## www.mathcentre.ac.uk/.../Pascal.mp4

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A binomial expression is the
sum or difference of two
• 0:08 - 0:14
terms. So for example 2X
plus three Y.
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Is an example of a binomial
• 0:17 - 0:22
expression. Because it's the sum
of the term 2X and the term 3 Y
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is the sum of these two terms.
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Some of the terms could be just
numbers, so for example X plus
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one. Is the sum of the term X
and the term one, so that's two
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is a binomial expression.
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A-B is the difference of
the two terms A&B, so
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that too is a binomial
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work you have seen many binomial
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expressions and you have raised
them to different powers. So you
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have squared them, cube them and
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so on. You probably already be
very familiar with working with
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the binomial expression like X
Plus One and squaring it.
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And you have done that by
remembering that when we want to
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square a bracket when
multiplying the bracket by
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itself. So X Plus One squared is
X Plus One multiplied by X plus
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one. And we remove the brackets
by multiplying all the terms in
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the first bracket by all the
terms in the SEC bracket, so
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they'll be an X multiplied by X.
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Which is X squared.
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X multiplied by one which
is just X.
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1 multiplied by X, which is
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another X. And 1 * 1, which
is just one.
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So to tide you all that up X
Plus One squared is equal to X
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squared. As an X plus another X
which is 2 X.
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Plus the one at the end.
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Note in particular that we
have two X here and that came
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from this X here and another X
there. I'll come back to that
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point later on and will see
why that's important.
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Now suppose we want to raise a
binomial expression to our power
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that's higher than two. So
suppose we want to cube it,
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raise it to the power four or
five or even 32.
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The process of removing the
brackets by multiplying term by
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term over and over again is very
very cumbersome. I mean, if we
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wanted to workout X plus one to
the Seven, you wouldn't really
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want to multiply a pair of
brackets by itself several
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times. So what we want is a
better way. Better way of doing
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that. And one way of doing it
is by means of a triangle of
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numbers, which is called
Pascals Triangle.
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Pascal was a 17th century French
mathematician and he derived
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this triangle of numbers that
will repeat for ourselves now,
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and this is how we form the
triangle. We start by writing
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down the number one.
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Then we form a new row and on
this nuro we have a one.
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And another one.
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We're going to build up a
triangle like this and each nuro
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that we write down will start
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with a one. And will end with a
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one. So my third row is
going to begin with the
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one and end with a one.
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And in a few minutes,
we'll write a number
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in there in the gap.
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The next row will begin with a
one and end with a one and will
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write a number in there and a
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number in there. And in this way
we can build a triangle of
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numbers and we can build it as
big as we want to.
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How do we find this number in
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here? Well, the number that
goes in here we find by
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looking on the row above.
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And looking above to the left
and above to the right.
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And adding what we find, there's
a one here. There's a one there.
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We add them one and one gives 2
and we write the result in
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there. So there's two.
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On the 3rd row has come by
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one together.
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Let's look at the next row down
the number that's going to go in
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here. Is found by looking
on the previous row.
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And we look above left which
gives us the one we look above
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to the right, which gives us
two, and we add the numbers
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that we find, so we're adding
a one and two which is 3 and
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we write that in there.
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Well again previous row above
to the left is 2 above to the
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right is one. We add what we
find 2 plus one is 3 and that
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goes in there.
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And we can carry on building
this triangle as big as we want
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to. Let's just do one more row.
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We start the row.
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With a one and we finished with
a one and we put some numbers in
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here and in here and in here.
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The number that's going
to go in here.
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Is found from the previous row
by adding the one and the three.
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So 1 + 3.
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Is 4 let me write that in there.
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The number that's going to go in
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here. Well, we look in the
previous row above left and
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above right. 3 + 3 is 6 and we
write that in there.
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And finally 3 Plus One is 4 when
we write that in there. So
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that's another row.
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And what you should do now is
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rows for yourself, and
altogether this triangle of
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numbers is called pascals.
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Pascal's triangle.
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OK.
Now we're going to use this
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triangle to expand binomial
expressions and will see that it
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can make life very easy for us.
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We'll start by
going back to
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the expression A+B.
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To the power 2.
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So we have binomial binomial
expression here, a I'd be and
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we're raising it to the power 2.
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Let's do it the old way. First
of all by multiplying A&B by
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itself. Because we're squaring
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A&B. Let's write down
what will get.
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A multiplied by a
gives us a squared.
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A multiplied by B will give us
a Times B or just a B.
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Be multiplied by a.
Gives us a BA.
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And finally, be multiplied by B.
Give us a B squared.
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And if we just tidy it, what we
found, there's a squared.
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There's an AB.
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And because BA is the same as a
bee, there's another a be here.
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So altogether there's two lots
of a B.
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And finally, AB squared
at the end.
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Now that's the sort of
expansion. This sort of removing
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brackets that you've seen many
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very familiar with, but what I
want to do is make some
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When we expanded A+B to the
power two, what we find is that
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as we successively move through
these terms that we've written
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down the power of a decreases,
it starts off here with an A
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squared. The highest power being
two corresponding to the power
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in the original binomial
expression, and then every
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subsequent term that power
drops. So it was 8 to the power
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2. There's A to the power one
in here, although we don't
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normally right the one in and
then know as at all, so the
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powers of a decrease as we
move from left to right.
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There's no bees in here.
There's a beta. The one in
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there, although we just
normally right B&AB to the
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power two there. So as we move
from left to right, the powers
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of B increase until we reach
the highest power B squared,
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and the squared corresponds to
the two in the original
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problem.
