Explainer: Pascalโ€™s Triangle and the Binomial Theorem

In this explainer, we will learn how to use Pascalโ€™s triangle to find the coefficients of the algebraic expansion of any binomial expression of the form (๐‘Ž+๐‘)๏Š.

We begin by considering the expansions of (๐‘Ž+๐‘)๏Š for consecutive powers of ๐‘›, starting with ๐‘›=0. Since any number raised to the power of zero equals 1 (note that we are using the convention that 0=1๏Šฆ),(๐‘Ž+๐‘)=1.๏Šฆ

Similarly, when ๐‘›=1, we have a somewhat trivial case: (๐‘Ž+๐‘)=๐‘Ž+๐‘.๏Šง

However, for ๐‘›โ‰ฅ2, things get a little more interesting. Using familiar algebra, we know (๐‘Ž+๐‘)=๐‘Ž+2๐‘Ž๐‘+๐‘.๏Šจ๏Šจ๏Šจ

We now consider ๐‘›=3 for this case. We will use our answer from ๐‘›=2 to write the expansions as follows: (๐‘Ž+๐‘)=(๐‘Ž+๐‘)(๐‘Ž+๐‘)=(๐‘Ž+๐‘)๏€น๐‘Ž+2๐‘Ž๐‘+๐‘๏….๏Šฉ๏Šจ๏Šจ๏Šจ

Expanding the parentheses, we have (๐‘Ž+๐‘)=๐‘Ž+2๐‘Ž๐‘+๐‘Ž๐‘+๐‘Ž๐‘+2๐‘Ž๐‘+๐‘=๐‘Ž+3๐‘Ž๐‘+3๐‘Ž๐‘+๐‘.๏Šฉ๏Šฉ๏Šจ๏Šจ๏Šจ๏Šจ๏Šฉ๏Šฉ๏Šจ๏Šจ๏Šฉ

Similarly, we can find the expansion of (๐‘Ž+๐‘)๏Šช using the expansion of (๐‘Ž+๐‘)๏Šฉ as follows: (๐‘Ž+๐‘)=(๐‘Ž+๐‘)(๐‘Ž+๐‘)=(๐‘Ž+๐‘)๏€น๐‘Ž+3๐‘Ž๐‘+3๐‘Ž๐‘+๐‘๏….๏Šช๏Šฉ๏Šฉ๏Šจ๏Šจ๏Šฉ

We can now expand the parentheses to get (๐‘Ž+๐‘)=๐‘Ž+3๐‘Ž๐‘+3๐‘Ž๐‘+๐‘Ž๐‘+๐‘Ž๐‘+3๐‘Ž๐‘+3๐‘Ž๐‘+๐‘=๐‘Ž+4๐‘Ž๐‘+6๐‘Ž๐‘+4๐‘Ž๐‘+๐‘.๏Šช๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช

As you can see, if we were to try to find the expansion of (๐‘Ž+๐‘)๏Šญ this way, it could take a serious amount of time and effort. What we need is a better method that generalizes to higher powers. To find such a method, we will first look for patterns which might help us. We begin by organizing the expansions of (๐‘Ž+๐‘)๏Š one above the other to see if we can see any patterns.

Some of the most obvious patterns we notice are related to the diagonals: the coefficients of the terms in the first diagonal only contain ones, whereas the coefficients in the second diagonal contain consecutive integers.

Furthermore, we can see there is reflectional symmetry about the center.

Furthermore, we notice that on any given row, the sum of the indices equals ๐‘›. For example, on the row representing ๐‘›=4, the second term is 4๐‘Ž๐‘๏Šฉ. The index of ๐‘Ž is 3 and the index of ๐‘ is 1. Hence, their sum is equal to 4.

Finally, we see there is a relation between the coefficients on consecutive rows: if we add the two coefficients in the row above, we get the coefficient in the following row.

The triangle which makes up the binomial coefficients is generally referred to as Pascalโ€™s triangle.

Pascalโ€™s Triangle

Pascalโ€™s triangle is a triangular array of the binomial coefficients. The rows are enumerated from the top such that the first row is numbered ๐‘›=0. Similarly, the elements of each row are enumerated from ๐‘˜=0 up to ๐‘›. The first eight rows of Pascalโ€™s triangle are shown below.

Although, in much of the Western world, the triangle is named after the French mathematician Blaise Pascal, it was, in fact, well known to mathematicians centuries before in places such as China, Persia, and India.

Pascalโ€™s triangle is easy to reproduce for small ๐‘› and is therefore extremely helpful in expanding binomials with moderate powers. Later, we will see how its properties give us a method to expand general binomials.

