Binomial Theorem

Learning Objectives

Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)

Objective 1: Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)

Pascal’s triangle helps us find the coefficients of the terms in the expansion of a binomial.

To find the coefficients of the terms, we write our expansion again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients. The array to the right is called Pascal’s Triangle.

A plus b to the power of 0 equals 1. The top level of Pascal’s Triangle is 1. A plus b to the power of 1 equals 1 a plus 1 b. The second level of Pascal’s Triangle is 1, 1. A plus b to the power of 2 equals 1 a to the power of 2 plus 2 a b plus 1 b to the power of 2. The third level of Pascal’s Triangle is 1, 2, 1. A plus b to the power of 3 equals 1 a to the power of 3 plus 3 a to the power of 2 b plus 3 a b to the power of 2 plus 1 b to the power of 3. The fourth level of Pascal’s Triangle is 1,3,3,1. A plus b to the power of 4 equals 1 a to the power of 4 plus 4 a to the power of 3 b plus 6 a to the power of 2 b to the power of 2 plus 4 a b to the power of 3 plus 1 b to the power of 4. The fifth level of Pascal’s Triangle is 1, 4, 6, 4, 1. A plus b to the power of 5 equals 1 a to the power of 5 plus 5 a to the power of 4 b plus 10 a to the power of 3 b to the power of 2 plus 10 a to the power of 2 b to the power of 3. The sixth row of the Pascal’s Triangle is 1, 5, 10, 10, 5, 1.

Notice that in each expansion the powers of a in each term decrease from n to 0, and the powers of b increase from 0 to n.

Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers.

To find the coefficients of the expansion of the binomial ${(a + b)}^{n}$ , go to the row that has the value n as a second entry.

This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.

Example 1

Use Pascal’s Triangle to expand ${(x + y)}^{6}$ .

Solution

.
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 6.
Write the expansion with the coefficients.
Fill in the variable with the power of x decreasing from 6 to 0, and the power of y increasing from 0 to 6.
Binomial expansion of ${(x + y)}^{6}$ .

Example 2

Use Pascal’s Triangle to expand ${(x + 3)}^{5}$ .

Solution

.
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 5.
Write the expansion with the coefficients.
Fill in the variable with the power of x decreasing from 5 to 0, and the power of 3 increasing from 0 to 5.
Binomial expansion of ${(x + 3)}^{5}$ .

Example 3

Use Pascal’s Triangle to expand ${(3 x - 2)}^{4}$ .

Solution

.
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 4.
Write the expansion with the coefficients.
Fill in the variable with the power of (3x) decreasing from 4 to 0, and the power of (-2) increasing from 0 to 4.
Binomial expansion of ${(3 x - 2)}^{4}$ .

Practice Makes Perfect

Use Pascal’s Triangle to expand a binomial.

Use Pascal’s Triangle to expand ${(a + b)}^{4}$ .

Use Pascal’s Triangle to expand ${(y + 3)}^{5}$ .

Use Pascal’s Triangle to expand ${(2 x - 5)}^{3}$ .

A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find ${(x + y)}^{n}$ without multiplying the binomial by itself $n$ times.

Identifying Binomial Coefficients

In Counting Principles, we studied combinations. In the shortcut to finding

{(x + y)}^{n},

we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation

(\begin{matrix} n \\ r \end{matrix})

instead of

C (n, r),

but it can be calculated in the same way. So

(\begin{matrix} n \\ r \end{matrix}) = C (n, r) = \frac{n!}{r! (n - r)!}

The combination $(\begin{matrix} n \\ r \end{matrix})$ is called a binomial coefficient. An example of a binomial coefficient is $(\begin{matrix} 5 \\ 2 \end{matrix}) = C (5, 2) = 10.$

Example 4

Finding Binomial Coefficients

Find each binomial coefficient.

ⓐ $(\begin{matrix} 5 \\ 3 \end{matrix})$
ⓑ $(\begin{matrix} 9 \\ 2 \end{matrix})$
ⓒ $(\begin{matrix} 9 \\ 7 \end{matrix})$

Solution

Use the formula to calculate each binomial coefficient. You can also use the $n C_{r}$ function on your calculator.

