down using

the triangle instead of
multiplying those brackets out

over and over again.

Now, sometimes either or both of
the terms in the binomial

expression might be negative.

So let's have a look at an
example where one of the terms

is negative. So suppose we want

to expand.

3A.
Minus 2B, so I've got a term

that's negative now, minus 2B,
and let's suppose we want this

to the power 5.

3A minus 2B all raised to
the power 5.

This is going to be a bit more
complicated this time, so let's

see how we get on with it.

As before. We
want to take our first term.

And raise it to the
highest power, the highest

power being 5.

So our first term will be 3A.

All raised to the power 5.

The next term will have a 3A

in it. And this time it will be
raised to the power 4.

There be another term with a 3A
in. It'll be 3A to the power 3.

Then 3A to the power 2.

Then 3A to the power one, and
then they'll be a final term

that doesn't have 3A in it at

all. That deals with this
first term.

Let's deal with
the minus 2B now.

In the first term here, there
won't be any minus two BS at

all, but there after the
powers of this term will

increase as we move from left
to right exactly as before. So

when we get to the second term
here will need a minus two

fee.

When we get to the next term
will leave minus 2B and we're

going to square it.

Minus 2B raised to the power 3.

Minus two be raised to the power

4. And the last term will be
minus two be raised to the power

5. The power five
corresponding to the highest

power in the original
problem.

We also need our coefficients.
The numbers in front of each of

these six terms.

The coefficients come from the
row beginning 15.

Because the problem has a
power five in it.

The coefficients
are one 5101051.

One 5101051 so we
want one of those.

Five of those.

Ten of

those. Ten of

those. Five of those, and
finally one of those you can see

now why I left a lot of space
when I was writing all this

down. There's a lot of things to
tidy up in here.

Just to tidy all this up, we
need to remember that when

we raise a negative number
to say the power two, the

results going to be positive
when we raise it to an even

even power, the result would
be positive. So this term is

going to be positive and the
minus 2B to the power four

will also become positive.

When we raise it to an odd power
like 3 or the five, the result

is going to be negative. So our
answer is going to have some

positive and some negative

numbers in it. Let's tidy
it all up.

Go to Calculator for this,
'cause I'm going to raise some

of these numbers to some powers.

First of all I want to raise 3
to the power 5.

3 to the power five is

243. So I have 243.

A to the power 5.

And it's all multiplied by
one which isn't going to

change anything.

Now here we've got a negative
number because this is minus 2

be raised to the power one is
going to be negative, so this

term is going to have a minus

sign at the front. We've got 3
to the power 4.

Well, I know 3 squared is 9 and
9, nine 481, so 3 to the power

four is 81. Five 210

So I'm going to multiply 81 by
10, which is 810th.

There will be 8 to the power

4. And a single be.

So that's my next term.

Now what have we got left?
There's 3 to the power three

which is 3 cubed, which is 27.

Multiplied by two
squared, which is 4.

All multiplied by 10.

Which is 1080.

8 to the

power 3. B to
the power 2.

And here we have two cubed which

is 8. 3 squared which is 9.

9 eight 472 *
10 is 720.

There will be an A squared from

this term. And not be a B cubed
from the last time.

What about here?

Well, we've 2 to the power 4.

Which is 16.

5 three is a 15 here.

And 15 * 16 is 240. It'll
be positive because here we been

negative number to an even power
248 to the power one or just

a. B to the power 4.

And finally. There will be one
more term and that will be minus

2 to the Power 5, which is going

to be negative. 32 B to
the power 5.

And that's the expansion of this
rather complicated expression,

which had both positive and
negative quantities in it. And

again, we've used pascals
triangle to do that.

We can use exactly the same
method even if there are

fractions involved, so let's
have a look at an example where

there's some fractions. Suppose
we want to expand.

This time 1 + 2 over X, so
I've deliberately put a fraction

in there all to the power 3.

Let's see what happens.

1 + 2 over
X to the power

3. Well. We start
with one raised to the highest

power which is 1 to the power 3.

Which is still 1.

And once at, any power will
still be one's remove all the

way through the calculation.

Will have two over X raised
first of all to the power one.

Two over X to
the power 2.

And two over X to the power
three and we stop there. When we

reached the highest power.

