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Algebra 2 Binomial Theorem

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Section 11: Sequences and Series: Lecture 7 | 48:30 min

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Lecture Comments (6)

	0 answers Post by Suhani Pant on July 27, 2013 Thank you so much for this amazing series on Algebra II. I greatly appreciate your hard-work and effort.
	1 answer Last reply by: Dr Carleen Eaton Mon Apr 8, 2013 4:35 PM Post by Kenneth Montfort on March 7, 2013 You are super great teacher - thank you so much for making algebra 2 super understandable :) it was an awesome course.
	1 answer Last reply by: Dr Carleen Eaton Sat Jan 5, 2013 9:55 AM Post by Angel La Fayette on January 2, 2013 The "2" was left out @ 38:13.
	0 answers Post by Timothy miranda on July 3, 2010 Thanks for all the help! This was a great course!

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Binomial Theorem

The expansion of (a + b)ⁿ has n + 1 terms.
In the kth term of (a + b)ⁿ , the exponent of b is k – 1 and the exponent of is n – (k – 1).
In the kth term, the coefficient has k – 1 factors in the numerator and the denominator.
You can always find the coefficients of the expansion or of a particular term by writing out Pascal’s triangle and using row n.
The coefficients of the expansion are symmetrical around the middle term. So once you have found the first half of the coefficients, you already have the coefficients for the rest of the terms. They are the same ones you have found but listed in reverse order.

Binomial Theorem

Expand (x + y)⁵

You can expand this problem using several properties of the binomial theorem.
Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
(x + y)⁵ = ___(xy) + ___(xy) + ___(xy) + ___(xy) + ____(xy) + ____(xy)
Step 2 - Distribute the power n, notice how you start at 5 with x, and you start at 0 with y
The sum of the powers of x and y always equal n, in this case 5
(x + y)⁵ = ___(xy) + ___(xy) + ___(xy) + ___(xy) + ____(xy) + ____(xy)
(x + y)⁵ = ___(x⁵y⁰) + ___(x⁴y¹) + ___(x³y²) + ___(x²y³) + ____(x¹y⁴) + ____(x⁰y⁵)
Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
(x + y)⁵ = 1(x⁵y⁰) + 5(x⁴y¹) + 10(x³y²) + 10(x²y³) + 5(x¹y⁴) + 1(x⁰y⁵)
Step 4 : Simplify. Eliminate any powers of zero.

(x + y)⁵ = x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵

Expand (2 + y)⁵

You can expand this problem using several properties of the binomial theorem.
Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
(2 + y)⁵ = ___(2y) + ___(2y) + ___(2y) + ___(2y) + ____(2y) + ____(2y)
Step 2 - Distribute the power n, notice how you start at 5 with 2, and you start at 0 with y
The sum of the powers of 2 and y always equal n, in this case 5
(2 + y)⁵ = ___(2y) + ___(2y) + ___(2y) + ___(2y) + ____(2y) + ____(2y)
(2 + y)⁵ = ___(2⁵y⁰) + ___(2⁴y¹) + ___(2³y²) + ___(2²y³) + ____(2¹y⁴) + ____(2⁰y⁵)
Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
(2 + y)⁵ = 1(2⁵y⁰) + 5(2⁴y¹) + 10(2³y²) + 10(2²y³) + 5(2¹y⁴) + 1(2⁰y⁵)
Step 4 : Simplify the powers of 2
(2 + y)⁵ = 1(32y⁰) + 5(16y¹) + 10(8y²) + 10(4y³) + 5(2y⁴) + 1(1y⁵)
Step 5 - Simplify completely
(2 + y)⁵ = 32 + 80y + 80y² + 40y³ + 10y⁴ + y⁵
or (2 + y)⁵ = y⁵ + 10y⁴ + 40y³ + 80y² + 80y + 32

(2 + y)⁵ = y⁵ + 10y⁴ + 40y³ + 80y² + 80y + 32

Expand (2x + 2y)⁵

You can expand this problem using several properties of the binomial theorem.
Step 1) Set - up the expansion. There will be n + 1 terms in total. Leave a space to enter the coefficients derived from Pascal's Triangle
(2x + 2y)⁵ = ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ____(( 2x )( 2y )) + ____(( 2x )( 2y ))
Step 2 - Distribute the power n, notice how you start at 5 with (2x), and you start at 0 with (2y)
The sum of the powers of (2x) and (2y) always equal n, in this case 5
(2x + 2y)⁵ = ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ___(( 2x )( 2y )) + ____(( 2x )( 2y )) + ____(( 2x )( 2y ))
(2x + 2y)⁵ = ___(( 2x )⁵( 2y )⁰) + ___(( 2x )⁴( 2y )¹) + ___(( 2x )³( 2y )²) + ___(( 2x )²( 2y )³) + ____(( 2x )¹( 2y )⁴) + ____(( 2x )⁰( 2y )⁵)
Step 3 - Enter the coefficients of Pascal's Triangle when n = 5
(2x + 2y)⁵ = 1(( 2x )⁵( 2y )⁰) + 5(( 2x )⁴( 2y )¹) + 10(( 2x )³( 2y )²) + 10(( 2x )²( 2y )³) + _5(( 2x )¹( 2y )⁴) + 1(( 2x )⁰( 2y )⁵)
Step 4 : Distribute the powers using properties of exponents.
(2x + 2y)⁵ = 1(( 2⁵x⁵ )( 2⁰y⁰ )) + 5(( 2⁴x⁴ )( 2y )) + 10(( 2³x³ )( 2²y² )) + 10(( 2²x² )( 2³y³ )) + _5(( 2x )( 2⁴y⁴ )) + 1(( 2⁰x⁰ )( 2⁵y⁵ ))
Step 5 - Simplify the powers of 2
(2x + 2y)⁵ = 1(( 32x⁵ )( 1y⁰ )) + 5(( 16x⁴ )( 2y )) + 10(( 8x³ )( 4y² )) + 10(( 4x² )( 8y³ )) + _5(( 2x )( 16y⁴ )) + 1(( 1x⁰ )( 32y⁵ ))
Step 6 - Multiply the coefficients of Pascal's Triangle, with the coefficients of x and y.

