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1 Two hungry Math 160 students are waiting in line at Chipotle
Two hungry Math 160 students are waiting in line at Chipotle. First math student: β€œOh, no! I forgot my wallet! What should I do?” Second math student: β€œπ‘Ž+𝑏”

2 Packet #31 The Binomial Theorem
Math 160 Packet #31 The Binomial Theorem

3 How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛
How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛? Let’s try 𝑛=1, 2, 3, 4,… and find patterns. π‘Ž+𝑏 1 = π‘Ž+𝑏 π‘Ž+𝑏 2 = π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 π‘Ž+𝑏 3 = π‘Ž 3 +3 π‘Ž 2 𝑏+3π‘Ž 𝑏 2 + 𝑏 3 π‘Ž+𝑏 4 = π‘Ž 4 +4 π‘Ž 3 𝑏+6 π‘Ž 2 𝑏 2 +4π‘Ž 𝑏 3 + 𝑏 4 π‘Ž+𝑏 5 = π‘Ž 5 +5 π‘Ž 4 𝑏+10 π‘Ž 3 𝑏 π‘Ž 2 𝑏 3 +5π‘Ž 𝑏 4 + 𝑏 5 Notice the exponent pattern (π‘Ž goes from 𝑛 to 0, and 𝑏 goes from 0 to 𝑛).

4 How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛
How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛? Let’s try 𝑛=1, 2, 3, 4,… and find patterns. π‘Ž+𝑏 1 = π‘Ž+𝑏 π‘Ž+𝑏 2 = π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 π‘Ž+𝑏 3 = π‘Ž 3 +3 π‘Ž 2 𝑏+3π‘Ž 𝑏 2 + 𝑏 3 π‘Ž+𝑏 4 = π‘Ž 4 +4 π‘Ž 3 𝑏+6 π‘Ž 2 𝑏 2 +4π‘Ž 𝑏 3 + 𝑏 4 π‘Ž+𝑏 5 = π‘Ž 5 +5 π‘Ž 4 𝑏+10 π‘Ž 3 𝑏 π‘Ž 2 𝑏 3 +5π‘Ž 𝑏 4 + 𝑏 5 Notice the exponent pattern (π‘Ž goes from 𝑛 to 0, and 𝑏 goes from 0 to 𝑛).

5 How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛
How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛? Let’s try 𝑛=1, 2, 3, 4,… and find patterns. π‘Ž+𝑏 1 = π‘Ž+𝑏 π‘Ž+𝑏 2 = π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 π‘Ž+𝑏 3 = π‘Ž 3 +3 π‘Ž 2 𝑏+3π‘Ž 𝑏 2 + 𝑏 3 π‘Ž+𝑏 4 = π‘Ž 4 +4 π‘Ž 3 𝑏+6 π‘Ž 2 𝑏 2 +4π‘Ž 𝑏 3 + 𝑏 4 π‘Ž+𝑏 5 = π‘Ž 5 +5 π‘Ž 4 𝑏+10 π‘Ž 3 𝑏 π‘Ž 2 𝑏 3 +5π‘Ž 𝑏 4 + 𝑏 5 Notice the exponent pattern (π‘Ž goes from 𝑛 to 0, and 𝑏 goes from 0 to 𝑛).

6 How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛
How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛? Let’s try 𝑛=1, 2, 3, 4,… and find patterns. π‘Ž+𝑏 1 = π‘Ž+𝑏 π‘Ž+𝑏 2 = π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 π‘Ž+𝑏 3 = π‘Ž 3 +3 π‘Ž 2 𝑏+3π‘Ž 𝑏 2 + 𝑏 3 π‘Ž+𝑏 4 = π‘Ž 4 +4 π‘Ž 3 𝑏+6 π‘Ž 2 𝑏 2 +4π‘Ž 𝑏 3 + 𝑏 4 π‘Ž+𝑏 5 = π‘Ž 5 +5 π‘Ž 4 𝑏+10 π‘Ž 3 𝑏 π‘Ž 2 𝑏 3 +5π‘Ž 𝑏 4 + 𝑏 5 Notice the exponent pattern (π‘Ž goes from 𝑛 to 0, and 𝑏 goes from 0 to 𝑛).