Which corresponds to the power
in the original problem.

We need the coefficients of each
of these terms from pascals

triangle and the row in the
triangle beginning 13.

Those numbers are 1331, so
there's one of these three of

those. Three of those.

I'm one of those.

And all we need to do now is
tidy at what we've got.

So there's once.

Two over X to the power one
is just two over X. We're

going to multiply it by
three, so 3 twos are six will

have 6 divided by X.

Here there's a 2 squared, which
is 4. Multiply it by three so we

have 12 divided by X to the
power 2 divided by X squared.

And finally, there's 2 to the

power 3. Which is 8.

And this time it's divided by X
to the power 34X cubed.

So that's a simple example which
illustrates how we can apply

exactly the same technique even
when the refraction is involved.

Now, that's not quite
the end of the

story. The problem is, supposing
I were to ask you to expand a

binomial expression to a very
large power, suppose I wanted

one plus X to the power 32 or
one plus X to the power 127. You

have an awful lot of rows of
pascals triangle to generate if

you wanted to do it this way.

Fortunately, there's an
alternative way, and it involves

a theorem called the binomial

theorem. So let's just have
a look at what the binomial

theorem says.

The binomial theorem allows us
to develop an expansion of

the binomial expression A+B
raised to the power N.

And it allows us to get an
expansion in terms of

decreasing powers of a,
exactly as we've seen before.

And increasing powers of B
exactly as we've seen before.

And it I'm going to quote the
theorem for the case when N is a

positive whole number.

This theorem will actually work
when is negative and when it's a

fraction, but only under
exceptional circumstances, which

we're not going to discuss here.
So in all these examples, N will

be a positive whole number.

Now what
the theorem

says is

this. A+B to the power N is
given by the following expansion

A to the power N.

Now that looks familiar, doesn't
it? Because as in all the

examples we've seen before,
we've taken the first term and

raised it to the highest power.
The power in the original

question 8 to the power N.

Then there's a next term, and
the next term will have an A to

the power N minus one.

And a B in it. That's exactly as
we've seen before, because we're

starting to see the terms
involving be appear and the

powers event at the powers of a

a decreasing. We want a
coefficient in here and the

binomial theorem tells us that
the coefficient is NTH.

The next term.

As an A to the power
N minus two in it.

Along with the line we had
before of decreasing the powers

and increasing the power of be
will give us a B squared.

And the binomial theorem
tells us the coefficient to

right in here and the
coefficient this time is NN

minus one over 2 factorial.

In case you don't know what this
notation means, 2 factorial

means 2 * 1.

That's called 2 factorial.

And this series goes on and on
and on. The next term will be

NN minus one and minus two over
3 factorial, and there's a

pattern developing here. You
see, here we had an N&NN minus

one. And minus one and minus 2.

With a 3 factorial at the bottom
where we had a two factor at the

bottom before 3 factorial means
3 * 2 * 1.

The power of a will be 1 less
again, which this time will be A

to the N minus three.

And we want to power of bee
which is B to the power 3.

So all the way through this
theorem you'll see the powers of

a are decreasing. And the powers
of B are increasing. Now this

series goes on and on and on
until we reach the term B to the

power N. When it stops. So this
is a finite series. It stops

after a finite number of terms.

Now, the theorems often quoted
in this form, but it's also

often quoted in a slightly

simpler form. And it's quoted in
the form for which a is the

simple value of just one.

And B is X. Now when a is one,
all of these A to the power ends

or A to the N minus one A to the
N minus two. Each one of those

terms will just simplify to the
number one, so the whole thing

looks simpler. So let's write
down the binomial theorem again

for the special case when a is
one and these X.

This time will get one
plus X raised to the

power N. Is equal to.

1.

Plus N.
X.

Plus NN minus one over 2
factorial X squared, and you can

see what's happening. This
second term X is starting to

appear and and its powers
increasing as we move from left

to right. So even X&X squared
the next time will have an X

cubed in it, one to any power is
still one, so I don't actually

need to write it down.

The next term will
be NN minus one and

minus two over 3
factorial X cubed.

The next term will be NN minus
one and minus 2 N minus three

over 4 factorial X to the four,
and this will go on and on until

eventually you'll get to the
stage where you get to the last

term raised to the highest power
you'll get to X to the power N,

and the series will stop.