(2x + 2y)⁵ = 32x⁵ + 160x⁴y + 320x³y² + 320x²y³ + 160xy⁴ + 32y⁵

Find the 6th term of (5x + 2y)⁷

Recall that the binomial theorem is defined as
(a + b)ⁿ = ∑_{k = 0}ⁿ (
n
k
)a^{n − k}b^k; (
n
k
) = [n!/(k!(n − k)!)]
This particular problem would then be defined as
(5x + 2y)⁷ = ∑_{k = 0}⁷ (
7
k
)(5x)^{7 − k}(2y)^k; where (
7k
) = [7!/(k!(7 − k)!)]
To find any term in the binomial expansion, all you need to solve is for k.
The relationship between Term# and k is the following
Term=k+1 Term# = k + 1
6 = k + 1
k = 5
Use k = 5, to find the 6th term
(
7
5
)(5x)^{7 − 5}(2y)⁵ = (
7
5
)(5x)²(2y)⁵ = [7!/(5!(7 − 5)!)](5x)²(2y)⁵ = [7*6*5!/5!(2)!](5x)²(2y)⁵ =
[7*6*5!/5!(2)!](5x)²(2y)⁵ = [7*6*/(2)!](5x)²(2y)⁵ = [42/2](5x)²(2y)⁵ = 21(25x²)(32y⁵)

The sixth term of (5x + 2y)⁷ = 16800x²y⁵

Find the 30th term of (x + y)⁵⁰

Recall that the binomial theorem is defined as
(a + b)ⁿ = ∑_{k = 0}ⁿ (
n
k
)a^{n − k}b^k; (
n
k
) = [n!/(k!(n − k)!)]
This particular problem would then be defined as
(x + y)⁵⁰ = ∑_{k = 0}⁵⁰ (
50
k
)x^{50 − k}y^k; where (
50
k
) = [50!/(k!(50 − k)!)]
To find any term in the binomial expansion, all you need to solve is for k.
The relationship between Term# and k is the following
Term=k+1
Term# = k + 1
30 = k + 1
k = 29
Use k = 29, to find the 30th term
(
50
29
)(x)^{50 − 29}(y)²⁹ = (
50
29
)(x)²¹(y)²⁹ = [50!/(29!(50 − 29)!)](x)²¹(y)²⁹ = [50!/29!(21)!](x)²¹(y)²⁹ =
[50*49*48...28*29!/29!(21)!](x)²¹(y)²⁹ = [50*49*48...28/(21)!](x)²¹(y)²⁹ = [50*49*48...28/(21)!](x)²¹(y)²⁹ =

The 30th term of (x + y)⁵⁰ = 67,327,446,062,800x²¹y²⁹

Find the 4th term of (6x + y)⁷

Recall that the binomial theorem is defined as
(a + b)ⁿ = ∑_{k = 0}ⁿ (
n
k
)a^{n − k}b^k; (
n
k
) = [n!/(k!(n − k)!)]
This particular problem would then be defined as
(6x + y)⁷ = ∑_{k = 0}⁷ (
7
k
)( 6x )^{7 − k}y^k; where (
7
k
) = [7!/(k!(7 − k)!)]
To find any term in the binomial expansion, all you need to solve is for k.
The relationship between Term# and k is the following
Term# = k + 1
4 = k + 1
k = 3
Use k = 3, to find the 4th term
(
7
3
)(6x)^{7 − 3}(y)³ = (
7
3
)(6x)⁴(y)³ = [7!/(3!(7 − 3)!)](6x)⁴(y)³ = [7!/3!(4)!](6x)⁴(y)³ =
[7*6*5*4!/3!(4)!](6x)⁴(y)³ = [7*6*5*/3!](6x)⁴(y)³ = 35(6x)⁴y³ = 35(6⁴x⁴)y³ = 35(1296x⁴)y³ = 45360x⁴y³

The 4th term of (6x + y)⁷ = 45360x⁴y³

Find the 4th term of (2x − 4y)⁷

Recall that the binomial theorem is defined as
(a + b)ⁿ = ∑_{k = 0}ⁿ (
n
k
)a^{n − k}b^k; (
n
k
) = [n!/(k!(n − k)!)]
This particular problem would then be defined as
(2x + ( − 4y))⁷ = ∑_{k = 0}⁷ (
7
k
)( 2x )^{7 − k}( − 4y)^k; where (
7
k
) = [7!/(k!(7 − k)!)]
To find any term in the binomial expansion, all you need to solve is for k.
The relationship between Term# and k is the following
Term# = k + 1
4 = k + 1
k = 3
Use k = 3, to find the 4th term
(
7
3
)(2x)^{7 − 3}( − 4y)³ = (
7
3
)(2x)⁴( − 4y)³ = [7!/(3!(7 − 3)!)](2x)⁴( − 4y)³ = [7!/3!(4)!](2x)⁴( − 4y)³ =
[7*6*5*4!/3!(4)!](2x)⁴( − 4y)³ = [7*6*5*/3!](2x)⁴( − 4y)³ = 35(2x)⁴( − 4y)³ = 35(2⁴x⁴)( − 4)³y³ = 35(16x⁴)( − 64)y³ = − 35840x⁴y³