(\begin{matrix} n \\ r \end{matrix}) = C (n, r) = \frac{n!}{r! (n - r)!}

ⓐ $(\begin{matrix} 5 \\ 3 \end{matrix}) = \frac{5!}{3! (5 - 3)!} = \frac{5 \cdot 4 \cdot 3!}{3! 2!} = 10$
ⓑ $(\begin{matrix} 9 \\ 2 \end{matrix}) = \frac{9!}{2! (9 - 2)!} = \frac{9 \cdot 8 \cdot 7!}{2! 7!} = 36$
ⓒ $(\begin{matrix} 9 \\ 7 \end{matrix}) = \frac{9!}{7! (9 - 7)!} = \frac{9 \cdot 8 \cdot 7!}{7! 2!} = 36$

Analysis

Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.

(\begin{matrix} n \\ r \end{matrix}) = (\begin{matrix} n \\ n - r \end{matrix})

Using the Binomial Theorem

When we expand ${(x + y)}^{n}$ by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand ${(x + y)}^{52},$ we might multiply $(x + y)$ by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

\begin{array}{l} {(x + y)}^{2} = x^{2} + 2 x y + y^{2} \\ {(x + y)}^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} \\ {(x + y)}^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4} \end{array}

First, let’s examine the exponents. With each successive term, the exponent for $x$ decreases and the exponent for $y$ increases. The sum of the two exponents is $n$ for each term.

Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

(\begin{matrix} n \\ 0 \end{matrix}), (\begin{matrix} n \\ 1 \end{matrix}), (\begin{matrix} n \\ 2 \end{matrix}), ..., (\begin{matrix} n \\ n \end{matrix}) .

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

\begin{array}{l} {(x + y)}^{n} & = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k} \\ = x^{n} + (\begin{matrix} n \\ 1 \end{matrix}) x^{n - 1} y + (\begin{matrix} n \\ 2 \end{matrix}) x^{n - 2} y^{2} + ... + (\begin{matrix} n \\ n - 1 \end{matrix}) x y^{n - 1} + y^{n} \end{array}

Another way to see the coefficients is to examine the expansion of a binomial in general form, $x + y,$ to successive powers 1, 2, 3, and 4.

\begin{array}{l} {(x + y)}^{1} = x + y \\ {(x + y)}^{2} = x^{2} + 2 x y + y^{2} \\ {(x + y)}^{3} = x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} \\ {(x + y)}^{4} = x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4} \end{array}

Can you guess the next expansion for the binomial ${(x + y)}^{5} ?$

See Figure 1, which illustrates the following:

There are $n + 1$ terms in the expansion of ${(x + y)}^{n} .$
The degree (or sum of the exponents) for each term is $n .$
The powers on $x$ begin with $n$ and decrease to 0.
The powers on $y$ begin with 0 and increase to $n .$
The coefficients are symmetric.

To determine the expansion on ${(x + y)}^{5},$ we see $n = 5,$ thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of $x,$ the pattern is as follows:

Introduce $x^{5},$ and then for each successive term reduce the exponent on $x$ by 1 until $x^{0} = 1$ is reached.
Introduce $y^{0} = 1,$ and then increase the exponent on $y$ by 1 until $y^{5}$ is reached.
$x^{5}, x^{4} y, x^{3} y^{2}, x^{2} y^{3}, x y^{4}, y^{5}$

The next expansion would be

{(x + y)}^{5} = x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + y^{5} .

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle, shown in Figure 2. Pascal didn't invent the triangle. The underlying principles had been developed and written about for over 1500 years, first by the Indian mathematician (and poet) Pingala in the second century BCE. Others throughout Asia and Europe worked with the concepts throughout, and the triangle was first published in its graphical form by Omar Khayyam, an Iranian mathematician and astronomer, for whom the triangle is named in Iran. French mathematician Blaise Pascal repopularized it when he republished it and used it to solve a number of probability problems.

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3^rd row, flank the ends of the rows with 1’s, and add $1 + 1$ to find the middle number, 2. In the $n th$ row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.

To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.

Pascal's Triangle expanded to show the values of the triangle as x and y terms with exponents

The Binomial Theorem

The Binomial Theorem is a formula that can be used to expand any binomial.

\begin{array}{l} {(x + y)}^{n} & = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k} \\ = x^{n} + (\begin{matrix} n \\ 1 \end{matrix}) x^{n - 1} y + (\begin{matrix} n \\ 2 \end{matrix}) x^{n - 2} y^{2} + ... + (\begin{matrix} n \\ n - 1 \end{matrix}) x y^{n - 1} + y^{n} \end{array}

Example 5

Expanding a Binomial

Write in expanded form.

ⓐ ${(x + y)}^{5}$
ⓑ ${(3 x - y)}^{4}$

Solution

ⓐSubstitute $n = 5$ into the formula. Evaluate the $k = 0$ through $k = 5$ terms. Simplify.
$\begin{array}{l} {(x + y)}^{5} & = (\begin{matrix} 5 \\ 0 \end{matrix}) x^{5} y^{0} + (\begin{matrix} 5 \\ 1 \end{matrix}) x^{4} y^{1} + (\begin{matrix} 5 \\ 2 \end{matrix}) x^{3} y^{2} + (\begin{matrix} 5 \\ 3 \end{matrix}) x^{2} y^{3} + (\begin{matrix} 5 \\ 4 \end{matrix}) x^{1} y^{4} + (\begin{matrix} 5 \\ 5 \end{matrix}) x^{0} y^{5} \\ {(x + y)}^{5} & = x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + y^{5} \end{array}$
ⓑSubstitute $n = 4$ into the formula. Evaluate the $k = 0$ through $k = 4$ terms. Notice that $3 x$ is in the place that was occupied by $x$ and that $- y$ is in the place that was occupied by $y .$ So we substitute them. Simplify.
$\begin{array}{l} {(3 x - y)}^{4} & = (\begin{matrix} 4 \\ 0 \end{matrix}) {(3 x)}^{4} {(- y)}^{0} + (\begin{matrix} 4 \\ 1 \end{matrix}) {(3 x)}^{3} {(- y)}^{1} + (\begin{matrix} 4 \\ 2 \end{matrix}) {(3 x)}^{2} {(- y)}^{2} + (\begin{matrix} 4 \\ 3 \end{matrix}) {(3 x)}^{1} {(- y)}^{3} + (\begin{matrix} 4 \\ 4 \end{matrix}) {(3 x)}^{0} {(- y)}^{4} \\ {(3 x - y)}^{4} & = 81 x^{4} - 108 x^{3} y + 54 x^{2} y^{2} - 12 x y^{3} + y^{4} \end{array}$

Analysis

Notice the alternating signs in part b. This happens because $(- y)$ raised to odd powers is negative, but $(- y)$ raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.

Using the Binomial Theorem to Find a Single Term

Expanding a binomial with a high exponent such as ${(x + 2 y)}^{16}$ can be a lengthy process.

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

Note the pattern of coefficients in the expansion of ${(x + y)}^{5} .$

{(x + y)}^{5} = x^{5} + (\begin{matrix} 5 \\ 1 \end{matrix}) x^{4} y + (\begin{matrix} 5 \\ 2 \end{matrix}) x^{3} y^{2} + (\begin{matrix} 5 \\ 3 \end{matrix}) x^{2} y^{3} + (\begin{matrix} 5 \\ 4 \end{matrix}) x y^{4} + y^{5}

The second term is $(\begin{matrix} 5 \\ 1 \end{matrix}) x^{4} y .$ The third term is $(\begin{matrix} 5 \\ 2 \end{matrix}) x^{3} y^{2} .$ We can generalize this result.

(\begin{matrix} n \\ r \end{matrix}) x^{n - r} y^{r}

Example 6

Writing a Given Term of a Binomial Expansion

Find the tenth term of ${(x + 2 y)}^{16}$ without fully expanding the binomial.

Solution

Because we are looking for the tenth term, $r + 1 = 10,$ we will use $r = 9$ in our calculations.

(\begin{matrix} n \\ r \end{matrix}) x^{n - r} y^{r}

(\begin{matrix} 16 \\ 9 \end{matrix}) x^{16 - 9} {(2 y)}^{9} = 5, 857, 280 x^{7} y^{9}

Key Equations

..
Binomial Theorem	${(x + y)}^{n} = \sum_{k - 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k}$
$(r + 1) t h$ term of a binomial expansion	$(\begin{matrix} n \\ r \end{matrix}) x^{n - r} y^{r}$

Key Concepts

$(\begin{matrix} n \\ r \end{matrix})$ is called a binomial coefficient and is equal to $C (n, r) .$ See Example 4.
The Binomial Theorem allows us to expand binomials without multiplying. See Example 5.
We can find a given term of a binomial expansion without fully expanding the binomial. See Example 6.

Section Exercises

Verbal

Exercise 1

What is a binomial coefficient, and how it is calculated?

Solution

A binomial coefficient is an alternative way of denoting the combination $C (n, r).$ It is defined as $(\begin{matrix} n \\ r \end{matrix}) = C (n, r) = \frac{n!}{r! (n - r)!} .$

Exercise 2

What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?

Exercise 3

What is the Binomial Theorem and what is its use?

Solution

The Binomial Theorem is defined as ${(x + y)}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k}$ and can be used to expand any binomial.

Exercise 4

When is it an advantage to use the Binomial Theorem? Explain.

Algebraic

For the following exercises, evaluate the binomial coefficient.

Exercise 5

$(\begin{matrix} 6 \\ 2 \end{matrix})$

Solution

Exercise 6

$(\begin{matrix} 5 \\ 3 \end{matrix})$

Exercise 7

$(\begin{matrix} 7 \\ 4 \end{matrix})$

Solution

Exercise 8

$(\begin{matrix} 9 \\ 7 \end{matrix})$

Exercise 9

$(\begin{matrix} 10 \\ 9 \end{matrix})$

Solution

Exercise 10

$(\begin{matrix} 25 \\ 11 \end{matrix})$

Exercise 11

$(\begin{matrix} 17 \\ 6 \end{matrix})$

Solution

12,376

Exercise 12

$(\begin{matrix} 200 \\ 199 \end{matrix})$

For the following exercises, use the Binomial Theorem to expand each binomial.

Exercise 13

${(4 a - b)}^{3}$

Solution

$64 a^{3} - 48 a^{2} b + 12 a b^{2} - b^{3}$

Exercise 14

${(5 a + 2)}^{3}$

Exercise 15

${(3 a + 2 b)}^{3}$

Solution

$27 a^{3} + 54 a^{2} b + 36 a b^{2} + 8 b^{3}$

Exercise 16

${(2 x + 3 y)}^{4}$

Exercise 17

${(4 x + 2 y)}^{5}$

Solution

$1024 x^{5} + 2560 x^{4} y + 2560 x^{3} y^{2} + 1280 x^{2} y^{3} + 320 x y^{4} + 32 y^{5}$

Exercise 18

${(3 x - 2 y)}^{4}$

Exercise 19

${(4 x - 3 y)}^{5}$

Solution

$1024 x^{5} - 3840 x^{4} y + 5760 x^{3} y^{2} - 4320 x^{2} y^{3} + 1620 x y^{4} - 243 y^{5}$

Exercise 20

${(\frac{1}{x} + 3 y)}^{5}$

Exercise 21

${(x^{- 1} + 2 y^{- 1})}^{4}$

Solution

$\frac{1}{x^{4}} + \frac{8}{x^{3} y} + \frac{24}{x^{2} y^{2}} + \frac{32}{x y^{3}} + \frac{16}{y^{4}}$

Exercise 22

${(\sqrt{x} - \sqrt{y})}^{5}$

For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.

Exercise 23

${(a + b)}^{17}$

Solution

$a^{17} + 17 a^{16} b + 136 a^{15} b^{2}$

Exercise 24

${(x - 1)}^{18}$

Exercise 25

${(a - 2 b)}^{15}$

Solution

$a^{15} - 30 a^{14} b + 420 a^{13} b^{2}$

Exercise 26

${(x - 2 y)}^{8}$

Exercise 27

${(3 a + b)}^{20}$

Solution

$3, 486, 784, 401 a^{20} + 23, 245, 229, 340 a^{19} b + 73, 609, 892, 910 a^{18} b^{2}$

Exercise 28

${(2 a + 4 b)}^{7}$

Exercise 29

${(x^{3} - \sqrt{y})}^{8}$

Solution

$x^{24} - 8 x^{21} \sqrt{y} + 28 x^{18} y$

For the following exercises, find the indicated term of each binomial without fully expanding the binomial.

Exercise 30

The fourth term of ${(2 x - 3 y)}^{4}$

Exercise 31

The fourth term of ${(3 x - 2 y)}^{5}$

Solution

$- 720 x^{2} y^{3}$

Exercise 32

The third term of ${(6 x - 3 y)}^{7}$

Exercise 33

The eighth term of ${(7 + 5 y)}^{14}$

Solution

$220, 812, 466, 875, 000 y^{7}$

Exercise 34

The seventh term of ${(a + b)}^{11}$

Exercise 35

The fifth term of ${(x - y)}^{7}$

Solution

$35 x^{3} y^{4}$

Exercise 36

The tenth term of ${(x - 1)}^{12}$

Exercise 37

The ninth term of ${(a - 3 b^{2})}^{11}$

Solution

$1, 082, 565 a^{3} b^{16}$

Exercise 38

The fourth term of ${(x^{3} - \frac{1}{2})}^{10}$

Exercise 39

The eighth term of ${(\frac{y}{2} + \frac{2}{x})}^{9}$

Solution

$\frac{1152 y^{2}}{x^{7}}$

Graphical

For the following exercises, use the Binomial Theorem to expand the binomial $f (x) = {(x + 3)}^{4} .$ Then find and graph each indicated sum on one set of axes.

Exercise 40

Find and graph $f_{1} (x),$ such that $f_{1} (x)$ is the first term of the expansion.

Exercise 41

Find and graph $f_{2} (x),$ such that $f_{2} (x)$ is the sum of the first two terms of the expansion.

Solution

$f_{2} (x) = x^{4} + 12 x^{3}$

Exercise 42

Find and graph $f_{3} (x),$ such that $f_{3} (x)$ is the sum of the first three terms of the expansion.

Exercise 43

Find and graph $f_{4} (x),$ such that $f_{4} (x)$ is the sum of the first four terms of the expansion.

Solution

$f_{4} (x) = x^{4} + 12 x^{3} + 54 x^{2} + 108 x$

Exercise 44

Find and graph $f_{5} (x),$ such that $f_{5} (x)$ is the sum of the first five terms of the expansion.

Extensions

Exercise 45

In the expansion of ${(5 x + 3 y)}^{n},$ each term has the form $(\begin{matrix} n \\ k \end{matrix}) a^{n - k} b^{k}$ , where $k$ successively takes on the value $0, 1, 2, ..., n .$ If $(\begin{matrix} n \\ k \end{matrix}) = (\begin{matrix} 7 \\ 2 \end{matrix}),$ what is the corresponding term?

Solution

$590, 625 x^{5} y^{2}$

Exercise 46

In the expansion of ${(a + b)}^{n},$ the coefficient of $a^{n - k} b^{k}$ is the same as the coefficient of which other term?

Exercise 47

Consider the expansion of ${(x + b)}^{40} .$ What is the exponent of $b$ in the $k th$ term?

Solution

$k - 1$

Exercise 48

Find $(\begin{matrix} n \\ k - 1 \end{matrix}) + (\begin{matrix} n \\ k \end{matrix})$ and write the answer as a binomial coefficient in the form $(\begin{matrix} n \\ k \end{matrix}) .$ Prove it. Hint: Use the fact that, for any integer $p,$ such that $p \geq 1, p! = p (p - 1)! .$

Exercise 49

Which expression cannot be expanded using the Binomial Theorem? Explain.

$(x^{2} - 2 x + 1)$
${(\sqrt{a} + 4 \sqrt{a} - 5)}^{8}$
${(x^{3} + 2 y^{2} - z)}^{5}$
${(3 x^{2} - \sqrt{2 y^{3}})}^{12}$

Solution

The expression ${(x^{3} + 2 y^{2} - z)}^{5}$ cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.

binomial coefficient

the number of ways to choose r objects from n objects where order does not matter; equivalent to

C (n, r),

denoted

(\begin{matrix} n \\ r \end{matrix})

binomial expansion

the result of expanding

{(x + y)}^{n}

by multiplying

Binomial Theorem

a formula that can be used to expand any binomial