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What else can we observe if we
look at the coefficients of
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these terms now the coefficients
are the numbers in front of each
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of these terms. Well, there's a
one in here, although we
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wouldn't normally write it in,
there's a two there, and there's
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a one inference of the B
squared, although we wouldn't
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normally write it in.
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So the coefficients
are 1, two and one.
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Now let me remind you again
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copy of the triangle here so we
can refer to it. If we look at
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pascals triangle here will see
that one 2 one is the numbers
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that's in the 3rd row of pascals
triangle. 121 other numbers that
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occur in the expansion of A+B to
the power 2.
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There's something else I want to
point out that this 2A. B in
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here. Came from a term here 1A B
on one BA in there and together
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the one plus the one gave the
two in exactly the same way as
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the two in pascals triangle came
from adding the one and the one
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in the previous row.
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So Pascal's triangle will give
us an easy way of evaluating a
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binomial expression when we want
to raise it to an even higher
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power. Let me look at what
happens if we want a plus B to
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the power three and will see
that we can do this almost
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straight away. What we note is
that the highest power now is 3.
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A to the power 3.
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Each successive term that power
of a will reduce, so they'll be
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a term in a squared.
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That'll be a term in A.
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And then they'll be a term
without any Asian at all.
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So as we move from left to
right, the powers of a decrease.
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Similarly, as we move from left
to right, we want the powers of
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be to increase just as they did
here. There will be no bees in
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the first term. ABB to the power
one or just B.
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In the second term.
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B to the power two in the next
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term. And then finally there
will be a B to the power
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three and we stop it be to
the power three that highest
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power corresponding to the
power in the original
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binomial expression.
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We need some coefficients.
That's the numbers in front of
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each of these terms.
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And the numbers come from the
relevant row in pascals
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triangle, and we want the row
that begins 1, three and the
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reason why we want the row
beginning 1. Three is because
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three is the power in the
original expression. So I go
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back to my pascals triangle and
I look for the robe beginning 1
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three, which is 1331.
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So these numbers are the
coefficients that I need.
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In this expansion I want one.
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331
And just to tidy that up a
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little bit 1A Cube would
normally just be written as a
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cubed. 3A squared
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B. 3A B
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squared. And finally 1B cubed
which would normally write as
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just be cubed.
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Now, I hope you'll agree that
using pascals triangle to expand
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A+B to the power 3.
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Is much simpler than multiplying
A+B Times A+B times A+B?
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What I want to do for just
before we go on is just actually
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go and do it the long way, just
to point something out.
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Let's go back to
a plus B.
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To the power three and work
it out the long way by noting
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that we can work this out as
a plus B multiplied by a plus
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B or squared.
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the power two, so let's write
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that down. Well remember A+B to
the power two we've already seen
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is A squared.
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2-AB And
B squared.
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Now to expand this,
everything in the first
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bracket must multiply
everything in the SEC
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bracket, so we've been a
multiplied by a squared
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which is a cubed.
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A multiplied by two AB.
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Which is 2.
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A squared B.
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A multiplied by
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B squared. Which
is a B squared.
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We be multiplied by a squared.
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She's BA squared.
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We be multiplied by two AB which
is 2A B squared.
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And finally, be multiplied by AB
squared is AB cubed.
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To tidy this up as a
cubed and then notice there's a
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squared B terms in here.
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And there's also an A squared B
turn there, one of them, so
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we've into there and a one there
too, and the one gives you three
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lots of A squared fee.
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There's an AB squared.
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Here, and there's more AB
squared's there. There's one
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there, two of them there so
altogether will have three lots
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of AB squared.
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And finally, the last term at
the end B cubed.
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That's working out the expansion
the long way. Why have I done
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that? Well, I've only done that
just to point out something to
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you and I want to point out that
the three in here in the three A
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squared B came from adding a 2
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here. And a one in there 2 plus
the one gave you the three.
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Similarly, this three here came
from a one lot of AB squared
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there and two lots of AB squared
there. So the one plus the two
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gave you the three, and that
mirrors exactly what we had when
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we generated the triangle,
because the three here came back
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from adding the one in the two
in the row above and the three
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here came from adding two and
one in the row above.
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Let's have a look at another
example and see if we can just
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straightaway. Suppose we want to
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expand A+B or raised to the
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power 4. Well, this is
straightforward to do. We know
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that when we expand this, our
highest power of a will be 4
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because that's the power in
the original expression.
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And thereafter every subsequent
term will have a power reduced
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by one each time. So there will
be an A cubed.
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And a squared and A and then, no
worries at all.
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As we move from left to
right, the powers of B will
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increase. There will be none
at all in the first term.
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And they'll be a big to the one.
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Or just be. A bit of the two.
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Beta three will be cubed and
finally the last term will be to
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the four and again the highest
power corresponding to the power
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four in the original expression.
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And all we need now are the
coefficients. The coefficients
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come from the appropriate role
in the triangle and this time
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because we're looking at power
four, we want to look at the Roo
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beginning 14. The row beginning
1 four is 14641.
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Those are the coefficients that
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will need.
14641
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And just to tidy it up, we
wouldn't normally right the one
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in there and the one in there so
A&B to the four is 8 to 4
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four A cubed B that's that.
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6A squared, B squared.
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4A B
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cubed. And finally, be
to the power 4.
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OK, so I hope you'll agree that
using pascals triangle to get
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this expansion was much simpler
than multiplying this bracket
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over and over by itself. Lots
and lots of times that way is
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also prone to error, so if you
can get used to using pascals
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triangle. We can use the same
technique even when we have
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slightly more complicated
expressions. Let's do another
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example. Suppose we want to
expand 2X plus Y all to
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the power 3.
