### Video Transcript

In this video, weβll learn how to
use Pascalβs triangle to find the coefficients of the algebraic expansion of any
binomial expression of the form π plus π to the πth power, where π is a positive
integer. Weβre going to begin by looking for
some patterns by distributing a few expressions.

Letβs begin by distributing the
expression π plus π to the zeroth power. Now, of course, any number raised
to the power of zero is one. So we can say that π plus π to
the zeroth power is simply one. And what about π plus π to the
first power? Well, of course, any number raised
to the power of one is simply the original number, so we get π plus π. The coefficients of both π and π
are one. Then, for exponents greater than or
equal to two, things start to get a little interesting. π plus π squared is π plus π
times π plus π. And distributing these parentheses
either using the FOIL or grid method and we get π squared plus two ππ plus π
squared. This time, the coefficients of π
squared and π squared are one and then we have a third coefficient. The coefficient of ππ is two.

Next, we consider the case where
the exponent is three. π plus π cubed is the same as
multiplying π plus π squared by π plus π. And distributing gives us π cubed
plus three π squared π plus three ππ squared plus π cubed. This time, our coefficients are
one, three, three, and then one. Weβll look at one more example
where the exponent is equal to four. This time, thatβs the result of
multiplying π plus π cubed by π plus π. When we expand these parentheses,
we get π to the fourth power plus four π cubed π plus six π squared π squared
plus four ππ cubed plus π to the fourth power. The coefficients are one, four,
six, four, and one.

Now, of course, should we wish to
distribute the expression π plus π to the 10th power, weβd probably be here all
day. And so instead, we need to find a
shortcut. Letβs list our expressions out one
on top of the other and look for some patterns. So what do you notice? Firstly, letβs look at the outer
edges of our triangle. These are quite
straightforward. The terms all have a coefficient of
one. And on the left edge, we simply
have the term π raised to the power of the exponent. On the right, we still have the
coefficient of one. And, this time, itβs the π raised
to the power of the exponent. So what else is happening? Well, we might look for some more
patterns by looking at different diagonals, or we can look at the terms in each
individual expression.

Letβs take π plus π to the fourth
power. Notice how the exponent of π
reduces by one each time. The exponent of π, however,
increases by one each time. And we might also notice that the
sum of their exponents is equal to four, which is the exponent weβre raising π plus
π to the power of. But whatβs happening with the
coefficients? Well, this bit is really
interesting. Letβs have a look at listing the
coefficients themselves in a triangle. Once again, we spot that the outer
edges of the triangle are all one, but we can see that the other digits are made up
by finding the sum of the two immediately above them. So, for example, one plus two is
equal to three, three plus three is equal to six, and so on. This triangle has a special
name. Itβs called Pascalβs triangle, and
itβs named after a French mathematician.

Letβs find the next row on the
triangle by using the patterns weβve seen. We know that weβre going to have
one on the outer edges. And then we find the other values
by adding the numbers directly above. One plus four is five, four plus
six is 10, six plus four is 10, and four plus one is five. Now, since Pascalβs triangle is
fairly easy to list out for small values of π, it can be really useful in helping
us to distribute parentheses of the form π plus π to the πth power. Letβs combine everything weβve seen
to find a general method for this.

To distribute parentheses of the
form π plus π to the πth power for positive integer values of π, we move through
the terms from left to right. The exponent of π begins with π
and then reduces by one each time until we get to π to the power of zero. Now, of course, π to the power of
zero is simply one. We do the reverse for the π. The exponent of π increases by one
each time, and we start off with π to the power of zero and we finish at π to the
πth power. Then we know that the coefficients
of each term are the numbers which appear in the π equals π row of Pascalβs
triangle. Now youβll see in a moment why I
havenβt written πth row because itβs not technically the πth row.

Letβs have a look at how we might
use this approach to expand a binomial expression.

Use Pascalβs triangle to evaluate
π₯ plus four to the fifth power.

Remember, the process for
distributing binomial expressions of the form π plus π to the πth power for
positive integer values of π is as follows. We move through the terms from left
to right. The exponent of π begins with π
and decreases by one each time until we get π to the power of zero. Then, for the π part, we begin
with an exponent of zero and we increase by one each time until we reach π to the
πth power. Then we find the coefficients for
each term by looking for the numbers which appear in the π equals π row of
Pascalβs triangle. If we compare the general form with
our binomial, we see that weβre going to let π be equal to π₯, π be equal to four,
and then we let π be equal to five. And in fact, weβre going to use a
table to ensure that we donβt lose any terms.

