

Binomial Theorem
Presentation
•
Mathematics
•
11th Grade
•
Hard
Joseph Anderson
FREE Resource
32 Slides • 38 Questions
1
Binomial Theorem

2
Introduction
Bino=2
3
Pascal's Triangle!
The secret to expanding binomials!
4
5
Notes
Index = n
Total no of terms = n+1
From left to right, a↓, b ↑
6
Pascal's Triangle
7
Pascal's Triangle
8
Continuation...
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9
Binomial Theorem:
10
Mastering the Binomial Theorem
To find the 5th coefficient of the expansion form of 2𝑎+3𝑏20, we use the formula 𝑛𝑘−1𝑎𝑛−(𝑘−1)𝑏(𝑘−1). In this case, 𝑛=20 and 𝑘=5. Plugging these values into the formula, we get 20𝑎15𝑏4. Therefore, the 5th coefficient is 20.
11
Multiple Choice
What is the 5th coefficient of the expansion form of 2𝑎+3𝑏20?
20𝑎15𝑏4
20𝑎14𝑏5
20𝑎16𝑏3
20𝑎13𝑏6
12
5th Coefficient:
Trivia: The 5th coefficient of the expansion form of 2𝑎+3𝑏20 is 20𝑎15𝑏4. This means that when the expression is expanded, the term with the 5th coefficient will have 20𝑎 raised to the power of 15 and 3𝑏 raised to the power of 4. It's interesting how the coefficients and exponents combine to form the expanded expression!
13
Mastering the Binomial Theorem
14
Multiple Choice
What is the expansion of 𝑎+4𝑏³ using the binomial theorem?
𝑎⁴+4𝑎³𝑏+6𝑎²𝑏²+4𝑎𝑏³+𝑏⁴
𝑎⁴+4𝑎³𝑏+6𝑎²𝑏²+4𝑎𝑏³
𝑎⁴+4𝑎³𝑏+6𝑎²𝑏²+4𝑎𝑏³+𝑏⁴+4𝑏⁴
𝑎⁴+4𝑎³𝑏+6𝑎²𝑏²+4𝑎𝑏³+𝑏⁴+4𝑏⁴+4𝑏⁴
15
Binomial Theorem:
16
Mastering the Binomial Theorem
17
Multiple Choice
What is the expanded form of the expression 𝑎+4𝑏3 using the binomial theorem?
𝑎3+12𝑎2𝑏+48𝑎𝑏2+64𝑏3𝑎+𝑏3 = 30𝑎3−0𝑏0+31𝑎3−1𝑏1+32𝑎3−2𝑏2+33𝑎
𝑎3+12𝑎2𝑏+48𝑎𝑏2+64𝑏3𝑎+𝑏3 = 30𝑎3−0𝑏0+31𝑎3−1𝑏1+32𝑎3−2𝑏2+33𝑎
𝑎3+12𝑎2𝑏+48𝑎𝑏2+64𝑏3𝑎+𝑏3 = 30𝑎3−0𝑏0+31𝑎3−1𝑏1+32𝑎3−2𝑏2+33𝑎
𝑎3+12𝑎2𝑏+48𝑎𝑏2+64𝑏3𝑎+𝑏3 = 30𝑎3−0𝑏0+31𝑎3−1𝑏1+32𝑎3−2𝑏2+33𝑎
18
Binomial Theorem:
19
Mastering the Binomial Theorem
To find the 5th coefficient of the expansion form of 2𝑎+3𝑏20, we use the formula 𝑛𝑘−1𝑎𝑛−(𝑘−1)𝑏(𝑘−1). In this case, 𝑛=20 and 𝑘=5. Plugging these values into the formula, we get 20𝑎15𝑏4. Therefore, the 5th coefficient is 20.
20
Multiple Choice
What is the 5th coefficient of the expansion form of 2𝑎+3𝑏20?
20𝑎15𝑏4
20𝑎14𝑏5
20𝑎16𝑏3
20𝑎13𝑏6
21
5th Coefficient:
Trivia: The 5th coefficient of the expansion form of 2𝑎+3𝑏20 is 20𝑎15𝑏4. This means that when the expression is expanded, the term with the 5th coefficient will have 20𝑎 raised to the power of 15 and 3𝑏 raised to the power of 4. It's interesting how the coefficients and exponents combine to form the expanded expression!
