Mastering the Binomial Theorem: Pascal's Triangle, Polynomial Expansion, and Coefficients

Expanding binomials can be difficult if the exponent is large. Considering that there is a formula for the square of a binomial, it might be possible to find similar expressions for greater exponents. This lesson aims to show an array of numbers that can be used to expand binomials.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Polynomial
Binomial
Factor Theorem
Difference of Squares
Difference of Cubes
Square of a Binomial

Explore

Recognizing Patterns

In the following diagram, an incomplete triangle partially filled with numbers is displayed. Complete the triangle by inserting the numbers listed at the bottom into their correct place within the triangle.

Considering the given numbers, does the triangle follow a pattern? If so, what is it?

Example

Difference of Squares

Two good friends, Diego and Maya, like to challenge each other on math related topics. One summer afternoon, Diego came up with an interesting problem.

Help Maya prove that the difference of the squares of any two consecutive integers is odd.

Answer

See solution.

Hint

Let $n$ be an arbitrary integer. Find the difference of the squares of $n$ and its consecutive integer.

Remember that the division of an even number by two always result on an integer.

Solution

To prove that the difference of the squares of any two consecutive integers is odd, let

n

be an integer number. The second of the consecutive integers is obtained by adding

1

n .

Therefore, the consecutive integer to

n

n + 1 .

Each number will be squared.

Numbers n and n + 1 Squares of the Numbers n^{2} and (n + 1)^{2}

Next, the difference of the squared numbers can be calculated. For simplicity the smaller integer is subtracted from the larger, but it should be noted that either way works.

(n + 1)^{2} - n^{2}

▼

Simplify

ExpandPosPerfectSquare

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

n^{2} + 2 n (1) + 1^{2} - n^{2}

IdPropMult

Identity Property of Multiplication

n^{2} + 2 n + 1^{2} - n^{2}

BaseOne

$1^{a} = 1$

n^{2} + 2 n + 1 - n^{2}

SubTerm

Subtract term

2 n + 1

To determine whether this difference is an odd number, the expression

2 n + 1

can be divided by

2 .

If the result is not an integer, the difference is odd.

\frac{2 n + 1}{2}

▼

Simplify

WriteSumFrac

Write as a sum of fractions

\frac{2 n}{2} + \frac{1}{2}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

\frac{2}{2} n + \frac{1}{2}

QuotOne

$\frac{a}{a} = 1$

1 n + \frac{1}{2}

IdPropMult

Identity Property of Multiplication

n + \frac{1}{2}

Since

n

is an integer number and

\frac{1}{2}

is not, the number

n + \frac{1}{2}

is not an integer. This means that the number

2 n + 1

is odd for any integer

n .

Therefore, the difference of the squares of any two consecutive integers odd.

Example

Difference of Cubes

After solving Diego's challenge, Maya now creates one for Diego. She asks him to prove that the difference of the cubes of any two consecutive integers is also an odd number.

Help Diego prove this statement!

Answer

See solution.

Hint

To prove this statement, consider an even number $2 n,$ where $n$ is any integer. Its consecutive number is $2 n + 1$ and its previous number is $2 n - 1 .$ Consider the difference of the cubes of $2 n + 1$ and $n,$ and the difference of $n$ and $2 n - 1 .$

Solution

Any even number can be written as

2 n,

where

n

is an integer. Consider the cube of this number.

(2 n)^{3} = 8 n^{3}

Now, when considering two consecutive integers, one is even and the other is odd. The odd number can be before or after

2 n .

These cases can be expressed by subtracting

1

from or adding

1

2 n,

respectively.

Number Before n : Number After n : 2 n - 1 2 n + 1

Recall the formula for the cube of a binomial.

(a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}

By using this formula, the cubes of

2 n - 1

and

2 n + 1

can be found. These formulas are obtained by substituting

2 n

for

a

and

1

for

b .