7 How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛
How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛? Let’s try 𝑛=1, 2, 3, 4,… and find patterns. π‘Ž+𝑏 1 = π‘Ž+𝑏 π‘Ž+𝑏 2 = π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 π‘Ž+𝑏 3 = π‘Ž 3 +3 π‘Ž 2 𝑏+3π‘Ž 𝑏 2 + 𝑏 3 π‘Ž+𝑏 4 = π‘Ž 4 +4 π‘Ž 3 𝑏+6 π‘Ž 2 𝑏 2 +4π‘Ž 𝑏 3 + 𝑏 4 π‘Ž+𝑏 5 = π‘Ž 5 +5 π‘Ž 4 𝑏+10 π‘Ž 3 𝑏 π‘Ž 2 𝑏 3 +5π‘Ž 𝑏 4 + 𝑏 5 Notice the exponent pattern (π‘Ž goes from 𝑛 to 0, and 𝑏 goes from 0 to 𝑛).

8 How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛
How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛? Let’s try 𝑛=1, 2, 3, 4,… and find patterns. π‘Ž+𝑏 1 = π‘Ž+𝑏 π‘Ž+𝑏 2 = π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 π‘Ž+𝑏 3 = π‘Ž 3 +3 π‘Ž 2 𝑏+3π‘Ž 𝑏 2 + 𝑏 3 π‘Ž+𝑏 4 = π‘Ž 4 +4 π‘Ž 3 𝑏+6 π‘Ž 2 𝑏 2 +4π‘Ž 𝑏 3 + 𝑏 4 π‘Ž+𝑏 5 = π‘Ž 5 +5 π‘Ž 4 𝑏+10 π‘Ž 3 𝑏 π‘Ž 2 𝑏 3 +5π‘Ž 𝑏 4 + 𝑏 5 Notice the exponent pattern (π‘Ž goes from 𝑛 to 0, and 𝑏 goes from 0 to 𝑛).

9 How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛
How can we find π‘Ž+𝑏 𝑛 for any natural number 𝑛? Let’s try 𝑛=1, 2, 3, 4,… and find patterns. π‘Ž+𝑏 1 = π‘Ž+𝑏 π‘Ž+𝑏 2 = π‘Ž 2 +2π‘Žπ‘+ 𝑏 2 π‘Ž+𝑏 3 = π‘Ž 3 +3 π‘Ž 2 𝑏+3π‘Ž 𝑏 2 + 𝑏 3 π‘Ž+𝑏 4 = π‘Ž 4 +4 π‘Ž 3 𝑏+6 π‘Ž 2 𝑏 2 +4π‘Ž 𝑏 3 + 𝑏 4 π‘Ž+𝑏 5 = π‘Ž 5 +5 π‘Ž 4 𝑏+10 π‘Ž 3 𝑏 π‘Ž 2 𝑏 3 +5π‘Ž 𝑏 4 + 𝑏 5 Notice the exponent pattern (π‘Ž goes from 𝑛 to 0, and 𝑏 goes from 0 to 𝑛).

10 The coefficient pattern is what’s known as Pascal’s triangle:
1 1 1

11 Ex 1. Find the expansion of π‘Ž+𝑏 7 using Pascal’s triangle. Ex 2
Ex 1. Find the expansion of π‘Ž+𝑏 7 using Pascal’s triangle. Ex 2. Find the expansion of 2βˆ’3π‘₯ 5 using Pascal’s triangle.

12 Ex 1. Find the expansion of π‘Ž+𝑏 7 using Pascal’s triangle. Ex 2
Ex 1. Find the expansion of π‘Ž+𝑏 7 using Pascal’s triangle. Ex 2. Find the expansion of 2βˆ’3π‘₯ 5 using Pascal’s triangle.

13 To figure out the 15th row of Pascal’s triangle, you first need to figure out rows 1 through 14. When expanding π‘Ž+𝑏 𝑛 for a large 𝑛, there’s a more direct way to get the coefficients. But first…

14 To figure out the 15th row of Pascal’s triangle, you first need to figure out rows 1 through 14. When expanding π‘Ž+𝑏 𝑛 for a large 𝑛, there’s a more direct way to get the coefficients. But first…

15 Definitions: 𝑛. =𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛. is read β€œπ’ factorial”) 0
Definitions: 𝑛!=𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛! is read β€œπ’ factorial”) 0!=1 𝑛 π‘Ÿ = 𝑛! π‘Ÿ! π‘›βˆ’π‘Ÿ ! ( 𝑛 π‘Ÿ is called the binomial coefficient) The Binomial Theorem: π‘Ž+𝑏 𝑛 = 𝑛 0 π‘Ž 𝑛 + 𝑛 1 π‘Ž π‘›βˆ’1 𝑏+ 𝑛 2 π‘Ž π‘›βˆ’2 𝑏 2 +…+ 𝑛 π‘›βˆ’1 π‘Ž 𝑏 π‘›βˆ’1 + 𝑛 𝑛 𝑏 𝑛