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So it's more complicated this
time because I just haven't got
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a single term here, but I've
actually got a 2X in there.
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The principle is
exactly the same.
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What will do is will write this
term down first. The whole of
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2X. And just like before, it
will be raised to the highest
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possible power which is 3 and
that corresponds to the three in
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the original problem.
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Every subsequent term will have
a 2X in it, but as we go from
• 19:11 - 19:16
left to right, the power of 2X
will decrease, so the next term
• 19:16 - 19:18
will have a 2X or squared.
• 19:19 - 19:24
The next term will have a 2X to
the power one or just 2X, and
• 19:24 - 19:27
then there won't be any at all
in the last term.
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Powers of Y will increase as we
move from the left to the right,
• 19:34 - 19:36
so there won't be any in the
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first term. Then they'll be Y.
• 19:40 - 19:43
Then they'll be Y squared and
finally Y cubed.
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And then we remember the
coefficients. Where do we get
• 19:49 - 19:50
the coefficients from?
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Well, because we're looking at
power three, we go to pascals
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triangle and we look for the row
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beginning 13. You might even
remember those numbers now.
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We've seen it so many times. The
numbers are 1331. Those are the
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coefficients we require, 1331.
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So I want one of those
three of those three of
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those, one of those.
• 20:17 - 20:21
And there's just a bit
more tidying up to do to
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finish it off.
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Here we've got 2 to the Power 3,
two cubed that's eight.
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X cubed
And the one just is, one could
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just stay there 1. Multiply by
all that is not going to do
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anything else, just 8X cubed.
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2 squared, which is 4, and it's
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got to be multiplied by three.
So 4 threes are 12, so we have
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12. What about powers of X?
Well, there be an X squared.
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Why?
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In this term, we've just got
2X to the power one. That's
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just 2X, so this is just
three times 2X, which is 6X,
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and there's a Y squared.
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And finally, there's just the Y
cubed at the end. One Y cubed is
• 21:13 - 21:17
just Y cubed. So there we've
expanded the binomial expression
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2X plus Y to the power three in
just a couple of lines using
• 21:23 - 21:27
pascals triangle. Let's look at
another one. Suppose this time
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we want one plus P different
letter just for a change one
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plus P or raised to the power 4.
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In lots of ways, this is going
to be a bit simpler.
• 21:40 - 21:43
Because as we move through the
terms from left to right, we
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want powers of the first term,
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which is one. It won't want to
the Power 4 one to the Power 3,
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one to the power two and so on,
but want to any power is still
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one that's going to make life
• 21:54 - 21:58
easier for ourselves. So 1 to
the power four is just one.
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And then thereafter they'll be
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just one. All the way through.
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We want the powers of P to
increase. We don't want any
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peace in the first term.
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We want to be there.
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P squared there the next time
will have a P cubed in and the
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last term will have a Peter. The
four in these ones.
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Are the powers of the first term
one, so 1 to the 4th, one to
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three, 1 to the two, 1 to the
one which is just one?
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And no ones there at all.
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And finally, we want some
coefficients and the
• 22:36 - 22:39
coefficients come from pascals
triangle. This time the row
• 22:39 - 22:42
beginning 1, four. Because of
this powerful here.
• 22:43 - 22:49
So the numbers we
want our 14641.
• 22:50 - 22:51
1.
• 22:52 - 22:56
4.
6.
• 22:57 - 23:03
4. One, let's
just tidy it
• 23:03 - 23:05
up as one.
• 23:06 - 23:10
4 * 1 is just four P.
• 23:10 - 23:14
6 * 1 is 66
• 23:14 - 23:21
P squared. 4 * 1
is 4 P cubed.
• 23:21 - 23:26
And last of all, one times Peter
the four is just Peter the four.
• 23:26 - 23:31
Again, another example of a
binomial expression raised to a
• 23:31 - 23:35
power, and we can almost write
• 23:35 - 23:39
multiplying those brackets out
• 23:39 - 23:40
over and over again.
• 23:40 - 23:47
Now, sometimes either or both of
the terms in the binomial
• 23:47 - 23:49
expression might be negative.
• 23:50 - 23:54
So let's have a look at an
example where one of the terms
• 23:54 - 23:56
is negative. So suppose we want
• 23:56 - 23:57
to expand.
• 23:58 - 24:04
3A.
Minus 2B, so I've got a term
• 24:04 - 24:07
that's negative now, minus 2B,
and let's suppose we want this
• 24:07 - 24:09
to the power 5.
• 24:10 - 24:15
3A minus 2B all raised to
the power 5.
• 24:16 - 24:18
This is going to be a bit more
complicated this time, so let's
• 24:18 - 24:19
see how we get on with it.
• 24:20 - 24:25
As before. We
want to take our first term.
• 24:26 - 24:29
And raise it to the
highest power, the highest
• 24:29 - 24:30
power being 5.
• 24:31 - 24:34
So our first term will be 3A.
• 24:34 - 24:36
All raised to the power 5.
• 24:37 - 24:41
The next term will have a 3A
• 24:41 - 24:45
in it. And this time it will be
raised to the power 4.
• 24:49 - 24:54
There be another term with a 3A
in. It'll be 3A to the power 3.
• 24:55 - 24:58
Then 3A to the power 2.
• 25:00 - 25:04
Then 3A to the power one, and
then they'll be a final term
• 25:04 - 25:07
that doesn't have 3A in it at
• 25:07 - 25:11
all. That deals with this
first term.
• 25:13 - 25:15
Let's deal with
the minus 2B now.