We might notice that there will be
one more column in our table than the value of π. So here weβre looking for six
columns. We move from left to right, and we
begin with π to the πth power. Well, π is π₯ and π is five. So we have π₯ to the fifth
power. We then reduce this power by one
each time, giving us π₯ to the fourth power, π₯ cubed, π₯ squared, π₯ to the power
of one, and π₯ to the power of zero. Now, π₯ to the power of one is just
π₯ and π₯ to the power of zero is one. And so we now have the π₯-part of
each term. Weβll now look at the four
part. We begin with π to the power of
zero. So here, thatβs four to the power
of zero. Then we increase that power by one
each time, and that gives us four to the power of one, four squared, four cubed,
four to the fourth power, and four to the fifth power.

Now, once again, we can, in fact,
replace four to the zeroth power with one and four to the power of one is just
four. This third row Iβve called π, and
thatβs the coefficient for each term. So letβs list out the first few
rows of Pascalβs triangle. Pascalβs triangle looks like
this. The first row is when π is equal
to zero; we have one. When π is equal to one, we have
the second row; we have one, one. When π is equal to two, we have
one, two, one. And we continue by keeping ones on
the outside and then adding the numbers above to get the next one. So one plus four is equal to five
and so on.

Notice that we stop at π is equal
to five, since thatβs our exponent. But this isnβt the fifth row. It is in fact the sixth row. And so, of course, itβs worth being
really careful when we define Pascalβs triangle. The coefficients are therefore one,
five, 10, 10, five, and one. Then, to find the terms in our
expansion, we multiply the terms in each individual column. And so the first term is π₯ to the
fifth power times one times one, which is simply π₯ to the fifth power. Next, we multiply each term in the
second column. So thatβs π₯ to the fourth power
times four times five, which is 20π₯ to the fourth power.

We continue in this manner, so our
third term is π₯ cubed times four squared times 10, which is 160π₯ cubed. Then we have π₯ squared times four
cubed times 10, which is 640π₯ squared. Multiplying the terms in our second
to last column gives us 1280π₯. And then our sixth and final term
is 1024. And so we find π₯ plus four to the fifth power can be written as π₯ to the
fifth power plus 20π₯ to the fourth power plus 160π₯ cubed plus 640π₯ squared plus
1280π₯ plus 1024.

Weβll now consider a second
example.

Use Pascalβs triangle to expand the
expression π₯ plus one over π₯ to the fourth power.

Remember, to distribute parentheses
of the form π plus π to the πth power for positive integer values of π, we move
from left to right. Our exponent of π begins with π
and reduces by one each time until we reach π to the power of zero. We do the reverse with π. We increase the exponent by one
each time, starting with π to the power of zero until we get to π to the πth
power. Then the coefficients of each term
are the numbers that appear in the π equals π row of Pascalβs triangle. And we remember that this is
actually the π plus oneth row.

Comparing our expression with the
general form, weβre going to let π be equal to π₯, π be equal to one over π₯, and
π, our exponent, is equal to four. Now weβre going to represent this
in table form, and we also recall that there will be π plus one terms in the
expansion. And so weβre going to need five
columns in our table. Letβs begin by dealing with the
π₯. We know we start with π₯ to the
πth power, so π₯ to the fourth power. Then we reduce that exponent by one
each time, remembering of course that π₯ to the power of one is π₯ and π₯ to the
power of zero is one.

Then, for our one over π₯ term, we
start off with an exponent of zero and then increase that by one each time. Once again, we know that the power
of zero just leaves us with one and one over π₯ to the first power is one over
π₯. We can then distribute our exponent
over each term in our fractions and we get one over π₯ squared, one over π₯ cubed,
and one over π₯ to the fourth power.

Letβs now consider the
coefficients. Weβre going to draw out the first
few rows of Pascalβs triangle. In fact, weβre going to draw the π
equals four row, which will actually be the four plus one, the fifth row, in the
triangle. We recall that Pascalβs triangle
looks like this. We have oneβs going down the outer
edges. And then to find the remaining
values, we add the two numbers directly above. So one plus three is equal to four,
three plus three is equal to six, and so on. And then the coefficients are one,
four; six, four; and one. Note that we know that weβve
probably chosen the correct row because we donβt have a surplus of numbers, nor do
we have any gaps.