We need to be careful to differentiate between referring to rows using ordinals, such as first row and second row, and referring to them using the row number ๐‘›: when we say the second row, we are referring to the row for which ๐‘›=1.

Example 1: Using Pascalโ€™s Triangle to Find Binomial Coefficients

Daniel has been exploring the relationship between Pascalโ€™s triangle and the binomial expansion. He has noticed that each row of Pascalโ€™s triangle can be used to determine the coefficients of the binomial expansion of (๐‘ฅ+๐‘ฆ)๏Š, as shown in the figure. For example, the fifth row of Pascalโ€™s triangle can be used to determine the coefficients of the expansion of (๐‘ฅ+๐‘ฆ)๏Šช.

  1. By calculating the next row of Pascalโ€™s triangle, find the coefficients of the expansion of (๐‘ฅ+๐‘ฆ)๏Šฌ.
  2. Daniel now wants to calculate the coefficients for each of the terms of the expansion (2๐‘ฅ+๐‘ฆ)๏Šช. By substituting 2๐‘ฅ into the expression above, or otherwise, calculate all of the coefficients of the expansion.

Answer

Part 1

To calculate the seventh row of Pascalโ€™s triangle, we start by writing out the sixth row. Then, since all rows start with the number 1, we can write this down. We can then add each consecutive pair of elements of the sixth row and write their sum in the gap beneath them. We will demonstrate this process below.

Starting with the first pair of terms, 1 and 5, we add them together to get 6 and place it into the space gap beneath them.

Moving onto the next pair of terms, we have 5+10=15, which we similarly add to the row.

Now we consider the middle terms 10+10=20.

Finally, we can use the symmetry of Pascalโ€™s triangle to write the rest of the row.

Since the elements of Pascalโ€™s triangle are the binomial coefficients, we can state that the coefficients of the terms of the expansion of (๐‘ฅ+๐‘ฆ)๏Šฌ will be 1, 6, 15, 20, 15, 6, and 1 respectively.

Part 2

Since (๐‘ฅ+๐‘ฆ)=๐‘ฅ+4๐‘ฅ๐‘ฆ+6๐‘ฅ๐‘ฆ+4๐‘ฅ๐‘ฆ+๐‘ฆ,๏Šช๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช we can substitute 2๐‘ฅ for ๐‘ฅ and write (2๐‘ฅ+๐‘ฆ)=(2๐‘ฅ)+4(2๐‘ฅ)๐‘ฆ+6(2๐‘ฅ)๐‘ฆ+4(2๐‘ฅ)๐‘ฆ+๐‘ฆ.๏Šช๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช

Simplifying, we have (2๐‘ฅ+๐‘ฆ)=2๐‘ฅ+4ร—2๐‘ฅ๐‘ฆ+6ร—2๐‘ฅ๐‘ฆ+4ร—2๐‘ฅ๐‘ฆ+๐‘ฆ=16๐‘ฅ+32๐‘ฅ๐‘ฆ+24๐‘ฅ๐‘ฆ+8๐‘ฅ๐‘ฆ+๐‘ฆ.๏Šช๏Šช๏Šช๏Šฉ๏Šฉ๏Šจ๏Šจ๏Šจ๏Šฉ๏Šช๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช

Therefore, the coefficients for each of the terms of the expansion (2๐‘ฅ+๐‘ฆ)๏Šช are 16, 32, 24, 8, and 1.

Example 2: Using Pascalโ€™s Triangle to Find Binomial Coefficients

Michael knows that he can use the 6th row of Pascalโ€™s triangle to calculate the coefficients of the expansion (๐‘Ž+๐‘)๏Šซ.

  1. Calculate the numbers in the 6th row of Pascalโ€™s triangle and, hence, write out the coefficients of the expansion (๐‘Ž+๐‘)๏Šซ.
  2. Now, by considering the different powers of ๐‘Ž and ๐‘ and using Pascalโ€™s triangle, work out the coefficients of the expansion (2๐‘Žโˆ’2๐‘)๏Šซ.

Answer

Part 1

Recall that we can write out the rows of Pascalโ€™s triangle by pairwise adding the terms in the previous rows. Therefore, starting from the first and second rows, which only contain ones, we can create the third rows by adding consecutive terms, as shown in the figure below.

Similarly, we can write the other rows using the same method, until we get to the sixth row.

Since the elements of Pascalโ€™s triangle are the binomial coefficients, we can state that the coefficients of the terms of the expansion of (๐‘Ž+๐‘)๏Šซ will be 1, 5, 10, 10, 5, and 1 respectively.