• 25:17 - 25:21
In the first term here, there
won't be any minus two BS at
• 25:21 - 25:24
all, but there after the
powers of this term will
• 25:24 - 25:28
increase as we move from left
to right exactly as before. So
• 25:28 - 25:33
when we get to the second term
here will need a minus two
• 25:33 - 25:33
fee.
• 25:36 - 25:40
When we get to the next term
will leave minus 2B and we're
• 25:40 - 25:41
going to square it.
• 25:43 - 25:47
Minus 2B raised to the power 3.
• 25:49 - 25:52
Minus two be raised to the power
• 25:52 - 25:58
4. And the last term will be
minus two be raised to the power
• 25:58 - 26:02
5. The power five
corresponding to the highest
• 26:02 - 26:03
power in the original
problem.
• 26:07 - 26:11
We also need our coefficients.
The numbers in front of each of
• 26:11 - 26:13
these six terms.
• 26:13 - 26:18
The coefficients come from the
row beginning 15.
• 26:19 - 26:21
Because the problem has a
power five in it.
• 26:22 - 26:27
The coefficients
are one 510-1051.
• 26:28 - 26:32
One 510-1051 so we
want one of those.
• 26:34 - 26:36
Five of those.
• 26:37 - 26:40
Ten of
• 26:40 - 26:44
those. Ten of
• 26:44 - 26:47
those. Five of those, and
finally one of those you can see
• 26:47 - 26:51
now why I left a lot of space
when I was writing all this
• 26:51 - 26:54
down. There's a lot of things to
tidy up in here.
• 26:55 - 26:59
Just to tidy all this up, we
need to remember that when
• 26:59 - 27:02
we raise a negative number
to say the power two, the
• 27:02 - 27:06
results going to be positive
when we raise it to an even
• 27:06 - 27:10
even power, the result would
be positive. So this term is
• 27:10 - 27:13
going to be positive and the
minus 2B to the power four
• 27:13 - 27:15
will also become positive.
• 27:16 - 27:20
When we raise it to an odd power
like 3 or the five, the result
• 27:20 - 27:23
is going to be negative. So our
answer is going to have some
• 27:23 - 27:24
positive and some negative
• 27:24 - 27:28
numbers in it. Let's tidy
it all up.
• 27:29 - 27:31
Go to Calculator for this,
'cause I'm going to raise some
• 27:31 - 27:32
of these numbers to some powers.
• 27:33 - 27:36
First of all I want to raise 3
to the power 5.
• 27:38 - 27:42
3 to the power five is
• 27:42 - 27:45
243. So I have 243.
• 27:46 - 27:49
A to the power 5.
• 27:49 - 27:51
And it's all multiplied by
one which isn't going to
• 27:51 - 27:52
change anything.
• 27:53 - 27:56
Now here we've got a negative
number because this is minus 2
• 27:56 - 27:59
be raised to the power one is
going to be negative, so this
• 27:59 - 28:01
term is going to have a minus
• 28:01 - 28:05
sign at the front. We've got 3
to the power 4.
• 28:07 - 28:12
Well, I know 3 squared is 9 and
9, nine 481, so 3 to the power
• 28:12 - 28:15
four is 81. Five 210
• 28:15 - 28:21
So I'm going to multiply 81 by
10, which is 810th.
• 28:22 - 28:26
There will be 8 to the power
• 28:26 - 28:29
4. And a single be.
• 28:29 - 28:31
So that's my next term.
• 28:31 - 28:36
Now what have we got left?
There's 3 to the power three
• 28:36 - 28:38
which is 3 cubed, which is 27.
• 28:39 - 28:42
Multiplied by two
squared, which is 4.
• 28:45 - 28:47
All multiplied by 10.
• 28:48 - 28:51
Which is 1080.
• 28:51 - 28:54
8 to the
• 28:54 - 29:00
power 3. B to
the power 2.
• 29:01 - 29:04
And here we have two cubed which
• 29:04 - 29:07
is 8. 3 squared which is 9.
• 29:08 - 29:14
9 eight 472 *
10 is 720.
• 29:14 - 29:18
There will be an A squared from
• 29:18 - 29:24
this term. And not be a B cubed
from the last time.
• 29:25 - 29:27
• 29:28 - 29:30
Well, we've 2 to the power 4.
• 29:30 - 29:32
Which is 16.
• 29:32 - 29:35
5 three is a 15 here.
• 29:35 - 29:42
And 15 * 16 is 240. It'll
be positive because here we been
• 29:42 - 29:49
negative number to an even power
248 to the power one or just
• 29:49 - 29:53
a. B to the power 4.
• 29:55 - 29:59
And finally. There will be one
more term and that will be minus
• 29:59 - 30:02
2 to the Power 5, which is going
• 30:02 - 30:07
to be negative. 32 B to
the power 5.
• 30:07 - 30:12
And that's the expansion of this
rather complicated expression,
• 30:12 - 30:16
negative quantities in it. And
• 30:16 - 30:20
again, we've used pascals
triangle to do that.
• 30:21 - 30:27
We can use exactly the same
method even if there are
• 30:27 - 30:33
fractions involved, so let's
have a look at an example where
• 30:33 - 30:38
there's some fractions. Suppose
we want to expand.
• 30:38 - 30:44
This time 1 + 2 over X, so
I've deliberately put a fraction
• 30:44 - 30:48
in there all to the power 3.
• 30:49 - 30:50
Let's see what happens.
• 30:51 - 30:58
1 + 2 over
X to the power
• 30:58 - 31:03
3. Well. We start
with one raised to the highest
• 31:03 - 31:06
power which is 1 to the power 3.
• 31:06 - 31:08
Which is still 1.