(a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}

SubstituteII

$a = 2 n$ , $b = 1$

(2 n \pm 1)^{3} = (2 n)^{3} \pm 3 (2 n)^{2} (1) + 3 (2 n) (1)^{2} \pm (1)^{3}

▼

Simplify

CalcPow

Calculate power

(2 n \pm 1)^{3} = 8 n^{3} \pm 3 (4 n^{2}) (1) + 3 (2 n) (1) \pm 1

MultByOne

$a \cdot 1 = a$

(2 n \pm 1)^{3} = 8 n^{3} \pm 3 (4 n^{2}) + 3 (2 n) \pm 1

Multiply

(2 n \pm 1)^{3} = 8 n^{3} \pm 12 n^{2} + 6 n \pm 1

There are two possible situations. The consecutive numbers can be

2 n

and

2 n + 1,

2 n - 1

and

2 n .

These two situations will be considered one at a time.

$2 n$ and $2 n - 1$

If the consecutive integers are

2 n

and

2 n - 1,

then their cubes are

8 n^{3}

and

8 n^{3} - 12 n^{2} + 6 n - 1,

respectively. Next, find their difference. For the result to be positive, the cube of

2 n - 1

is subtracted from the cube of

2 n .

8 n^{3} - (8 n^{3} - 12 n^{2} + 6 n - 1)

▼

Simplify

Distr

Distribute $- 1$

8 n^{3} - 8 n^{3} + 12 n^{2} - 6 n + 1

SubTerm

Subtract term

12 n^{2} - 6 n + 1

Now, divide this expression by

2 .

An even number is always divisible by two. Therefore, if dividing the above expression by 2 results in an integer, the expression represents an even number.

\frac{1 2 n ^{2} - 6 n + 1}{2}

▼

Simplify

WriteSumFrac

Write as a sum of fractions

\frac{1 2 n ^{2}}{2} - \frac{6 n}{2} + \frac{1}{2}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

\frac{1 2}{2} n^{2} - \frac{6}{2} n + \frac{1}{2}

CalcQuot

Calculate quotient

6 n^{2} - 3 n + \frac{1}{2}

This expression will be examined closely to determine if it is an integer or not. Each term will be examined individually, gradually building the expression of

\frac{1 2 n ^{2} - 6 n + 1}{2} .

Expression	Is it an Integer?	Reason
$n$	Yes	It is defined this way.
$n^{2}$	Yes	Closure Property of Multiplication
$6 n^{2}$	Yes	Closure Property of Multiplication
$3 n$	Yes	Closure Property of Multiplication
$6 n^{2} - 3 n$	Yes	Closure Property of Subtraction
$6 n^{2} - 3 n + \frac{1}{2}$	No	$\frac{1}{2}$ is not an integer

Since dividing $12 n^{2} - 6 n + 1$ by $2$ did not result in an integer number, the expression $12 n^{2} - 6 n + 1,$ which is the difference of the cubes of $2 n$ and $2 n - 1,$ is odd.

$2 n + 1$ and $2 n$

If the consecutive integers are

2 n + 1

and

2 n,

then their cubes are

8 n^{3} + 12 n^{2} + 6 n + 1

and

8 n^{3},

respectively. Their difference can be calculated.

(8 n^{3} + 12 n^{2} + 6 n + 1) - 8 n^{3}

▼

Simplify

RemovePar

Remove parentheses

8 n^{3} + 12 n^{2} + 6 n + 1 - 8 n^{3}

SubTerm

Subtract term

12 n^{2} + 6 n + 1

Just like in the previous case, the expression that represents the difference will be divided by

2

to determine if the result is even.

\frac{1 2 n ^{2} + 6 n + 1}{2}

▼

Simplify

WriteSumFrac

Write as a sum of fractions

\frac{1 2 n ^{2}}{2} + \frac{6 n}{2} + \frac{1}{2}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

\frac{1 2}{2} n^{2} + \frac{6}{2} n + \frac{1}{2}

CalcQuot

Calculate quotient

6 n^{2} + 3 n + \frac{1}{2}

Similar to the previous case,

6 n^{2} + 3 n

is an integer. Because of the

\frac{1}{2},

the sum

6 n^{2} + 3 n + \frac{1}{2}

is not an integer. Therefore, the difference of the cubes is odd. Now that the two cases are considered, the statement has been proved.

