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.
A-B 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.
2-AB 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 510-1051.
One 510-1051 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.