• 31:09 - 31:12
And once at, any power will
still be one's remove all the
• 31:12 - 31:13
way through the calculation.
• 31:15 - 31:21
Will have two over X raised
first of all to the power one.
• 31:21 - 31:27
Two over X to
the power 2.
• 31:28 - 31:32
And two over X to the power
three and we stop there. When we
• 31:32 - 31:34
reached the highest power.
• 31:34 - 31:37
Which corresponds to the power
in the original problem.
• 31:39 - 31:44
We need the coefficients of each
of these terms from pascals
• 31:44 - 31:47
triangle and the row in the
triangle beginning 13.
• 31:48 - 31:55
Those numbers are 1331, so
there's one of these three of
• 31:55 - 31:57
those. Three of those.
• 31:57 - 31:59
I'm one of those.
• 32:00 - 32:03
And all we need to do now is
tidy at what we've got.
• 32:04 - 32:05
So there's once.
• 32:07 - 32:11
Two over X to the power one
is just two over X. We're
• 32:11 - 32:15
going to multiply it by
three, so 3 twos are six will
• 32:15 - 32:16
have 6 divided by X.
• 32:18 - 32:23
Here there's a 2 squared, which
is 4. Multiply it by three so we
• 32:23 - 32:28
have 12 divided by X to the
power 2 divided by X squared.
• 32:29 - 32:32
And finally, there's 2 to the
• 32:32 - 32:34
power 3. Which is 8.
• 32:35 - 32:41
And this time it's divided by X
to the power 34X cubed.
• 32:41 - 32:45
So that's a simple example which
illustrates how we can apply
• 32:45 - 32:48
exactly the same technique even
when the refraction is involved.
• 32:49 - 32:57
Now, that's not quite
the end of the
• 32:57 - 33:03
story. The problem is, supposing
I were to ask you to expand a
• 33:03 - 33:06
binomial expression to a very
large power, suppose I wanted
• 33:06 - 33:11
one plus X to the power 32 or
one plus X to the power 127. You
• 33:11 - 33:14
have an awful lot of rows of
pascals triangle to generate if
• 33:14 - 33:17
you wanted to do it this way.
• 33:17 - 33:20
Fortunately, there's an
alternative way, and it involves
• 33:20 - 33:22
a theorem called the binomial
• 33:22 - 33:27
theorem. So let's just have
a look at what the binomial
• 33:27 - 33:27
theorem says.
• 33:29 - 33:36
The binomial theorem allows us
to develop an expansion of
• 33:36 - 33:43
the binomial expression A+B
raised to the power N.
• 33:44 - 33:49
And it allows us to get an
expansion in terms of
• 33:49 - 33:52
decreasing powers of a,
exactly as we've seen before.
• 33:53 - 33:57
And increasing powers of B
exactly as we've seen before.
• 33:57 - 34:02
And it I'm going to quote the
theorem for the case when N is a
• 34:02 - 34:03
positive whole number.
• 34:04 - 34:09
This theorem will actually work
when is negative and when it's a
• 34:09 - 34:11
fraction, but only under
exceptional circumstances, which
• 34:11 - 34:16
we're not going to discuss here.
So in all these examples, N will
• 34:16 - 34:18
be a positive whole number.
• 34:19 - 34:26
Now what
the theorem
• 34:26 - 34:29
says is
• 34:29 - 34:35
this. A+B to the power N is
given by the following expansion
• 34:35 - 34:37
A to the power N.
• 34:37 - 34:40
Now that looks familiar, doesn't
it? Because as in all the
• 34:40 - 34:43
examples we've seen before,
we've taken the first term and
• 34:43 - 34:45
raised it to the highest power.
The power in the original
• 34:45 - 34:47
question 8 to the power N.
• 34:48 - 34:52
Then there's a next term, and
the next term will have an A to
• 34:52 - 34:53
the power N minus one.
• 34:54 - 34:58
And a B in it. That's exactly as
we've seen before, because we're
• 34:58 - 35:01
starting to see the terms
involving be appear and the
• 35:01 - 35:03
powers event at the powers of a
• 35:03 - 35:07
a decreasing. We want a
coefficient in here and the
• 35:07 - 35:10
binomial theorem tells us that
the coefficient is NTH.
• 35:12 - 35:16
The next term.
• 35:16 - 35:19
As an A to the power
N minus two in it.
• 35:21 - 35:24
Along with the line we had
before of decreasing the powers
• 35:24 - 35:28
and increasing the power of be
will give us a B squared.
• 35:28 - 35:32
And the binomial theorem
tells us the coefficient to
• 35:32 - 35:36
right in here and the
coefficient this time is NN
• 35:36 - 35:38
minus one over 2 factorial.
• 35:39 - 35:44
In case you don't know what this
notation means, 2 factorial
• 35:44 - 35:46
means 2 * 1.
• 35:46 - 35:49
That's called 2 factorial.
• 35:50 - 35:58
And this series goes on and on
and on. The next term will be
• 35:58 - 36:04
NN minus one and minus two over
3 factorial, and there's a
• 36:04 - 36:10
pattern developing here. You
see, here we had an N&NN minus
• 36:10 - 36:13
one. And minus one and minus 2.
• 36:14 - 36:19
With a 3 factorial at the bottom
where we had a two factor at the
• 36:19 - 36:23
bottom before 3 factorial means
3 * 2 * 1.
• 36:24 - 36:29
The power of a will be 1 less
again, which this time will be A
• 36:29 - 36:31
to the N minus three.
• 36:31 - 36:35
And we want to power of bee
which is B to the power 3.