The 4th term of (2x − 4y)⁷ = − 35840x⁴y³

Find the 4th term of (2x³ − 4y³)⁷

Recall that the binomial theorem is defined as
(a + b)ⁿ = ∑_{k = 0}ⁿ (
n
k
)a^{n − k}b^k; (
n
k
) = [n!/(k!(n − k)!)]
This particular problem would then be defined as
(2x³ + ( − 4y³))⁷ = ∑_{k = 0}⁷ (
7
k
)( 2x³ )^{7 − k}( − 4y³)^k; where (
7
k
) = [7!/(k!(7 − k)!)]
To find any term in the binomial expansion, all you need to solve is for k.
The relationship between Term# and k is the following
Term=k+1
Term# = k + 1
4 = k + 1
k = 3
Use k = 3, to find the 4th term
(
7
3
)(2x³)^{7 − 3}( − 4y)³ = (
7
3
)(2x)⁴( − 4y)³ = [7!/(3!(7 − 3)!)](2x³)⁴( − 4y³)³ = [7!/3!(4)!](2x³)⁴( − 4y³)³ =
[7*6*5*4!/3!(4)!](2x³)⁴( − 4y³)³ = [7*6*5*/3!](2x³)⁴( − 4y³)³ = 35(2⁴x¹²)( − 4³)(y⁹) = 35(16x¹²)( − 64)(y⁹) = − 35840x¹²y⁹

The 4th term of (2x³ − 4y³)⁷ = − 35840x¹²y⁹

Find the 4th term of (x² + y³)¹⁴

Recall that the binomial theorem is defined as
(a + b)ⁿ = ∑_{k = 0}ⁿ (
n
k
)a^{n − k}b^k; (
n
k
) = [n!/(k!(n − k)!)]
This particular problem would then be defined as
(x² + y³)¹⁴ = ∑_{k = 0}¹⁴ (
14
k
)( x² )^{14 − k}(y³)^k; where (
14
k
) = [14!/(k!(14 − k)!)]
To find any term in the binomial expansion, all you need to solve is for k.
The relationship between Term# and k is the following
Term# = k + 1
4 = k + 1
k = 3
Use k = 3, to find the 4th term
(
14
3
)(x²)^{14 − 3}(y³)³ = (
14
3
)(x²)¹¹(y³)³ = [14!/(3!(14 − 3)!)](x²)¹¹(y³)³ = [14!/3!(11)!](x²²)(y⁹)
[14!/3!(11)!](x²²)(y⁹) = [14*13*12*/3!](x²²)(y⁹) = 364x²²y⁹

The 4th term of (x² + y³)¹⁴ = 364x²²y⁹

Find the 10th term of (x² + y³)¹⁴

Recall that the binomial theorem is defined as
(a + b)ⁿ = ∑_{k = 0}ⁿ (
n
k
)a^{n − k}b^k; (
nk
) = [n!/(k!(n − k)!)]
This particular problem would then be defined as
(x² + y³)¹⁴ = ∑_{k = 0}¹⁴ (
14
k
)( x² )^{14 − k}(y³)^k; where (
14
k
) = [14!/(k!(14 − k)!)]
To find any term in the binomial expansion, all you need to solve is for k.
The relationship between Term# and k is the following
Term=k+1
Term# = k + 1
10 = k + 1
k = 9
Use k = 9, to find the 10th term
(
14
9
)(x²)^{14 − 9}(y³)⁹ = (
14
9
)(x²)⁵(y³)⁹ = [14!/(9!(14 − 9)!)](x²)⁵(y³)⁹ = [14!/9!(5)!](x¹⁰)(y²⁷)
[14!/9!(5)!](x¹⁰)(y²⁷) = [14*13*12*11*10*/(5)!](x²²)(y⁹) = 2002x¹⁰y²⁷

The 10th term of (x² + y³)¹⁴ = 2002x¹⁰y²⁷

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Answer

Binomial Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Pascal's Triangle

Properties

Example: Properties of Binomials

Factorials

Binomial Theorem

Example: Binomial Theorem

Finding a Specific Term

Example: Specific Term

Example 1: Expand

Example 2: Fourth Term

Example 3: Five Terms

Example 4: Three Iterates

Intro 0:00
Pascal's Triangle 0:06

Expand Binomial
Pascal's Triangle

Properties 6:52

Example: Properties of Binomials

Factorials 9:11

Product
Example: Factorial

Binomial Theorem 11:08

Example: Binomial Theorem

Finding a Specific Term 18:36

Example: Specific Term

Example 1: Expand 24:39
Example 2: Fourth Term 30:26
Example 3: Five Terms 36:13
Example 4: Three Iterates 45:07

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Binomial Theorem

Welcome to Educator.com.0000

In today's lesson, we are going to be covering the binomial theorem.0002

We are going to start out by talking about what we mean when we say we are going to expand a binomial.0007

So, if you expand a binomial, this is what happens.0013

I have (a + b)ⁿ; and I am going to expand that for n = 1, 2, 3, 4.0019

I am going to let n equal 1; and I am going to then get (a + b)¹, which is just a + b.0027

If I let n equal 2, then I have(a + b)²; that is going to give me, if you will recall, a² + 2ab + b².0041

Let n equal 3: you get (a + b)³, which is a little bit more complicated to work out.0057

But if you figure that all out, (a + b)(a + b)(a + b), you would end up with a³ + 3a²b + 3ab² + b³.0064