Part 2

To find the coefficients of the terms in the expansion of (2๐‘Žโˆ’2๐‘)๏Šซ, we can first factor the 2 out of the parentheses as follows: (2๐‘Žโˆ’2๐‘)=2(๐‘Žโˆ’๐‘).๏Šซ๏Šซ๏Šซ

We can now substitute โˆ’๐‘ for ๐‘ in the expansion, (๐‘Ž+๐‘)=๐‘Ž+5๐‘Ž๐‘+10๐‘Ž๐‘+10๐‘Ž๐‘+5๐‘Ž๐‘+๐‘,๏Šซ๏Šซ๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช๏Šซ to get (๐‘Žโˆ’๐‘)=๐‘Ž+5๐‘Ž(โˆ’๐‘)+10๐‘Ž(โˆ’๐‘)+10๐‘Ž(โˆ’๐‘)+5๐‘Ž(โˆ’๐‘)+(โˆ’๐‘).๏Šซ๏Šซ๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช๏Šซ

We can simplify this expression as follows: (๐‘Žโˆ’๐‘)=๐‘Žโˆ’5๐‘Ž๐‘+(โˆ’1)10๐‘Ž๐‘+(โˆ’1)10๐‘Ž๐‘+(โˆ’1)5๐‘Ž๐‘+(โˆ’1)๐‘=๐‘Žโˆ’5๐‘Ž๐‘+10๐‘Ž๐‘โˆ’10๐‘Ž๐‘+5๐‘Ž๐‘โˆ’๐‘.๏Šซ๏Šซ๏Šช๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šช๏Šช๏Šซ๏Šซ๏Šซ๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช๏Šซ

Therefore, (2๐‘Žโˆ’2๐‘)=2๏€น๐‘Žโˆ’5๐‘Ž๐‘+10๐‘Ž๐‘โˆ’10๐‘Ž๐‘+5๐‘Ž๐‘โˆ’๐‘๏…=32๐‘Žโˆ’160๐‘Ž๐‘+320๐‘Ž๐‘โˆ’320๐‘Ž๐‘+160๐‘Ž๐‘โˆ’32๐‘.๏Šซ๏Šซ๏Šซ๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช๏Šซ๏Šซ๏Šช๏Šฉ๏Šจ๏Šจ๏Šฉ๏Šช๏Šซ

Hence, the coefficients for each of the terms of the expansion (2๐‘Žโˆ’2๐‘)๏Šซ are 32, โˆ’160, 320, โˆ’320, 160, and โˆ’32.

Although using Pascalโ€™s triangle can seriously simplify finding binomial expansions for powers of ๐‘› up to around 10, much beyond this point it becomes impractical. It would, therefore, be helpful to see if there is a connection between consecutive elements in the rows of Pascalโ€™s triangle.

As an example, let us consider the ninth row of Pascalโ€™s triangle (i.e., the row labeled ๐‘›=8). We consider the multipliers taking us from one element to the next. The figure represents this.

We can see clearly that there is a pattern linking one element to the next. In fact, we can express this in a general way as follows: to move from the (๐‘˜โˆ’1)th element to the ๐‘˜th, we multiply by ๐‘›โˆ’๐‘˜+1 and divide by ๐‘˜. This rule does not only apply to the ninth row but also generalizes to any row of Pascalโ€™s triangle. Using this fact, we can expand binomials with arbitrarily large exponents.

Connection between Consecutive Terms in the Same Row of Pascalโ€™s Triangle

The connection between consecutive elements in the (๐‘›+1)th row (which by convention we enumerate by ๐‘›) in Pascalโ€™s triangle is as follows: to move from the (๐‘˜โˆ’1)th element to the ๐‘˜th, we multiply by ๐‘›โˆ’๐‘˜+1๐‘˜.

The next couple of examples will demonstrate this fact.

Example 3: Using Pascalโ€™s Triangle to Find Binomial Expansions

Write the first 5 terms of the expansion of (2+๐‘ฅ)๏Šง๏Šฎ in ascending powers of ๐‘ฅ.

Answer

We will start by considering the coefficients of the first five terms of this expansion. The coefficients are given by the nineteenth row of Pascalโ€™s triangle, that is, the row we label ๐‘›=18. The first element in any row of Pascalโ€™s triangle is 1. Recall the connection between consecutive elements in a row in Pascalโ€™s triangle: to move from the (๐‘˜โˆ’1)th element to the ๐‘˜th, we multiply by ๐‘›โˆ’๐‘˜+1๐‘˜. Applying this rule, we can calculate the 1st element by multiplying the 0th element by 181. Then, to find the second element, we multiply by 172. Continuing this way, we can find the first five terms in the row, as demonstrated in the figure below.