• 36:37 - 36:40
So all the way through this
theorem you'll see the powers of
• 36:40 - 36:45
a are decreasing. And the powers
of B are increasing. Now this
• 36:45 - 36:51
series goes on and on and on
until we reach the term B to the
• 36:51 - 36:56
power N. When it stops. So this
is a finite series. It stops
• 36:56 - 36:58
after a finite number of terms.
• 36:59 - 37:02
Now, the theorems often quoted
in this form, but it's also
• 37:02 - 37:04
often quoted in a slightly
• 37:04 - 37:08
simpler form. And it's quoted in
the form for which a is the
• 37:08 - 37:10
simple value of just one.
• 37:10 - 37:16
And B is X. Now when a is one,
all of these A to the power ends
• 37:16 - 37:22
or A to the N minus one A to the
N minus two. Each one of those
• 37:22 - 37:26
terms will just simplify to the
number one, so the whole thing
• 37:26 - 37:30
looks simpler. So let's write
down the binomial theorem again
• 37:30 - 37:33
for the special case when a is
one and these X.
• 37:34 - 37:40
This time will get one
plus X raised to the
• 37:40 - 37:43
power N. Is equal to.
• 37:44 - 37:45
1.
• 37:46 - 37:50
Plus N.
X.
• 37:52 - 37:56
Plus NN minus one over 2
factorial X squared, and you can
• 37:56 - 38:00
see what's happening. This
second term X is starting to
• 38:00 - 38:04
appear and and its powers
increasing as we move from left
• 38:04 - 38:08
to right. So even X&X squared
the next time will have an X
• 38:08 - 38:13
cubed in it, one to any power is
still one, so I don't actually
• 38:13 - 38:15
need to write it down.
• 38:16 - 38:22
The next term will
be NN minus one and
• 38:22 - 38:27
minus two over 3
factorial X cubed.
• 38:28 - 38:34
The next term will be NN minus
one and minus 2 N minus three
• 38:34 - 38:39
over 4 factorial X to the four,
and this will go on and on until
• 38:39 - 38:44
eventually you'll get to the
stage where you get to the last
• 38:44 - 38:49
term raised to the highest power
you'll get to X to the power N,
• 38:49 - 38:51
and the series will stop.
• 38:52 - 38:55
So this is a slightly simpler
form of the theorem, and it's
• 38:55 - 38:57
often quoted in this form.
• 38:57 - 39:03
Now let's use it to examine some
binomial expressions that you're
• 39:03 - 39:04
• 39:04 - 39:09
Let's suppose we want to expand
one plus X or raised to the
• 39:09 - 39:13
power two. Now I've written down
the theorem again so we can
• 39:13 - 39:18
refer to it and this is printed
in the notes. If you want to use
• 39:18 - 39:20
the one in the notes.
• 39:21 - 39:27
So we've one plus X to the power
N. In our problem, we've got one
• 39:27 - 39:34
plus X to the power two, so all
we have to do is let NB two in
• 39:34 - 39:35
all of this formula through
• 39:35 - 39:40
here. So let's see what we get
or from the theorem.
• 39:41 - 39:44
The first thing will write down
is just the one.
• 39:46 - 39:53
Then we want NX, but N IS
two, so will just put plus 2X.
• 39:53 - 39:57
And then the next term we want
is going to be a term
• 39:57 - 40:00
involving X to the power two,
but that's the highest power
• 40:00 - 40:03
we want because we've got a
power to in here. We want to
• 40:03 - 40:07
stop when we get to X to the
power two, so we're actually
• 40:07 - 40:11
already at the end with the
next term, and we just want an
• 40:11 - 40:13
X to the power two on its own.
• 40:15 - 40:18
1 + 2 X plus X squared and
that's the expansion that
• 40:18 - 40:21
with, and you'll notice in it
• 40:21 - 40:25
that the powers of X increase
as we move through from left to
• 40:25 - 40:29
right, and there's powers of
one in there, but we don't see
• 40:29 - 40:32
them, and the one 2 one other
numbers in pascals triangle.
• 40:34 - 40:40
Let's look at the theorem for
the case when is 3, let's expand
• 40:40 - 40:44
one plus X to the power 3.
• 40:45 - 40:47
I'm going to use the theorem
again, but this time we're
• 40:47 - 40:49
going to let NB 3.
• 40:51 - 40:55
So we want 1 + 3
• 40:55 - 41:01
X. And then
we want 3.
• 41:02 - 41:08
3 - 1 three minus
one is 2.
• 41:08 - 41:11
All divided by 2 factorial.
• 41:12 - 41:15
Than an X squared.
• 41:16 - 41:20
And then the next term will be a
term involving X cubed, which is
• 41:20 - 41:23
the term that we stop with
because we're only working 2X to
• 41:23 - 41:27
the power three here. So the
last term will be just a plus X
• 41:27 - 41:32
cubed. We can tie this up to 1
+ 3 X.
• 41:33 - 41:37
2 factorial is 2 * 1, which is
just two little cancel with the
• 41:37 - 41:42
two at the top, so will be left
with just three X squared, and
• 41:42 - 41:44
finally an X cubed.
• 41:44 - 41:47
And again, that's something that
• 41:47 - 41:50
with. You'll notice the
coefficients, the 1331 other
• 41:50 - 41:55
numbers we've seen many times in
pascals triangle the powers of X
• 41:55 - 42:01
increase. As we move from the
left to the right, and this is a
• 42:01 - 42:05
finite series, it stops when we
get to the term involving X
• 42:05 - 42:07
cubed corresponding to this
highest power over there.
• 42:08 - 42:15
Now.
That suppose we want to look at
• 42:15 - 42:19
it and more complicated problem.