Letting n equal 4, (a + b)⁴ is going to give you a⁴ + 4a³b +0079

6a²b² + 4ab³ + b⁴.0091

One more: let n equal 5: this is going to give me (a + b)⁵.0099

This equals a⁵ + 5a⁴b + 10a³b² + 10a²b...0108

actually, this is going to be b³...+ 5ab⁴ + b⁵.0127

All right, before we go on to talk about Pascal's triangle, let's note a few things about this expansion.0136

And these will be summed up on the next slide.0144

A few things that you will notice are that the first term is aⁿ, and the last term is bⁿ.0146

The second thing is that the number of terms equals n + 1.0157

You will also notice a pattern: let's look at n = 4, (a + b)⁴.0179

The first term is aⁿ, so it is a⁴; in each subsequent term, the power that a is raised to (the exponent) decreases by 1.0185

So, I started out with a⁴; then here it's a³; a²; a; and then a is gone.0195

So, the exponent for a decreases by 1 for each term; b does the opposite.0203

There is no b here; this would be b⁰...it is 1; and then in my next term, I get b to the first power, b², b³, b⁴.0213

So, with each term, exponents for a are decreasing; exponents for b are increasing.0223

The next thing that you might notice is the coefficients; let's look at the coefficients.0230

The coefficient of the second term is equal to n; again, this is all summed up on the next slide, but this is just to introduce it here.0234

So, if I am looking at n = 3 (that is (a + b)³), and I expand that; then I get that the second term has a coefficient of 3.0242

The second term here has a coefficient of 4; the second term here has a coefficient of 5.0252

Pascal's triangle--we are getting to that now.0258

I can take the coefficient of these terms and use them to create an array that is called Pascal's triangle.0260

Looking at it, it actually starts out with 1; so if we let n equal 0, that is just going to give me (a + b)⁰, which is going to be 1.0266

Then, in my next row, I am going to have 1, 1, and 1; those are my coefficients.0277

In the next row, they are 1, 2, 1; in the next row, 1, 3, 3, 1.0289

Now, I can look up here to get the next row; but I don't even need to, because each number is the sum of the two numbers just above it.0302

So, 1 + 1 forms 2; 1 + 2 gives me 3; 2 + 1 gives me 3.0314

So, all I have to do, actually--these outside numbers are always 1; but I could just say 1 + 3...this is going to be 4.0325

3 + 3 is equal to 6; 3 + 1...that is equal to 4.0334

So, I look up here to verify that; and indeed, it is 1, 4, 6, 4 1.0340

In the next row, I would add 1, and I add 1 + 4; I am going to get 5; 4 + 6 gives me 10;0345

6 + 4 gives me 10; 4 + 1 gives me 5; and again, I have a 1 out here.0352

And that matches up with what I have for the expansion of (a + b)⁵.0357

So, Pascal's triangle is an interesting array that comes from the coefficients of the expansion of this binomial.0363

Something else to note is the symmetry around the middle term; you might have seen that up here, but it is even more obvious right here.0371

The middle term is 2; it is flanked by 1's; here I have two middle terms, because there is an even number of terms;0377

so the two middle terms are 3; they are flanked by 1's; now I have a middle term of 6; next to it are 4's; outside of that, 1's.0384

Two middle terms are 10; the next terms are 5; the next terms are 1.0392

And that is helpful, because if you are looking for the coefficients, you don't need to find all of them; you only need to find half of them.0397

And then, you can use this symmetry to know what the other half are.0403

Summing up these properties of the expansion of (a + b)ⁿ: there are n + 1 terms.0409

Let's use an example that we just talked about, (a + b)⁵, which gave me a⁵ +0418

5a⁴b + 10a³b² + 10a²b³ +0427

5ab⁴ + b⁴, to illustrate this.0437

There are n + 1 terms; so here, n = 5; therefore, I am going to have n + 1 terms, equals 5 + 1, so I am going to have 6 terms.0442

And I do have 1, 2, 3, 4, 5, 6 terms.0455

The first term is going to be aⁿ (that is a⁵), and the last term is bⁿ, which should be b⁵ right here.0458

In successive terms, the exponent of a decreases by 1, while that of b increases by 1.0470

I start out with a⁵, then a⁴, to the third, to the second, and then a.0475

b goes the opposite way: I start with b, b², b³, b⁴, and b⁵.0482

I pointed out before that the second term has a coefficient that is equal to n.0490

Something I also pointed out in the last slide is that there is symmetry around the middle term or terms,0498

which tells me that, if I figure out these middle coefficients, well, I already know that the first and last coefficients are 1.0504

And I know that the second (and, by symmetry, second-to-last) coefficients are equal to n.0511

So then, all I would have to figure out is one of these; and I know that the other one is the same.0519

Something else to be aware of is that the sum of the exponents in each term is n.0523

Let's look right here: for this term, 10a³b², if I add up 3 and 2, that is going to give me 5.0528

So, the sum of the exponents is equal to 5.0537

Or looking at this, 10a²b³, I get the same thing.0540

The sum of the exponents of each term is equal to n.0544

The next concept that we will learn requires that you remember how to work with factorials.0552

So, I am going to go ahead and review factorials: factorials are products, and the symbol that looks like an exclamation point is actually read as "factorial."0557

This is a product: n! is equal to n times n - 1 times n - 2 times n -3...and you continue on that way until you get all the way to 2, and then finally 1.0567

For example, 4 factorial: here, n = 4; this is going to be equal to 4, so here n = 4; n - 1...4 - 1 is 3, so that is 4 times 3.0586

4 - 2 gives me 2; 4 - 3 gives me 1; and this equals...4 times 3 is 12, times 2, times 1, is just 24.0599