Therefore, the first five terms are given by 2+18ร—2๐‘ฅ+153ร—2๐‘ฅ+816ร—2๐‘ฅ+3,060ร—2๐‘ฅ.๏Šง๏Šฎ๏Šง๏Šญ๏Šง๏Šฌ๏Šจ๏Šง๏Šซ๏Šฉ๏Šง๏Šช๏Šช

Simplifying, we have 262,144+2,359,296๐‘ฅ+10,027,008๐‘ฅ+26,738,688๐‘ฅ+50,135,040๐‘ฅ.๏Šจ๏Šฉ๏Šช

Example 4: Using Pascalโ€™s Triangle to Find Binomial Expansions

Fully expand the expression (2+3๐‘ฅ)๏Šง๏Šฆ.

Answer

We will begin by finding the binomial coefficient. The coefficients are given by the eleventh row of Pascalโ€™s triangle, which is the row we label ๐‘›=10. The first element in any row of Pascalโ€™s triangle is 1. Then, recall the connection between consecutive elements in a row in Pascalโ€™s triangle: to move from the (๐‘˜โˆ’1)th element to the ๐‘˜th, we multiply by ๐‘›โˆ’๐‘˜+1๐‘˜. Applying this rule, we can calculate the 1st element by multiplying the 0th element (equal to 1) by 101. Then, to find the second element, we multiply by 92. Continuing this way, we can find the first five terms in the row, as demonstrated in the figure below.

Notice that once we get to the middle term, we can simply appeal to the symmetry of Pascalโ€™s triangle and fill in the other entries.

Therefore, (๐‘Ž+๐‘)=๐‘Ž+10๐‘Ž๐‘+45๐‘Ž๐‘+120๐‘Ž๐‘+210๐‘Ž๐‘+252๐‘Ž๐‘+210๐‘Ž๐‘+120๐‘Ž๐‘+45๐‘Ž๐‘+10๐‘Ž๐‘+๐‘.๏Šง๏Šฆ๏Šง๏Šฆ๏Šฏ๏Šฎ๏Šจ๏Šญ๏Šฉ๏Šฌ๏Šช๏Šซ๏Šซ๏Šช๏Šฌ๏Šฉ๏Šญ๏Šจ๏Šฎ๏Šฏ๏Šง๏Šฆ

Setting ๐‘Ž=2 and ๐‘=3๐‘ฅ, we have (2+3๐‘ฅ)=2+10ร—2(3๐‘ฅ)+45ร—2(3๐‘ฅ)+120ร—2(3๐‘ฅ)+210ร—2(3๐‘ฅ)+252ร—2(3๐‘ฅ)+210ร—2(3๐‘ฅ)+120ร—2(3๐‘ฅ)+45ร—2(3๐‘ฅ)+10ร—2(3๐‘ฅ)+(3๐‘ฅ).๏Šง๏Šฆ๏Šง๏Šฆ๏Šฏ๏Šฎ๏Šจ๏Šญ๏Šฉ๏Šฌ๏Šช๏Šซ๏Šซ๏Šช๏Šฌ๏Šฉ๏Šญ๏Šจ๏Šฎ๏Šฏ๏Šง๏Šฆ

Finally, we can simplify the numerical terms as follows: (2+3๐‘ฅ)=1,024+15,360๐‘ฅ+103,680๐‘ฅ+414,720๐‘ฅ+1,088,640๐‘ฅ+1,959,552๐‘ฅ+2,449,440๐‘ฅ+2,099,520๐‘ฅ+1,180,980๐‘ฅ+393,660๐‘ฅ+59,049๐‘ฅ.๏Šง๏Šฆ๏Šจ๏Šฉ๏Šช๏Šซ๏Šฌ๏Šญ๏Šฎ๏Šฏ๏Šง๏Šฆ

Pascalโ€™s triangle has another important interpretation. Let us consider the first five binomial expansions arranged in a triangle. If we consider this as a graph and that each term represents a node, then moving to either of the two nodes below a given node represents a choice: choosing either an ๐‘Ž or a ๐‘. Hence, in the figure below, moving from the term 3๐‘Ž๐‘๏Šจ to either of the nodes below represents a choice of either an ๐‘Ž or a ๐‘ from the highlighted pair of parentheses.