To find each of our respective
terms in the expansion of π₯ plus one over π₯ to the fourth power, we multiply each
term in each column. So we do π₯ to the fourth power
times one times one, which is π₯ to the fourth power, then π₯ cubed times one over
π₯ times four, which simplifies to four π₯ squared. We then multiply the terms in the
third, the fourth, and the fifth column, and this simplifies quite nicely. π₯ plus one over π₯ to the fourth
power is therefore equal to π₯ to the fourth power plus four π₯ squared plus six
plus four over π₯ squared plus one over π₯ to the fourth power. Note that in this case, due to the
nature of π and π, we actually end up with a constant term of six despite the fact
that both π and π are algebraic expressions.

Now, we might not always want to
list out every single row of Pascalβs triangle. So instead, weβre going to find a
shortcut. We know that for a binomial π plus
π to the πth power, where π is a positive integer or a natural number, the
expansion becomes π naught times π to the πth power times π to the power of zero
plus π one times π to the power of π minus one times π to the power of one and
so on, all the way down to π π times π to the power of zero times π to the power
of π. In this case, this π subscript
πβs are from the π plus oneth row of Pascalβs triangle. We decrease the power of π each
time, and we increase the power of π. But we can actually redefine the
numbers π naught, π one, all the way up to π π.

Letβs go back to the expansion of
π plus π to the fourth power, for example. Thinking about this second term,
the coefficient four represents the number of ways we can choose a single π from
our four sets of parentheses π plus π. But we know we can define this as
four π one or four choose one. Similarly, if we think about this
third term, the coefficient of six here represents the number of ways we can choose
two πβs from our four sets of parentheses. And we could define that as four
choose two. And this means we can now redefine
each of our coefficients as π choose zero, π choose one, π choose two, all the
way up to π choose π, where π choose π is π factorial over π factorial times
π minus π factorial. Now, since π choose zero and π
choose π are simply one, we might rewrite this as shown. Letβs see how we can actually apply
this formula.

Expand the expression three plus π₯
to the fourth power.

Here we have a binomial raised to
an integer power, and so we can use the binomial theorem. This says that π plus π to the
πth power, where π is a positive integer, can be written as π to the power of π
plus π choose one π to the power of π minus one π plus π choose two times π to
the power of π minus two π squared, all the way up to π to the πth power. We notice that the power of π
reduces by one each time and the power of π increases by one each time. And so we compare our expression
with the general form. And we see weβre going to let π be
equal to three, weβre going to let π be equal to π₯, and finally π is equal to
four.

So letβs see if we can use this to
expand our expression. The first term is π to the πth
power. So thatβs three to the fourth
power. We then have π choose one, so
thatβs four choose one times three to the third power. And remember, weβre reducing these
exponents and we multiply that by π₯. Then, our next term is four choose
two. We reduce the power of three and
increase the power of π₯. So we get times three squared times
π₯ squared. Our fourth term is four choose
three times three times π₯ cubed. And actually, this is our final
term. So itβs π to the πth power, which
is π₯ to the fourth power.

Notice that the number of terms we
have will always be one more than the exponent. So we have five terms here. Letβs simplify these terms. Three to the fourth power is
81. Then, four choose one is four and
three cubed is 27. And so our next term is 108π₯. Four choose two is six, and so our
third term is six times nine times π₯ squared, which is 54π₯ squared. Four choose three is four, and so
our fourth term is 12π₯ cubed, and our fifth and final term is still π₯ to the
fourth power. Three plus π₯ to the fourth power
is 81 plus 108π₯ plus 54π₯ squared plus 12π₯ cubed plus π₯ to the fourth power.

Note that we can also identify a
specific term by using the general term of the binomial theorem. Itβs π choose π times π to the
power of π minus π times π to the power of π. And so say we want to define the
fourth term in this expansion. Weβd let π be equal to three. And we see once again we get four
choose three times three to the power of four minus three times π₯ cubed, which once
again gives us 12π₯ cubed. This can be a nice little
shortcut.

Weβll now consider the key points
from this lesson. In this lesson, we saw that we can
quickly reproduce Pascalβs triangle for small values of π to easily expand
binomials of the form π plus π to the πth power. Alternatively, especially when π
is large, we can use the binomial theorem. And that says for integer values of
π, π plus π to the πth power is π to the πth power plus π choose one π to
the power of π minus one π and so on, all the way up to π to the πth power. But if we want to identify a
specific term, we can use the general term, and that is π choose π times π to the
power of π minus π times π to the power of π.