So, 4! is just the product of 4 times 3 times 2 times 1.0613

If you are working with fractions that involve factorials in the numerator and denominator, you can often do a lot of canceling out to make things easier.0618

If you were working with 8!/6!,that gives you 8 times 7 times 6 times 5 times 4, and then on down to 1.0625

In the denominator, you have 6 times 5 times 4 times 3 times 2 times 1.0637

So, instead of multiplying all of this out and dividing, it makes a lot more sense to start canceling out all of these common terms.0644

1 through 6; 1 through 6; this just leaves me with 8 times 7, or 56.0657

Now, the reason we reviewed it is to work with this formula.0665

First, we are talking about the binomial theorem, and then the binomial formula.0668

The binomial theorem is really what we just discussed; and it is the idea that, when you expand (a + b)ⁿ,0672

you are going to get an equation with the properties we just discussed.0682

And those properties are that the first term is going to be aⁿ;0688

the second term is going to have some coefficient; and it is going to be raised to the power a^{n - 1}.0692

b is going to be to the first power; then you are going to get another term; and you are going to have a total of n terms.0699

You are going to have another term; it is going to have a coefficient that is going to be a^{n - 2}.0707

b is going b²; it is going to increase by 1.0711

Then, you are going to get another coefficient, a^{n - 3}; b is going to become b³.0714

And that is going to go on and continue on until you get ab^{n - 1}, bⁿ, and so on.0721

Excuse me, plus...that is a^{n - 1}, times b, plus bⁿ; the last term will be bⁿ; the first term is aⁿ.0731

Each of these terms can be given in the general form a^{n - k}b^k.0751

And we will look at some examples in a few minutes, and talk about what this means, as we relate it to the binomial formula.0760

This is the binomial formula; this is pronounced "n choose k."0769

Now, why are we even looking at this--what is this for?0781

Think about how we can find these coefficients.0785

We can use Pascal's triangle, but that could be really impractical and quite a bit of work,0788

if you have to go through that whole large array, if you have a really big expansion.0793

And if you just find the coefficient for a particular term, you are going to have to go through a lot of work to get there.0799

So, as usual, we have a formula that gets us directly there, without having to go through all that work.0804

This formula will give me the coefficients for a term that has the exponent a^{n - k}b^k.0812

So, the binomial formula allows me to find this coefficient.0823

Let's use an example of (a + b)⁷: I can find the coefficient, using this, of a particular term;0829

or I can expand the whole thing and find all of the terms.0839

Let's look at what this expansion would look like: it would look like this.0841

I know that the first term is going to be a⁷; recall that the second coefficient is going to be equal to n.0845

So this one I also know; it is 7; and then, I know that a...the exponent is going to decrease by 1; and b is going to appear, raised to the first power.0851

I also know that I am going to have 8 terms, because I am going to have n + 1 terms, or 7 + 1.0866

So, I have another term here; I don't know the coefficient, but I do know that it is going to be a⁵b².0872

I have another term with an unknown coefficient: a⁴b³.0878

I have another term, again with an unknown coefficient: a³b⁴.0884

Another term is a²b⁵; another term is ab⁶; plus b⁷.0890

I have 8 terms, so I am going to have these two middle terms the same; these two terms will be the same;0900

the second-to-last terms will be the same; so I know this is 7; and these two outer terms will be the same, 1 and 1.0907

I may be asked to find this expansion; I found this much; now, in order to finish it out, I need to find this coefficient and this coefficient.0917

If I have those, I have the rest, because I would have the first half, and then I just reflect for the second half.0927

Let's use this formula to find the coefficient for this third term.0935

In order to use this formula, I need to know two things: I need to know n, and I need to know k.0950

n is easy--we know n equals 7; k...if you look up here, k is the exponent of b for that term.0956

So, I am looking for the coefficient that goes right here, where n is 7 and k is 2.0969

Once I have that, it is a matter of using the formula: n!...this is going to be 7!/k!, which is 2!...times (7 - 2)!;0978

this equals 7! divided by 2!, times 5!.0995

I am going down here to complete this: 7 times 6 times 5 times 4 times 3, and so on.1002

In the denominator, I have 2!, 2 times 1, times 5 times 4 times 3 times 2 times 1 (factorials).1010

Again, I am going to cancel out common factors to make this much easier to work with.1023

This leaves me with 7 times 6, divided by 2; that is simply 42/2, or 21.1030

Therefore, the coefficient for the third term is actually 21.1043

Since I know that, and I know that we are going 1, 7, 21, starting from this end1052

(we are also going 1, 7, 21), I could do the same thing to find the fourth term coefficient.1057

The fourth term coefficient would have, again, an n equal to 7; but this time, k would be equal to 3.1067

Note one thing: instead of having to write all this out, to figure out what k is and everything,1077

you can just be aware that the term number equals k + 1.1082

Let's look at the fourth term: the term number is 4--that equals k + 1; k = 3.1092

For the fourth term, k = 3; so that is 1, 2, 3, 4; I know n = 7; and since this is the fourth term, I quickly know that k = 3.1101

So, you can use the binomial theorem, and specifically the binomial formula, to find a specific term of the expanded form of (a + b)ⁿ.1117

And I talked, in the last section, about how you can use that formula to find the whole expansion.1126

Or you may just need to find a certain coefficient within the expansion.1133

Recall that the binomial formula is equal to n!, divided by k!, times (n - k)!, where n is this n,1140

and k is the value of the exponent for the b in the term that you are looking for the coefficient of.1158

As an example, we are going to do something slightly more complex.1167

Instead of just a + b, we are actually going to say a - 2b; and we are going to expand that--it is going to be to the fifth.1169

So, we have to account for this -2--that the second term here is not just b; it is -2b.1178