Suppose we want to workout one
• 42:19 - 42:21
plus X to the power 32. Now you
• 42:21 - 42:25
would never. Use pascals
triangle to attempt this problem
• 42:25 - 42:29
because you'd have to generate
so many rows of the triangle,
• 42:29 - 42:31
but we can use the binomial
• 42:31 - 42:35
theorem. What I'm going to do
is I'm going to write down
• 42:35 - 42:38
the first three terms of the
series using the binomial
• 42:38 - 42:41
theorem, and I'm going to use
it with N being equal to 32.
• 42:43 - 42:50
So we're putting any 32 in. The
theorem will get 1 + 32 X.
• 42:50 - 42:53
That's the one plus the NX.
• 42:54 - 42:57
We want an which is 32.
• 42:58 - 43:01
And minus one which will be 31.
• 43:02 - 43:04
All over 2 factorial.
• 43:05 - 43:06
X squared
• 43:08 - 43:12
And we know that this series
will go on and on until we
• 43:12 - 43:16
reached the term, the last term
being X to the power 32.
• 43:17 - 43:21
But I only want to look at
the first three terms here in
• 43:21 - 43:25
this problem, so the first
three terms are just going to
• 43:25 - 43:29
be 1 + 32 X and we want to
simplify this. We've got 32 *
• 43:29 - 43:31
31 and then divided by two.
• 43:35 - 43:38
Which is 496.
• 43:38 - 43:43
And I just put some dots there
to show that this series goes on
• 43:43 - 43:47
a lot further than the terms
that I've just written down
• 43:47 - 43:53
there. I'm going to have a look
at a couple more examples with
• 43:53 - 43:58
some ingenuity. We can use the
theorem in a slightly different
• 43:58 - 44:03
form. Suppose we want to expand
this binomial expression this
• 44:03 - 44:09
time, I'm going to look at one
plus Y divided by 3. All raised
• 44:09 - 44:14
to the power 10 and suppose that
I'm interested. I'm interested
• 44:14 - 44:16
in generating the first.
• 44:17 - 44:23
Four terms. Let's see how we can
do that. Well, we've got our
• 44:23 - 44:29
theorem. I've written it down
again here for us in terms of
• 44:29 - 44:35
one plus X to the power N. We
can use it in this problem if we
• 44:35 - 44:37
replace every X.
• 44:37 - 44:40
In the theorem with a Y over 3.
• 44:41 - 44:45
So everywhere there's an X in
the theorem, I'm going to write
• 44:45 - 44:50
Y divided by three and then the
pattern will match exactly what
• 44:50 - 44:54
we have in the theorem ends
going to be 10 in this problem.
• 44:54 - 45:00
So let's see what we get will
have one plus Y over three
• 45:00 - 45:03
raised to the power 10 is equal
• 45:03 - 45:06
• 45:06 - 45:10
as always. Then we
• 45:10 - 45:14
want NX. And
• 45:14 - 45:19
it's 10. And we said that
instead of X, but replacing the
• 45:19 - 45:20
X with a Y over 3.
• 45:21 - 45:24
So we have a Y over three there.
• 45:25 - 45:30
What's the next term we want NN
minus one over 2 factorial?
• 45:31 - 45:33
Which is 10.
• 45:34 - 45:38
10 - 1 is 9 over 2
factorial.
• 45:40 - 45:45
And then we'd want an X squared.
So in this case we want X being
• 45:45 - 45:46
why over 3?
• 45:47 - 45:50
All square
• 45:52 - 45:55
I want to generate one more term
'cause I said I want to look for
• 45:55 - 45:59
four terms, so the next term is
going to be an which was 10.
• 46:00 - 46:02
N minus one which is 9.
• 46:03 - 46:07
And minus two, which is 8 and
this time over 3 factorial.
• 46:07 - 46:13
So I'm here NN minus one and
minus two over 3 factorial and
• 46:13 - 46:20
we want X cubed X is Y over
three, so we want why over 3
• 46:20 - 46:26
cubed. And the series goes on
and on. Let's just tidy up what
• 46:26 - 46:28
we've got. There's one.
• 46:29 - 46:33
That'll be 10, why over 3?
• 46:33 - 46:37
What if we got in here? Well,
there's a 3 square at the bottom
• 46:37 - 46:42
which is 9, and there's a 9 at
the top, so the three squared in
• 46:42 - 46:44
here is going to cancel with the
• 46:44 - 46:47
nine there. 2 factorial
• 46:47 - 46:48
Is just two.
• 46:49 - 46:55
And choosing to 10 is 5, so will
have five 5 squared.
• 46:56 - 47:02
And then this is a bit more
complicated. We've got a 3
• 47:02 - 47:05
factorial which is 3 * 2.
• 47:05 - 47:11
And three cubed. At the bottom
there, which is 3 * 3 * 3. Some
• 47:11 - 47:16
of this will cancel down. 3 * 3
will cancel, with the nine in
• 47:16 - 47:21
here. The two will cancel their
with the eight will have four
• 47:21 - 47:26
and let's see what we're left
with at the top will have 10 *
• 47:26 - 47:28
4, which is 40.
• 47:28 - 47:34
And at the bottom will have 3
* 3, which is 9.
• 47:35 - 47:38
And they'll be a Y cubed.
• 47:38 - 47:44
So altogether we've 1 + 10 Y
over 3 five Y squared, 40 over 9
• 47:44 - 47:50
Y cubed, and those are the first
four terms of a series which
• 47:50 - 47:55
will actually continue until you
get to a term involving the
• 47:55 - 48:01
highest power, which will be a Y
over 3 to the power 10.