So this is a slightly simpler
form of the theorem, and it's

often quoted in this form.

Now let's use it to examine some
binomial expressions that you're

already very familiar with.

Let's suppose we want to expand
one plus X or raised to the

power two. Now I've written down
the theorem again so we can

refer to it and this is printed
in the notes. If you want to use

the one in the notes.

So we've one plus X to the power
N. In our problem, we've got one

plus X to the power two, so all
we have to do is let NB two in

all of this formula through

here. So let's see what we get
or from the theorem.

The first thing will write down
is just the one.

Then we want NX, but N IS
two, so will just put plus 2X.

And then the next term we want
is going to be a term

involving X to the power two,
but that's the highest power

we want because we've got a
power to in here. We want to

stop when we get to X to the
power two, so we're actually

already at the end with the
next term, and we just want an

X to the power two on its own.

1 + 2 X plus X squared and
that's the expansion that

you're already very familiar
with, and you'll notice in it

that the powers of X increase
as we move through from left to

right, and there's powers of
one in there, but we don't see

them, and the one 2 one other
numbers in pascals triangle.

Let's look at the theorem for
the case when is 3, let's expand

one plus X to the power 3.

I'm going to use the theorem
again, but this time we're

going to let NB 3.

So we want 1 + 3

X. And then
we want 3.

3  1 three minus
one is 2.

All divided by 2 factorial.

Than an X squared.

And then the next term will be a
term involving X cubed, which is

the term that we stop with
because we're only working 2X to

the power three here. So the
last term will be just a plus X

cubed. We can tie this up to 1
+ 3 X.

2 factorial is 2 * 1, which is
just two little cancel with the

two at the top, so will be left
with just three X squared, and

finally an X cubed.

And again, that's something that
you're already very familiar

with. You'll notice the
coefficients, the 1331 other

numbers we've seen many times in
pascals triangle the powers of X

increase. As we move from the
left to the right, and this is a

finite series, it stops when we
get to the term involving X

cubed corresponding to this
highest power over there.

Now.
That suppose we want to look at

it and more complicated problem.
Suppose we want to workout one

plus X to the power 32. Now you

would never. Use pascals
triangle to attempt this problem

because you'd have to generate
so many rows of the triangle,

but we can use the binomial

theorem. What I'm going to do
is I'm going to write down

the first three terms of the
series using the binomial

theorem, and I'm going to use
it with N being equal to 32.

So we're putting any 32 in. The
theorem will get 1 + 32 X.

That's the one plus the NX.

We want an which is 32.

And minus one which will be 31.

All over 2 factorial.

X squared

And we know that this series
will go on and on until we

reached the term, the last term
being X to the power 32.

But I only want to look at
the first three terms here in

this problem, so the first
three terms are just going to

be 1 + 32 X and we want to
simplify this. We've got 32 *

31 and then divided by two.

Which is 496.

And I just put some dots there
to show that this series goes on

a lot further than the terms
that I've just written down

there. I'm going to have a look
at a couple more examples with

some ingenuity. We can use the
theorem in a slightly different

form. Suppose we want to expand
this binomial expression this

time, I'm going to look at one
plus Y divided by 3. All raised

to the power 10 and suppose that
I'm interested. I'm interested

in generating the first.

Four terms. Let's see how we can
do that. Well, we've got our

theorem. I've written it down
again here for us in terms of

one plus X to the power N. We
can use it in this problem if we

replace every X.

In the theorem with a Y over 3.

So everywhere there's an X in
the theorem, I'm going to write

Y divided by three and then the
pattern will match exactly what

we have in the theorem ends
going to be 10 in this problem.

So let's see what we get will
have one plus Y over three

raised to the power 10 is equal

to. Well, we start with a one

as always. Then we

want NX. And

it's 10. And we said that
instead of X, but replacing the

X with a Y over 3.

So we have a Y over three there.

What's the next term we want NN
minus one over 2 factorial?

Which is 10.

10  1 is 9 over 2
factorial.

And then we'd want an X squared.
So in this case we want X being

why over 3?

All square

I want to generate one more term
'cause I said I want to look for

four terms, so the next term is
going to be an which was 10.

N minus one which is 9.

And minus two, which is 8 and
this time over 3 factorial.