I know that the first term is going to be a⁵.1188

I know that the second term is going to have a coefficient of n, so it is a coefficient of 5, and that a is going to decrease by 1.1193

I have to be very careful and not just write b; I have to write -2b.1200

Then, I have an unknown coefficient; but I do know that I am going to get a³, because the exponent that a is raised to will decrease by 1.1206

And here, I just have to say that b is really -2b; that is going to be squared; it is going to increase by 1.1216

I have another unknown coefficient, and it becomes a³ and goes down to a²; -2b squared goes up to -2b cubed.1226

Then, there is another coefficient here, and I get a³, -2b to the fourth, and then finally...1237

actually, this is just a; it goes from a² to a; and then finally, a is gone, and I have -2b⁵.1247

By symmetry, I know the coefficient of these outer terms, right here, is actually 1; you can think of that as 1.1256

The two outer terms, then, have coefficients of 5.1266

All right, so back to finding a specific term: let's say that I want to find the coefficient for this third term.1270

n is going to equal 5; so in order to find the coefficient for this third term, I need to use the binomial formula; I need to have n, and I need to have k.1298

k...I could just say it is one less than the term number, so k is 2.1307

Or I could look up here and say it is whatever power this is raised to, which is 2.1311

Next, I insert these values into the formula; that is going to give me 5!, divided by 2!, times (n - 2)!.1318

This is equal to 5! divided by 2!, times 3!, equals 5 times 4 times 3 times 2 times 1, divided by 2! (2 times 1), times 3 times 2 times 1.1334

Cancel out common factors; that leaves me with 5 times 4, divided by 2 times 1, or 20 divided by 2, is 10.1354

Therefore, coefficient for the third term is equal to 10.1368

I can go back up and put this in here, but there is some simplifying that needs to be done.1374

So, the third term equals 10a³, (-2b)²; this equals 10a³...-2 times -2 is 4b².1379

This equals 10 times 4, a³, b², which equals 40a³b².1398

It is a little bit more complicated than just putting the 10 in there.1408

And then, something to be careful of, as well: then I can say that by symmetry, this is also 10.1411

But I need to work with what is in here, which is actually a²(-2b)³, so it is somewhat different.1416

So, let's look at what this fourth term is: 1, 2, 3,...the fourth term.1425

Again, I am going to use that coefficient, 10, but here I have a²(-2b)³.1431

So, it is 10a²...-2 times -2 is 4, times -2 is -8; so it is times -8b³.1439

I can rewrite this as 10 times -8a²b³ = -80a²b³.1454

When I simplify that fourth term, this is what it is going to look like.1466

That was using the binomial formula to find a specific term of the expanded form.1471

In the first example, we need to expand (x + 2)⁴.1481

The general form is (a + b)ⁿ; here, I do the same thing, just realizing that a is equal to x, and b is equal to 2.1486

I am going to get something that looks like this: (a + 2)⁴; I am going to have five terms.1497

I am going to have n equal 4; so, there are going to be 5 terms.1502

The first term is going to be a²; in this case, it is x, so it is x².1508

Excuse me, not squared; to the n--it is going to be a to the n, which is 4, so it is actually going to be x⁴.1517

In the next term, I have the coefficient; and the coefficient of the second term is equal to n.1525

So, that is going to be 4, times x...this exponent is going to decrease by 1...times 2.1531

The next term: I am going to have some unknown coefficient; x is going to go from cubed to squared;1543

2 is going to go from being the first power to the second power.1555

I am going to have another term, and then x² is just going to become x; 2 is going to go from being squared to being cubed;1560

and then, finally, x drops out, and I get, for my last term, 2ⁿ, or 2⁴.1572

By symmetry, I know that I am going to have a middle term (I have an odd number of terms--I have one middle term);1581

the two next to it are going to have the same coefficient--it is going to be 4.1587

I can simplify this later on, because I am going to have to deal with this 2³ and this 2, and multiply that times the coefficient.1591

But right now, let's just leave it like this.1600

I am almost done with my expansion; for this middle term, there are two ways I could go, because this is not that large of an expansion.1603

I could use the binomial formula, and we are going to get more practice with that in a minute.1612

But this time, I am actually going to use Pascal's triangle.1615

Remember that it starts out with (a + b)⁰, and you end up with 1 as the only coefficient, because this equals 1.1617

Then, I get (a + b)¹ = (a + b); that gives me 1 and 1 as coefficients.1632

Once I have those two, I can find the rest, because I know I have 1's out here.1639

And to find the other terms, I add the two above: 1 + 1 gives me 2.1645

So, for (a + b)², these are my coefficients; then I want (a + b)³.1651

And what I am finally looking for, to find this coefficient, is when (a + b) is expanded to the fourth.1659

So, I have a 1 out here and a 1 out here; 1 + 2 gives me 3; 2 + 1 gives me 3;1667

so the coefficients for this expansion (that is going to have 4 terms) are going to be 1, 3, 3, 1.1674

Finally, the one I am looking for: I have a 1 and a 1, which I have up here;1680

1 + 3 is 4; 1 + 3 is 4, which I already knew; this gives me my middle coefficient of 6.1685

Now, I could have used the binomial formula to find this; but I wanted1695

to just do it a little bit differently this time, since it was a fairly small expansion, and use Pascal's triangle.1699

Now, I have the expansion, but it is not really simplified; we need to simplify these various terms.1706

So, let's take...the first term is already simplified; the second term equals (here is the second term) 4 times x³ times 2; that equals 8x³.1712

The third term equals (that is 1, 2...the third term) 6x², 2², equals 6x² times 4, equals 24x².1733