16 Definitions: 𝑛. =𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛. is read β€œπ’ factorial”) 0
Definitions: 𝑛!=𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛! is read β€œπ’ factorial”) 0!=1 𝑛 π‘Ÿ = 𝑛! π‘Ÿ! π‘›βˆ’π‘Ÿ ! ( 𝑛 π‘Ÿ is called the binomial coefficient) The Binomial Theorem: π‘Ž+𝑏 𝑛 = 𝑛 0 π‘Ž 𝑛 + 𝑛 1 π‘Ž π‘›βˆ’1 𝑏+ 𝑛 2 π‘Ž π‘›βˆ’2 𝑏 2 +…+ 𝑛 π‘›βˆ’1 π‘Ž 𝑏 π‘›βˆ’1 + 𝑛 𝑛 𝑏 𝑛

17 Definitions: 𝑛. =𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛. is read β€œπ’ factorial”) 0
Definitions: 𝑛!=𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛! is read β€œπ’ factorial”) 0!=1 𝑛 π‘Ÿ = 𝑛! π‘Ÿ! π‘›βˆ’π‘Ÿ ! ( 𝑛 π‘Ÿ is called the binomial coefficient) The Binomial Theorem: π‘Ž+𝑏 𝑛 = 𝑛 0 π‘Ž 𝑛 + 𝑛 1 π‘Ž π‘›βˆ’1 𝑏+ 𝑛 2 π‘Ž π‘›βˆ’2 𝑏 2 +…+ 𝑛 π‘›βˆ’1 π‘Ž 𝑏 π‘›βˆ’1 + 𝑛 𝑛 𝑏 𝑛

18 Definitions: 𝑛. =𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛. is read β€œπ’ factorial”) 0
Definitions: 𝑛!=𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛! is read β€œπ’ factorial”) 0!=1 𝑛 π‘Ÿ = 𝑛! π‘Ÿ! π‘›βˆ’π‘Ÿ ! ( 𝑛 π‘Ÿ is called the binomial coefficient) The Binomial Theorem: π‘Ž+𝑏 𝑛 = 𝑛 0 π‘Ž 𝑛 + 𝑛 1 π‘Ž π‘›βˆ’1 𝑏+ 𝑛 2 π‘Ž π‘›βˆ’2 𝑏 2 +…+ 𝑛 π‘›βˆ’1 π‘Ž 𝑏 π‘›βˆ’1 + 𝑛 𝑛 𝑏 𝑛

19 𝑛!=𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹…1 (𝑛! is read β€œπ’ factorial”) 0!=1
Definitions: 𝑛!=𝑛⋅ π‘›βˆ’1 ⋅…⋅3β‹…2β‹… (𝑛! is read β€œπ’ factorial”) 0!=1 𝑛 π‘Ÿ = 𝑛! π‘Ÿ! π‘›βˆ’π‘Ÿ ! ( 𝑛 π‘Ÿ is called the binomial coefficient) The Binomial Theorem: π‘Ž+𝑏 𝑛 = 𝑛 0 π‘Ž 𝑛 + 𝑛 1 π‘Ž π‘›βˆ’1 𝑏+ 𝑛 2 π‘Ž π‘›βˆ’2 𝑏 2 +…+ 𝑛 π‘›βˆ’1 π‘Ž 𝑏 π‘›βˆ’1 + 𝑛 𝑛 𝑏 𝑛 (or, more compactly, π‘Ž+𝑏 𝑛 = π‘˜=0 𝑛 𝑛 π‘˜ π‘Ž π‘›βˆ’π‘˜ 𝑏 π‘˜ )

20 Note: In combinatorics, 𝑛 π‘˜ counts the number of ways to choose π‘˜ objects from 𝑛 objects, where order doesn’t matter. For example, the number of ways to choose a group of 4 students from a class of 36 students is 36 4 = 36! 4! 36βˆ’4 ! =58905 ways.

21 Note: In combinatorics, 𝑛 π‘˜ counts the number of ways to choose π‘˜ objects from 𝑛 objects, where order doesn’t matter. For example, the number of ways to choose a group of 4 students from a class of 36 students is 36 4 = 36! 4! 36βˆ’4 ! =58905 ways.