• 48:01 - 48:04
So you can still use the theorem
in slightly different form if
• 48:04 - 48:08
you use a bit of ingenuity. Want
to look at one final example
• 48:08 - 48:15
before we finish? And this time
I want to look at the example 3
• 48:15 - 48:16
- 5 Z.
• 48:16 - 48:20
To the power 40 again, it's an
example where you wouldn't want
• 48:20 - 48:24
to use pascals triangle because
the power for teens too high and
• 48:24 - 48:27
you have too many rose to
• 48:28 - 48:31
I'm going to use the original
form of the theorem, the One I
• 48:31 - 48:33
have here in terms of A+B to the
• 48:33 - 48:37
power N. A will be 3.
• 48:39 - 48:43
Now B is a negative number be
will be minus five said.
• 48:45 - 48:46
Ends going to be 14.
• 48:47 - 48:50
But we can still use the
theorem. Let's see what
• 48:50 - 48:54
happens and in this problem
I'm just going to generate
• 48:54 - 48:55
the first three terms.
• 48:56 - 49:03
OK so A is 3 and we
want to raise the three.
• 49:03 - 49:05
To the highest power which is
• 49:05 - 49:10
14. So my first term is 3 to
the power 14.
• 49:12 - 49:15
My second term is this one.
• 49:15 - 49:16
Begins with an N.
• 49:18 - 49:21
The power in the original
expression, which was 14.
• 49:22 - 49:27
Multiplied by A
to the power N
• 49:27 - 49:29
minus one AS 3.
• 49:31 - 49:36
And we want to raise it to the
power N minus 114 - 1 is 30.
• 49:36 - 49:38
And we want to be.
• 49:40 - 49:42
B is minus five set.
• 49:42 - 49:47
So our second term looking ahead
is going to be negative because
• 49:47 - 49:49
of that minus five in there.
• 49:51 - 49:56
My third term and I'll stop
after the third term, his N,
• 49:56 - 49:57
which is 14.
• 49:58 - 50:03
And minus one which is 13
all over 2 factorial.
• 50:05 - 50:10
A to the power N minus two will
be 3 to the power N minus two
• 50:10 - 50:13
will be 14 - 2 which is 12.
• 50:14 - 50:17
And finally AB, which
was minus five said.
• 50:18 - 50:20
Raised to the power 2.
• 50:21 - 50:25
And we know this goes on and on
until we reach term instead to
• 50:25 - 50:29
the power 14. But we've only
written down the first three
• 50:29 - 50:32
terms there. Perhaps we should
just tidy it up a little bit.
• 50:32 - 50:35
There's a 3 to the power 14 at
• 50:35 - 50:43
the beginning. There's a minus
five here, minus 5. Four
• 50:43 - 50:46
teens are minus 70.
• 50:46 - 50:49
As a 3 to the power 13, let me
just leave it like that for the
• 50:49 - 50:51
time being. And then
they'll be as Ed.
• 50:54 - 50:57
Over here there's a 3 to the
• 50:57 - 51:00
power 12. That's this term in
• 51:00 - 51:06
here. And I'm reaching for my
Calculator again because this is
• 51:06 - 51:09
a bit more complicated. We have
got a 14.
• 51:10 - 51:11
Multiplied by 13.
• 51:12 - 51:16
When multiplied by 5 squared,
which is 25.
• 51:18 - 51:24
And divided by the two factorial
that's divided by two, this will
• 51:24 - 51:26
be multiplied by two 275.
• 51:26 - 51:31
And this expression will be
positive because we've got a
• 51:31 - 51:33
minus 5 squared.
• 51:34 - 51:39
And we need to remember to
include zed squared in there.
• 51:39 - 51:42
OK, we observe as before that
the powers of zed are increasing
• 51:42 - 51:46
as we move from the left to the
right. Now we could leave it
• 51:46 - 51:50
like that. I'm just going to
tidy it up and write it in a
• 51:50 - 51:53
slightly different form because
this is often the way you see
• 51:53 - 51:56
answers in the back of textbooks
or people ask you to give an
• 51:56 - 52:00
answer in a particular form and
the form I'm going to write it
• 52:00 - 52:04
in is one obtained by taking out
a factor of 3 to the power 40.
• 52:05 - 52:09
If I take her three to 14 out
from the first term, I'll be
• 52:09 - 52:10
just left with one.
• 52:11 - 52:14
Now 3 to the 13. In the second
• 52:14 - 52:19
term. But if I multiply top and
bottom by three, I'll have a 3
• 52:19 - 52:21
to the 14th at the top, which I
• 52:21 - 52:25
can take out. But have
multiplied the bottom by
• 52:25 - 52:30
three as well, which will
leave me with minus 70 Zedd
• 52:30 - 52:31
divided by three.
• 52:33 - 52:35
Here with a 3 to the power 12.
• 52:35 - 52:38
And I want to take out a 3 to
• 52:38 - 52:43
14. If I multiply the top and
bottom by three squared or nine,
• 52:43 - 52:46
I affectively get a 3 to 14 in
• 52:46 - 52:51
this term. So I'm multiplying
top and bottom by 9, taking
• 52:51 - 52:56
the three to the 14 out, and
that will leave me here with
• 52:56 - 52:58
two 275 over 9.
• 52:59 - 53:04
continues. As I said before,
• 53:04 - 53:09
until you get to a term
involving zed to the power 14,
• 53:09 - 53:12
but those are the first three
terms of the series.
Title:
www.mathcentre.ac.uk/.../Pascal.mp4
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 mathcentre edited English subtitles for www.mathcentre.ac.uk/.../Pascal.mp4