That is the third term; the fourth term equals 4x times 2 cubed; that is 4x times 8; 8 times 4 is 32, so that is 32x.1750

The fifth term equals 2⁴; 2³ is 8, times 2 again gives me 16.1764

Rewriting it all here, the expansion of (x + 2)⁴ is1774

x⁴ + (second term) 8x³ + (third term) 24x² + (fourth term) 42x + (the final term) 16.1779

Again, I was able to expand this using my properties that I am aware of for the expansion of a binomial1794

to find what these exponents are, how many terms there are going to be, and what this coefficient and this coefficient are.1802

I was left with just finding one coefficient that I could have used Pascal's triangle for.1810

That is what I did; I found the 6 (or I could have used the binomial formula).1814

Once I had these coefficients, all I did was to simplify; and this is the expansion, right down here.1818

Here, I am asked to find the fourth term of the expansion of (a + 2b)⁸.1829

So, I am asked to find this specific term; and we know that we can use the binomial formula, n!/k! times (n - k)!.1834

I know that I have n; n = 4; what is k?1847

Well, the term number equals k + 1; the term number is 4, so this means 4 = k + 1.1853

This gives me k = 3; so I have n = 4; k = 3; I can find the coefficient.1866

I need to find the coefficient of this fourth term; it is going to have a coefficient.1875

For the first term, it is going to be a raised to some power; for the second component of it, it is going to be 2b (not just b), raised to some power.1884

How do I know what these powers are?1902

Well, recall that it is going to be equal to (let's give ourselves some more room)...1904

when you find a term used in a binomial formula, you also know that it is in the form n - k,1910

and then the second part is raised to the k power, because that is what k is--1919

it is whatever you are raising b to; and in this case, b = 2b.1926

I also know that the sum of the exponents has to equal n; the sum of the exponents of a term equals n.1934

This makes sense, because n - k + k needs to equal n; the sum of the exponents must equal n, and that works out; so this makes sense.1945

So, the fourth term: I need to find the coefficient using the binomial formula.1954

I know what the power a is going to be raised to is, n - k; and I know that the power that 2b is going to be raised to is k.1960

Let's go ahead and find the coefficient: the coefficient of the fourth term is going to be equal to n!, which is 4!...1967

excuse me, n is actually 8--a correction on that: we are finding the fourth term, but n is right here, so n is actually 8.1981

So, it is n! (that is 8!), divided by k! (that is 3!); n (which is 8), minus 3, factorial, equals 8!/3! times 5!.1993

That gives me 8! divided by 3 times 2 times 1, times 5!.2013

I do my canceling to make things easier to work with.2026

This leaves me with 8 times 7 times 6, divided by 3 times 2 times 1.2032

6 is equal to 3 times 2, so I have more common factors and more canceling.2040

That just leaves me with 8 times 7, divided by 1, which is 56; that was the hardest part here--finding the coefficient.2045

The coefficient of this fourth term is equal to 56.2053

To write that out completely, I am going to write it out as 56a^{n - k}; that is 8 - 3;2058

times (2b) raised to the k power (k is 3); this equals 56a⁵(2b)³.2071

Simplifying further: 56a⁵...2 cubed is 8, times b³.2087

If you multiply out 56 times 8, you will get 448; a⁵, b³.2095

This is the fourth term of this expanded form of (a + 2b)⁸.2103

It is a little bit complicated: the first thing I had to do is find the coefficient.2111

And I did that by using my binomial formula and knowing that n = 8 and k = the term number, plus 1.2115

The term number plus 1 is k, so k is one less than the term number: k is 3.2128

I was able to find, then, using this formula, that the coefficient is 56.2133

The second thing I had to do is figure out what power I should raise a to, and what power I should raise 2b to.2139

Well, when I am using the binomial formula, I am finding the coefficient for a term in this form, which is a^{n - k}2b^k.2145

So, I knew my n and my k; so I put my coefficient here and said, "OK, I have a^{8 - 3}(2b)³," using this.2153

Simplify, and then multiply to get 448a⁵b³.2163

We are asked to expand (2x + 3y)⁵...this binomial to the fifth.2172

And the general form is (a + b)ⁿ.2178

So, here we actually have that a is equal to 2x, and b is equal to 3y; so that makes things a little bit more complicated at the end.2181

We have to take these into account when we simplify our final answer.2190

Since this is the fifth power, I have an n equal to 5.2197

I am going to have a total of 6 terms.2202

It is going to look something like this at the end: (2x + 3y)⁵ =...my first term is 2x raised to the n power,2205

which is the fifth; plus the coefficient, and then 2x is going to decrease to the fourth power, and I am going to have my 3y here.2214

Plus...some other coefficient; 2x becomes cubed, and 3y increases from an exponent of 1 to an exponent of 2.2225

Another coefficient is going to bring me to 2x²3y³;2241

another coefficient...now I just have 2x, and 3y⁴; and then, finally, 3y⁵.2250

And I expect to have 6 terms--let's verify that I am not missing anything: 1, 2, 3, 4, 5, 6.2261

Down here at the bottom, I am going to start building up what our final answer is going to look like, because I have some simplifying that I need to do.2270

The first term, I can figure out: I already know that it is 2x⁵.2281

2³ is 8, times 2 is 16, times 2 is 32; so that is 32x⁵.2286

The second term: recall that the coefficient for the second term is equal to n--that is 5.2293

Also, remember that symmetry is involved: since there are 6 terms (that is an even number of terms), I am going to have two middle terms.2301

These two are going to have the same coefficient; the two next to those will have the same coefficients; and the two outer have the same coefficients.2309

