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.