So I'm here NN minus one and
minus two over 3 factorial and

we want X cubed X is Y over
three, so we want why over 3

cubed. And the series goes on
and on. Let's just tidy up what

we've got. There's one.

That'll be 10, why over 3?

What if we got in here? Well,
there's a 3 square at the bottom

which is 9, and there's a 9 at
the top, so the three squared in

here is going to cancel with the

nine there. 2 factorial

Is just two.

And choosing to 10 is 5, so will
have five 5 squared.

And then this is a bit more
complicated. We've got a 3

factorial which is 3 * 2.

And three cubed. At the bottom
there, which is 3 * 3 * 3. Some

of this will cancel down. 3 * 3
will cancel, with the nine in

here. The two will cancel their
with the eight will have four

and let's see what we're left
with at the top will have 10 *

4, which is 40.

And at the bottom will have 3
* 3, which is 9.

And they'll be a Y cubed.

So altogether we've 1 + 10 Y
over 3 five Y squared, 40 over 9

Y cubed, and those are the first
four terms of a series which

will actually continue until you
get to a term involving the

highest power, which will be a Y
over 3 to the power 10.

So you can still use the theorem
in slightly different form if

you use a bit of ingenuity. Want
to look at one final example

before we finish? And this time
I want to look at the example 3

 5 Z.

To the power 40 again, it's an
example where you wouldn't want

to use pascals triangle because
the power for teens too high and

you have too many rose to
generate in your triangle.

I'm going to use the original
form of the theorem, the One I

have here in terms of A+B to the

power N. A will be 3.

Now B is a negative number be
will be minus five said.

Ends going to be 14.

But we can still use the
theorem. Let's see what

happens and in this problem
I'm just going to generate

the first three terms.

OK so A is 3 and we
want to raise the three.

To the highest power which is

14. So my first term is 3 to
the power 14.

My second term is this one.

Begins with an N.

The power in the original
expression, which was 14.

Multiplied by A
to the power N

minus one AS 3.

And we want to raise it to the
power N minus 114  1 is 30.

And we want to be.

B is minus five set.

So our second term looking ahead
is going to be negative because

of that minus five in there.

My third term and I'll stop
after the third term, his N,

which is 14.

And minus one which is 13
all over 2 factorial.

A to the power N minus two will
be 3 to the power N minus two

will be 14  2 which is 12.

And finally AB, which
was minus five said.

Raised to the power 2.

And we know this goes on and on
until we reach term instead to

the power 14. But we've only
written down the first three

terms there. Perhaps we should
just tidy it up a little bit.

There's a 3 to the power 14 at

the beginning. There's a minus
five here, minus 5. Four

teens are minus 70.

As a 3 to the power 13, let me
just leave it like that for the

time being. And then
they'll be as Ed.

Over here there's a 3 to the

power 12. That's this term in

here. And I'm reaching for my
Calculator again because this is

a bit more complicated. We have
got a 14.

Multiplied by 13.

When multiplied by 5 squared,
which is 25.

And divided by the two factorial
that's divided by two, this will

be multiplied by two 275.

And this expression will be
positive because we've got a

minus 5 squared.

And we need to remember to
include zed squared in there.

OK, we observe as before that
the powers of zed are increasing

as we move from the left to the
right. Now we could leave it

like that. I'm just going to
tidy it up and write it in a

slightly different form because
this is often the way you see

answers in the back of textbooks
or people ask you to give an

answer in a particular form and
the form I'm going to write it

in is one obtained by taking out
a factor of 3 to the power 40.

If I take her three to 14 out
from the first term, I'll be

just left with one.

Now 3 to the 13. In the second

term. But if I multiply top and
bottom by three, I'll have a 3

to the 14th at the top, which I

can take out. But have
multiplied the bottom by

three as well, which will
leave me with minus 70 Zedd

divided by three.

Here with a 3 to the power 12.

And I want to take out a 3 to

14. If I multiply the top and
bottom by three squared or nine,

I affectively get a 3 to 14 in

this term. So I'm multiplying
top and bottom by 9, taking

the three to the 14 out, and
that will leave me here with

two 275 over 9.

Said squad And this series
continues. As I said before,

until you get to a term
involving zed to the power 14,

but those are the first three
terms of the series.