It is not the same value, once I simplify them; but the same coefficient.2319

So, since these two are the same, and these two are the same, this term also has a coefficient of 5.2323

Let's work this one out up here; this is the second term, and I know that it is equal to 5 times (2x)⁴ times 3y.2332

That is...2 times 2 is 4; 2 cubed is 8, times 2 is 16; so that is times 16x⁴, times 3y.2346

Multiplying 5 times 16 times 3 is going to give you 240x⁴y; I am going to put that in here; that is 240x⁴y.2358

I don't know what I have for these two middle ones, but it is going to be something here, and it is going to be x³y².2373

Something is here, then x²y³; I can figure this one out:2384

right here, this is the third term; it equals 5 times 2x, times (3y)⁴.2394

5 times 2x is just 10x; 3 times 3 is 9, times 3 is 27, times 3 is going to give me 81.2406

So, that is times 81y⁴; so this one is easy, because it is just 81 times 10, is 810xy⁴.2419

So, that was...actually not the third term: correction--we don't know the third term; the first, second, third, and fourth...2436

the two middle terms we don't know; we do know the fifth term--this is the fifth term.2448

All right, so the fifth term I can fill in: 810xy⁴.2456

The sixth term, right here, I can also fill in: the sixth term equals (3y)⁵.2465

3³ is 27; 3⁴ is 81; 3⁵, if you figure it out, is 243; 243y⁵.2475

Just using the properties of binomial expansions, I got most of the terms.2489

And since there is symmetry, the two middle terms will have the same coefficient inserted up here.2496

And then, I can work out the rest and finalize it.2505

So, I only need to find one coefficient, using the binomial formula.2507

The binomial formula, recall, is n!/k!, times (n - k)!.2511

I know that n = 5; for the first, second, third term, k is equal to 2; you could just say "Third term: 3 - 1 is 2";2521

or you could look up here and say, "OK, 3y is raised to the second power, so k = 2."2532

Therefore, I am going to have 5!/2!, times (5 - 2)!; this equals 5!/2! times 3!,2537

which equals 5 times 4 times 3 times 2 times 1; 2 times 1; 3 times 2 times 1.2559

1, 2, and 3 all cancel; this gives me 5 times 4, divided by 2; that is 20/2, or 10.2568

So, the missing coefficient here is 10; by symmetry, the missing coefficient right here is also 10; I am almost done.2578

So, for the first, second, third term, I have my coefficient; but I need to simplify that a bit.2588

My third term equals 10 times (2x)³ times (3y)².2595

This is going to be equal to 10 times...2 cubed is 8x³; 3 squared is 9y².2605

This is going to give me 80 times 9x³y², which is just 720x³y².2617

That is my third term, so I am going to put a 720 right here.2625

The fourth term: I also have a 10 right here.2630

What is slightly different, though: it is 10 times (2x)²(3y)³.2635

This is going to give me 10 times 4x²; 3 cubed is 27, so it is y³.2642

This is 40 times 27...let's put all of the constants together...x²y³.2651

40 times 27 actually comes out to 1080; you can multiply that out for yourself; x²y³.2663

The fourth term, therefore: I am going to put a 1080 right here.2673

So, this is quite a bit of work; but we actually only needed to find one coefficient using the binomial formula,2676

because by symmetry, these two middle ones were the same (the two middle coefficients that we will put in the blank);2682

and I knew that the second and the second-to-the-last had coefficients equal to n.2690

Once I found that a 10 went into these blanks, all I needed to do was work out what each of these terms would simplify to,2695

to end up with the expansion that I found down here.2703

We are asked to find a specific term this time, not the entire expansion.2710

I just want to find the fourth term of (x - 3y)⁷, using the binomial formula, n!/k!, times (n - k)!.2713

All right, n = 7; what does k equal? Well, it is the fourth term; k is going to be one less than that; 4 - 1...I am going to have k = 3.2725

This is going to give me the coefficient for a term in this form.2737

It is going to have a coefficient, and then it is going to have x raised to the n - k power, times -3y, raised to the k power.2741

This is going to become x^{7 - 3}, (-3y)³.2757

Therefore, I am going to get some coefficient (that I am going to find in a moment), times x⁴, times (-3y)³.2768

Let's go ahead and find the coefficient: this missing coefficient equals n!, 7!, divided by k!, times (7 - 3)!;2775

that equals 7!/3!, times 4!, equals (let's write this out)...divided by 3 times 2 times 1, times 4 times 3 times 2 times 1.2793

Do some canceling out: 1's, 2's, 3's, and 4's are the same.2812

This leaves me with 7 times 6 times 5, divided by 3 times 2 times 1.2819

6 has common factors here of 3 and 2; I can just cancel those out.2825

So, this is 7 times 5, divided by 1; so just 7 times 5 equals 35.2832

I know that my coefficient is, then, 21 that is what is going to go in this blank.2838

So, I am going to end up with...the fourth term is going to be 35 (I am going to put 35 right here in this blank)2842

x⁴(-3y)³, which equals 35x⁴...-3 squared is 9, times -3 again is -27.2849

So, it is -27y³; so it is time to get out your calculators and figure out 35 times -27, or multiply it out, if necessary.2864

And you will find that that equals -945x⁴y³.2875

OK, we found the fourth term of this expansion by using the binomial formula to find the coefficient,2881

and also, an awareness that the exponent of this x is going to be n - k; the exponent for the -3y is going to be k.2887

We figured out the coefficient (it is 35), and then simplified, using our rules for working with exponents.2896

That concludes this session on the binomial theorem here at Educator.com; thanks for visiting!2905

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Dr. Carleen Eaton

Binomial Theorem

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro