Lesson Video: Pascal’s Triangle and the Binomial Theorem Mathematics

In this video, we will learn how to use Pascal’s triangle to find the coefficients of the algebraic expansion of any binomial expression of the form (π‘Ž + 𝑏)^(𝑛).

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Video Transcript

In this video, we’ll learn how to use Pascal’s triangle to find the coefficients of the algebraic expansion of any binomial expression of the form π‘Ž plus 𝑏 to the 𝑛th power, where 𝑛 is a positive integer. We’re going to begin by looking for some patterns by distributing a few expressions.

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

Next, we consider the case where the exponent is three. π‘Ž plus 𝑏 cubed is the same as multiplying π‘Ž plus 𝑏 squared by π‘Ž plus 𝑏. And distributing gives us π‘Ž cubed plus three π‘Ž squared 𝑏 plus three π‘Žπ‘ squared plus 𝑏 cubed. This time, our coefficients are one, three, three, and then one. We’ll look at one more example where the exponent is equal to four. This time, that’s the result of multiplying π‘Ž plus 𝑏 cubed by π‘Ž plus 𝑏. When we expand these parentheses, we get π‘Ž to the fourth power plus four π‘Ž cubed 𝑏 plus six π‘Ž squared 𝑏 squared plus four π‘Žπ‘ cubed plus 𝑏 to the fourth power. The coefficients are one, four, six, four, and one.

Now, of course, should we wish to distribute the expression π‘Ž plus 𝑏 to the 10th power, we’d probably be here all day. And so instead, we need to find a shortcut. Let’s list our expressions out one on top of the other and look for some patterns. So what do you notice? Firstly, let’s 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‘Ž raised to the power of the exponent. On the right, we still have the coefficient of one. And, this time, it’s the 𝑏 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.

Let’s take π‘Ž plus 𝑏 to the fourth power. Notice how the exponent of π‘Ž reduces by one each time. The exponent of 𝑏, 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 we’re raising π‘Ž plus 𝑏 to the power of. But what’s happening with the coefficients? Well, this bit is really interesting. Let’s 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. It’s called Pascal’s triangle, and it’s named after a French mathematician.

Let’s find the next row on the triangle by using the patterns we’ve seen. We know that we’re 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 Pascal’s triangle is fairly easy to list out for small values of 𝑛, it can be really useful in helping us to distribute parentheses of the form π‘Ž plus 𝑏 to the 𝑛th power. Let’s combine everything we’ve seen to find a general method for this.

To distribute parentheses of the form π‘Ž plus 𝑏 to the 𝑛th power for positive integer values of 𝑛, we move through the terms from left to right. The exponent of π‘Ž begins with 𝑛 and then reduces by one each time until we get to π‘Ž to the power of zero. Now, of course, π‘Ž to the power of zero is simply one. We do the reverse for the 𝑏. The exponent of 𝑏 increases by one each time, and we start off with 𝑏 to the power of zero and we finish at 𝑏 to the 𝑛th power. Then we know that the coefficients of each term are the numbers which appear in the 𝑛 equals 𝑛 row of Pascal’s triangle. Now you’ll see in a moment why I haven’t written 𝑛th row because it’s not technically the 𝑛th row.

Let’s have a look at how we might use this approach to expand a binomial expression.

Use Pascal’s triangle to evaluate π‘₯ plus four to the fifth power.

Remember, the process for distributing binomial expressions of the form π‘Ž plus 𝑏 to the 𝑛th power for positive integer values of 𝑛 is as follows. We move through the terms from left to right. The exponent of π‘Ž begins with 𝑛 and decreases by one each time until we get π‘Ž to the power of zero. Then, for the 𝑏 part, we begin with an exponent of zero and we increase by one each time until we reach 𝑏 to the 𝑛th power. Then we find the coefficients for each term by looking for the numbers which appear in the 𝑛 equals 𝑛 row of Pascal’s triangle. If we compare the general form with our binomial, we see that we’re going to let π‘Ž be equal to π‘₯, 𝑏 be equal to four, and then we let 𝑛 be equal to five. And in fact, we’re going to use a table to ensure that we don’t lose any terms.