22 For the Binomial Theorem, when you multiply out π‘Ž+𝑏 𝑛 = π‘Ž+𝑏 π‘Ž+𝑏 π‘Ž+𝑏 β‹…β‹…β‹…(π‘Ž+𝑏), each term is the product of π‘Žβ€™s and 𝑏’s chosen from each factor. For example, if you don’t choose any 𝑏’s, then you’re just multiplying all of the π‘Žβ€™s to get π‘Ž 𝑛 , and there’s only one way to do this. There are 𝑛 1 =𝑛 ways to choose one 𝑏 and π‘›βˆ’1 π‘Žβ€™s (that is, the term π‘Ž π‘›βˆ’1 𝑏). In general, you can think of it like this: there are 𝑛 π‘˜ to choose π‘˜ 𝑏’s from the 𝑛 factors of π‘Ž+𝑏.

23 For the Binomial Theorem, when you multiply out π‘Ž+𝑏 𝑛 = π‘Ž+𝑏 π‘Ž+𝑏 π‘Ž+𝑏 β‹…β‹…β‹…(π‘Ž+𝑏), each term is the product of π‘Žβ€™s and 𝑏’s chosen from each factor. For example, if you don’t choose any 𝑏’s, then you’re just multiplying all of the π‘Žβ€™s to get π‘Ž 𝑛 , and there’s only one way to do this. There are 𝑛 1 =𝑛 ways to choose one 𝑏 and π‘›βˆ’1 π‘Žβ€™s (that is, the term π‘Ž π‘›βˆ’1 𝑏). In general, you can think of it like this: there are 𝑛 π‘˜ to choose π‘˜ 𝑏’s from the 𝑛 factors of π‘Ž+𝑏.

24 For the Binomial Theorem, when you multiply out π‘Ž+𝑏 𝑛 = π‘Ž+𝑏 π‘Ž+𝑏 π‘Ž+𝑏 β‹…β‹…β‹…(π‘Ž+𝑏), each term is the product of π‘Žβ€™s and 𝑏’s chosen from each factor. For example, if you don’t choose any 𝑏’s, then you’re just multiplying all of the π‘Žβ€™s to get π‘Ž 𝑛 , and there’s only one way to do this. There are 𝑛 1 =𝑛 ways to choose one 𝑏 and π‘›βˆ’1 π‘Žβ€™s (that is, the term π‘Ž π‘›βˆ’1 𝑏). In general, you can think of it like this: there are 𝑛 π‘˜ to choose π‘˜ 𝑏’s from the 𝑛 factors of π‘Ž+𝑏.

25 For the Binomial Theorem, when you multiply out π‘Ž+𝑏 𝑛 = π‘Ž+𝑏 π‘Ž+𝑏 π‘Ž+𝑏 β‹…β‹…β‹…(π‘Ž+𝑏), each term is the product of π‘Žβ€™s and 𝑏’s chosen from each factor. For example, if you don’t choose any 𝑏’s, then you’re just multiplying all of the π‘Žβ€™s to get π‘Ž 𝑛 , and there’s only one way to do this. There are 𝑛 1 =𝑛 ways to choose one 𝑏 and π‘›βˆ’1 π‘Žβ€™s (that is, the term π‘Ž π‘›βˆ’1 𝑏). In general, you can think of it like this: there are 𝑛 π‘˜ to choose π‘˜ 𝑏’s from the 𝑛 factors of π‘Ž+𝑏.

26 Ex = = =

27 Note: 𝑛 0 = 𝑛 1 =𝑛 𝑛 𝑛 =1

28 Ex 4. Use the Binomial Theorem to expand 2π‘₯βˆ’1 4 .

29 Note: The term that contains π‘Ž π‘›βˆ’π‘˜ in the expansion of π‘Ž+𝑏 𝑛 is: 𝒏 π’Œ 𝒂 π’βˆ’π’Œ 𝒃 π’Œ Ex 5. Find the term that contains π‘₯ 5 in the expansion of 2π‘₯+𝑦 20 . Ex 6. Find the coefficient of π‘₯ 8 in the expansion of π‘₯ π‘₯ 10 .

30 Note: The term that contains π‘Ž π‘›βˆ’π‘˜ in the expansion of π‘Ž+𝑏 𝑛 is: 𝒏 π’Œ 𝒂 π’βˆ’π’Œ 𝒃 π’Œ Ex 5. Find the coefficient of π‘₯ 4 in the expansion of 3π‘₯βˆ’2 6 . Ex 6. Find the coefficient of π‘₯ 8 in the expansion of π‘₯ π‘₯ 10 .

31 Note: The Binomial Theorem can be proved by induction
Note: The Binomial Theorem can be proved by induction! (see book for details)


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