Example

Square of the Sum of Two Consecutive Integers

Diego, engulfed by Maya's last challenge, asks if she could create another one for him. This time, Maya asks Diego to prove that the square of the sum of two consecutive integers is odd.

Help Diego solve Maya's challenge!

Answer

See solution.

Hint

Let $n$ and $n + 1$ be two consecutive integers. Find their sum, calculate its square, and prove that the obtained result is odd.

Remember that the division of an even number by two always result on an integer.

Solution

Let

n

and

n + 1

be two consecutive integer numbers. To prove the given statement, the numbers first must be added.

n + (n + 1) = 2 n + 1

The sum resulted in a binomial. Then, the square of the sum is the square of a binomial.

(2 n + 1)^{2}

ExpandPosPerfectSquare

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

(2 n)^{2} + 2 (2 n) (1) + 1^{2}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

4 n^{2} + 2 (2 n) (1) + 1^{2}

BaseOne

$1^{a} = 1$

4 n^{2} + 2 (2 n) (1) + 1

Multiply

4 n^{2} + 4 n + 1

An even number is always divisible by

2 .

This means that, if dividing the above expression by

2

results in an integer, the expression represents an even number. Otherwise, it represents an odd number.

\frac{4 n ^{2} + 4 n + 1}{2}

WriteSumFrac

Write as a sum of fractions

\frac{4 n ^{2}}{2} + \frac{4 n}{2} + \frac{1}{2}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

\frac{4}{2} n^{2} + \frac{4}{2} n + \frac{1}{2}

CalcQuot

Calculate quotient

2 n^{2} + 2 n + \frac{1}{2}

Since

n

is an integer, the expression

2 n^{2} + 2 n

also represents an integer. Adding a non-integer to an integer results in a non-integer. Therefore, the expression

2 n^{2} + 2 n + \frac{1}{2}

is not an integer. This means that

4 n^{2} + 4 n + 1

is an odd number, proving the statement.

Discussion

Pascal's Triangle and the Binomial Theorem

The Pascal's triangle is a triangular representation of the coefficients in the expansion of a binomial expression. In other words, it contains the coefficients that result when expanding the expression $(a + b)^{n},$ with $n = 0, 1, 2, \dots$

The rows are numbered from top to bottom starting with $0 .$ The number in each cell equals the sum of the numbers in the two neighboring cells above.

The numbers in the Pascal's Triangle have an interesting application. Consider the following binomial expansions.

Looking at the coefficients of the binomial expansions, it can be noted that every number of a row in the Pascal's Triangle is the same as the coefficients of a binomial expansion.

This is expressed in the Binomial Theorem.

Binomial Theorem

Consider a binomial raised to the power of $n,$ where $n$ is a positive integer. Let $a$ and $b$ be real numbers. Expanding the binomial $(a + b)^{n}$ results in the following expression.

In this expansion, the numbers

P_{0},

P_{1},

\dots,

P_{n}

are the numbers in the

n

th row of Pascal's Triangle.

Example

The Volume of a Room

Maya, feeling like she challenged Diego well, remembers that her mom asked her to measure her room for new wallpaper. She has to take a phone call, so she asks Diego if he can take the measurements. The room is in the shape of a cube. Diego hands over the measurements to her — he wrote them as a challenge!

Diego wrote that each side of the room has a length of

(x - 3)^{2} .

Write the expanded polynomial for the volume of the room.

Hint

Use the Binomial Theorem to expand the binomial.

Solution

The room has the shape of a room with side length

(x - 3)^{2} .

To find its volume, it is necessary to use the formula for the volume of a cube.

V = s^{3}

In this formula,

V

is the volume of the cube and

s

its side length. Substituting

(x - 3)^{2}

for

s

gives the formula for the volume of the room.

V = s^{3}

Substitute

$s = (x - 3)^{2}$

V = ((x - 3)^{2})^{3}

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

V = (x - 3)^{6}

Instead of multiplying the binomial by itself

6

times, the Binomial Theorem can be used to expand the binomial. This theorem indicates that the coefficients of a binomial to the sixth power coincide with the numbers in the

sixth

row

of Pascal's Triangle.

Each number in the sixth row corresponds to a coefficient of the binomial expansion, considering that the result is written in standard form.

V = 1 x^{6} + 6 x^{5} (- 3) + 15 x^{4} (- 3)^{2} + 20 x^{3} (- 3)^{3} + 15 x^{2} (- 3)^{4} + 6 x (- 3)^{5} + 1 (- 3)^{6}

▼

Simplify right-hand side

IdPropMult

Identity Property of Multiplication

V = x^{6} + 6 x^{5} (- 3) + 15 x^{4} (- 3)^{2} + 20 x^{3} (- 3)^{3} + 15 x^{2} (- 3)^{4} + 6 x (- 3)^{5} + (- 3)^{6}

CalcPow

Calculate power

V = x^{6} + 6 x^{5} (- 3) + 15 x^{4} (9) + 20 x^{3} (- 27) + 15 x^{2} (81) + 6 x (- 243) + (729)

Multiply

V = x^{6} - 18 x^{5} + 135 x^{4} - 540 x^{3} + 1215 x^{2} - 1458 x + 729

Example

Finding a Certain Coefficient

It's Diego's birthday, and without fail, Maya has gotten him a cool present. Diego is so excited to open it, but there is a problem. The gift has a lock that asks for a combination that he does not have!

Diego thinking about how to open the gift

Maya, of course, wants to have a bit of fun and tells Diego that the combination is the fifth term in the expansion

(f + w)^{9} .

Help Diego open his gift!

Hint

Consider the Binomial Theorem.

Solution

The Binomial Theorem indicates that the coefficients in the expansion of $(f + w)^{9}$ coincide with the numbers in the ninth row of Pascal's Triangle.

The fifth term in the expansion of the binomial raised to the ninth power matches the fifth number in the ninth row of Pascal's Triangle.

The terms of the polynomial are counted starting from the term $f^{9} w^{0},$ this being the first one. The variables for each consecutive term are obtained by subtracting one from the exponent of $f^{9}$ and adding one to the exponent of $w .$ The coefficients are the same as the numbers in the ninth row of Pascal's Triangle.

$n th$ term
Number of Term	Coefficient	Variables	Value of Term
$1$	$1$	$f^{9} w^{0}$	$f^{9}$
$2$	$9$	$f^{8} w^{1}$	$9 f^{8} w$
$3$	$36$	$f^{7} w^{2}$	$36 f^{7} w^{2}$
$4$	$84$	$f^{6} w^{3}$	$84 f^{6} w^{3}$
$5$	$126$	$f^{5} w^{4}$	$126 f^{5} w^{4}$
$6$	$126$	$f^{4} w^{5}$	$126 f^{4} w^{5}$
$7$	$84$	$f^{3} w^{6}$	$84 f^{3} w^{6}$
$8$	$36$	$f^{2} w^{7}$	$36 f^{2} w^{7}$
$9$	$9$	$f^{1} w^{8}$	$9 f w^{8}$
$10$	$1$	$f^{0} w^{9}$	$w^{9}$

Therefore, the fifth term is $126 f^{5} w^{4} .$

Closure

Using the Binomial Theorem to Solve a Problem

To finish this lesson, a more complex exercise will be solved using the Binomial Theorem. Consider the following expression.

(2 a + 3 x)^{6}

It is known that the coefficient in the

x^{3} -

term in the expansion of the above expression is the same as the coefficient in the

x^{5} -

term. Find the possible values of

a .

Write the exact answer in its simplest form, with no radicals in denominators.

Hint

Use the Binomial Theorem to find the coefficients of every term.

Solution

The Binomial Theorem indicates that the coefficients in the expansion of $(2 a + 3 x)^{6}$ are the numbers in the sixth row of Pascal's Triangle.

Now the terms for $x^{5}$ and $x^{3}$ have to be identified. To identify these terms, every term of the binomial expansion is determined using the Binomial Theorem.

Number of Term	Coefficient	Variables	Value of Term
$1$	$1$	$(2 a)^{6} (3 x)^{0}$	$64 a^{6}$
$2$	$6$	$(2 a)^{5} (3 x)^{1}$	$576 a^{5} x$
$3$	$15$	$(2 a)^{4} (3 x)^{2}$	$2160 a^{4} x^{2}$
$4$	$20$	$(2 a)^{3} (3 x)^{3}$	$4320 a^{3} x^{3}$
$5$	$15$	$(2 a)^{2} (3 x)^{4}$	$4860 a^{2} x^{4}$
$6$	$6$	$(2 a)^{1} (3 x)^{5}$	$2916 a x^{5}$
$7$	$1$	$(2 a)^{0} (3 x)^{6}$	$729 x^{6}$

It is given that the coefficients of the

x^{5} -

and the

x^{3} -

terms are the same. Therefore,

2916 a

must be equal to

4320 a^{3} .

This can be written as an equation which can be solved for

a .

2916 a = 4320 a^{3}

▼

Solve for

a

DivEqn

$LHS / 108 a = RHS / 108 a$

\frac{2 9 1 6 a}{1 0 8 a} = \frac{4 3 2 0 a ^{3}}{1 0 8 a}

WriteProdFrac

Write as a product of fractions

\frac{2 9 1 6}{1 0 8} \cdot \frac{a}{a} = \frac{4 3 2 0}{1 0 8} \cdot \frac{a ^{3}}{a}

CalcQuot

Calculate quotient

27 \cdot \frac{a}{a} = 40 \cdot \frac{a ^{3}}{a}

QuotOne

$\frac{a}{a} = 1$

27 (1) = 40 \cdot \frac{a ^{3}}{a}

IdPropMult

Identity Property of Multiplication

27 = 40 \cdot \frac{a ^{3}}{a}

$\frac{a ^{m}}{a} = a^{m - 1}$

27 = 40 a^{2}

DivEqn

$LHS / 40 = RHS / 40$

\frac{2 7}{4 0} = a^{2}

RearrangeEqn

Rearrange equation

a^{2} = \frac{2 7}{4 0}

SqrtEqn

$LHS = RHS$

a^{2} = \frac{2 7}{4 0}

$a^{2} = \pm a$

a = \pm \frac{2 7}{4 0}

▼

Simplify right-hand side

SplitIntoFactors

Split into factors

a = \pm \frac{9 ( 3 )}{4 ( 1 0 )}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

a = \pm \frac{9 ( 3 )}{4 ( 1 0 )}

SqrtProd

$a \cdot b = a \cdot b$

a = \pm \frac{3 3}{2 1 0}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 1 0}{b \cdot 1 0}$

a = \pm \frac{3 3 \cdot 1 0}{2 1 0 \cdot 1 0}

ProdSqrt

$a \cdot b = a \cdot b$

a = \pm \frac{3 3 0}{2 1 0 0}

CalcRoot

Calculate root

a = \pm \frac{3 3 0}{2 ( 1 0 )}

Multiply

a = \pm \frac{3 3 0}{2 0}

Therefore, the possible values for

a

are

\frac{3 3 0}{2 0}

and

- \frac{3 3 0}{2 0} .

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Catch-Up and Review

Answer

Hint

Solution

Answer

Hint

Solution

2n and 2n−1

2n+1 and 2n

Answer

Hint

Solution

Binomial Theorem

Hint

Solution

Hint

Solution

Hint

Solution

$2 n$ and $2 n - 1$

$2 n + 1$ and $2 n$