We might notice that there will be one more column in our table than the value of 𝑛. So here we’re looking for six columns. We move from left to right, and we begin with π‘Ž to the 𝑛th power. Well, π‘Ž is π‘₯ and 𝑛 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. We’ll now look at the four part. We begin with 𝑏 to the power of zero. So here, that’s 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 I’ve called 𝑐, and that’s the coefficient for each term. So let’s list out the first few rows of Pascal’s triangle. Pascal’s triangle looks like this. The first row is when 𝑛 is equal to zero; we have one. When 𝑛 is equal to one, we have the second row; we have one, one. When 𝑛 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 𝑛 is equal to five, since that’s our exponent. But this isn’t the fifth row. It is in fact the sixth row. And so, of course, it’s worth being really careful when we define Pascal’s 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 that’s 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.

We’ll now consider a second example.

Use Pascal’s triangle to expand the expression π‘₯ plus one over π‘₯ to the fourth power.

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

Comparing our expression with the general form, we’re going to let π‘Ž be equal to π‘₯, 𝑏 be equal to one over π‘₯, and 𝑛, our exponent, is equal to four. Now we’re going to represent this in table form, and we also recall that there will be 𝑛 plus one terms in the expansion. And so we’re going to need five columns in our table. Let’s begin by dealing with the π‘₯. We know we start with π‘₯ to the 𝑛th 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.

Let’s now consider the coefficients. We’re going to draw out the first few rows of Pascal’s triangle. In fact, we’re going to draw the 𝑛 equals four row, which will actually be the four plus one, the fifth row, in the triangle. We recall that Pascal’s triangle looks like this. We have one’s 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 we’ve probably chosen the correct row because we don’t 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‘Ž and 𝑏, we actually end up with a constant term of six despite the fact that both π‘Ž and 𝑏 are algebraic expressions.

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

Let’s go back to the expansion of π‘Ž plus 𝑏 to the fourth power, for example. Thinking about this second term, the coefficient four represents the number of ways we can choose a single 𝑏 from our four sets of parentheses π‘Ž plus 𝑏. But we know we can define this as four 𝑐 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 𝑏’s 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 𝑛 choose zero, 𝑛 choose one, 𝑛 choose two, all the way up to 𝑛 choose 𝑛, where 𝑛 choose π‘Ÿ is 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. Now, since 𝑛 choose zero and 𝑛 choose 𝑛 are simply one, we might rewrite this as shown. Let’s 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‘Ž plus 𝑏 to the 𝑛th power, where 𝑛 is a positive integer, can be written as π‘Ž to the power of 𝑛 plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 plus 𝑛 choose two times π‘Ž to the power of 𝑛 minus two 𝑏 squared, all the way up to 𝑏 to the 𝑛th power. We notice that the power of π‘Ž reduces by one each time and the power of 𝑏 increases by one each time. And so we compare our expression with the general form. And we see we’re going to let π‘Ž be equal to three, we’re going to let 𝑏 be equal to π‘₯, and finally 𝑛 is equal to four.

So let’s see if we can use this to expand our expression. The first term is π‘Ž to the 𝑛th power. So that’s three to the fourth power. We then have 𝑛 choose one, so that’s four choose one times three to the third power. And remember, we’re 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 it’s 𝑏 to the 𝑛th 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. Let’s 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. It’s 𝑛 choose π‘Ÿ times π‘Ž to the power of 𝑛 minus π‘Ÿ times 𝑏 to the power of π‘Ÿ. And so say we want to define the fourth term in this expansion. We’d let π‘Ÿ 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.

We’ll now consider the key points from this lesson. In this lesson, we saw that we can quickly reproduce Pascal’s triangle for small values of 𝑛 to easily expand binomials of the form π‘Ž plus 𝑏 to the 𝑛th power. Alternatively, especially when 𝑛 is large, we can use the binomial theorem. And that says for integer values of 𝑛, π‘Ž plus 𝑏 to the 𝑛th power is π‘Ž to the 𝑛th power plus 𝑛 choose one π‘Ž to the power of 𝑛 minus one 𝑏 and so on, all the way up to 𝑏 to the 𝑛th power. But if we want to identify a specific term, we can use the general term, and that is 𝑛 choose π‘Ÿ times π‘Ž to the power of 𝑛 minus π‘Ÿ times 𝑏 to the power of π‘Ÿ.

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