22
Fill in the Blank
Type answer...
23
Multiple Choice
(2x+5)4
2x4 + 40x3 + 300x2 + 1000x +625
16x4 + 160x3 +600x2 +1000x + 625
16x4 + 1000x3 + 600x2 + 160x +625
24
25
Combination refers to a list of numbers where the order doesn't matter at all.
26
27
28
29
30
31
32
33
34
35
36
37
38
Multiple Choice
Expand the expression using the Binomial Theorem.
(2x+5)4
2x4 + 40x3 + 300x2 + 1000x +625
16x4 + 160x3 +600x2 +1000x + 625
16x4 + 1000x3 + 600x2 + 160x +625
39
Multiple Choice
(2x+5)4
40
Multiple Choice
41
Multiple Choice
Expand: (a+b)2
a2+ab+b2
a2+2ab+b2
a2−ab+b2
a2−2ab+b2
42
Multiple Choice
Choose the right Pascal's triangle
43
Multiple Choice
Expand: (a−b)5
a5+5a4b+10a3b2+10a2b3+5ab4+b5
a5−5a4b+10a3b2−10a2b3+5ab4−b5
a5−5a4b+10a2b3−10a3b2+5ab4−b5
a5−5a4b−10a3b2−10a2b3−5ab4−b5
44
Multiple Choice
Use the binomial theorem to expand (4−2z)3
64−96z+48z2+8z3
64−96z+48z2−8z3
64+96z+48z2+8z3
64−64z+144z2−8z3
45
Multiple Choice
(y - 3)(y + 7)
46
Multiple Choice
Find the product:
(3x2 – 1)(3x2 + 5x+4)
6x2 + 4x + 5
9x4 + 15x3 - 4x
6x2 + 15x3 - 20x2 - 5x + 4
9x4 + 15x3 - 9x2 - 5x - 4
47
Multiple Choice
What would be the 5th term of the expansion of (n+4)4?
257
256
n4
259
48
Multiple Choice
What would be the 2nd term of the expansion of (y+4)4?
253y
16y3
64y
96y2
49
Multiple Choice
What would be the 4th term of the expansion of (a+3)4?
108a
12a3
81
27a
50
Multiple Choice
Number of Combination
1. A person is going to a candy shop where there are 7 types of flavors, if this person is only going to buy 3, define every combination possible.
C = 35
C = 84
51
Multiple Choice
Number of Combination
2. Two girls will go to a party, if between the two, they have 4 pairs of fancy shoes, define the combination of shoes this two girls can wear
C = 6
C = 10
52
Multiple Choice
Number of Combination
3. A man will go on a trip for 3 days, so he will take with him 3 shirts, if he has 7 shirts, how many combination of shirts can he take?
C = 35
C = 84
53
Multiple Choice
Number of Combination
5. A sportsman goes to the store to buy 4 pairs of shoes, if at the store there are a lot of shoes in 7 available colors, how many combination of colors can this man buy.
C = 35
C = 210
54
Fill in the Blank
Type answer...
55
Fill in the Blank
56
Multiple Choice
Students falling in line during the flag ceremony
Combination
Permutation
57
Multiple Choice
Batting order in a baseball game
Permutation
Combination
58
Multiple Choice
Evaluate 10C8
10
45
90
1 814 400
59
60
Multiple Choice
Rhina's mom gave her seven keychains bought from Baguio. If she decided to put five of them in her backpack, in how many ways can she do it?
21
42
1260
2520
61
Multiple Choice
If Marie has 10 pre-loved clothes how many ways can she choose five of them to give to her younger cousin.
25
125
252
30 240
62
Multiple Choice
You need to bring 4 from 8 different shoes in your friends pageant. How many possible combination of shoes can you bring?
16
56
70
1680
63
Multiple Choice
Evaluate: 6C2
15
30
360
720
64
Multiple Choice
Evaluate: 3C1
1
2
3
4
65
Multiple Choice
What is the formula for Combination?
n!
n!/(n-r)!
n!/(n-r)!r!
n!/r!(n-r)!r!
66
Multiple Choice
Which of the following situation involves combination?
Arranging chairs in a circular table
Friends sitting on the front row of movie theater
Choosing three students in class committee
Creating a 4-digit gcash MPIN passcode
67
Multiple Choice
It is the number of ways of selecting from a set when the order
is not important
Combination
Fundamental Counting Principle
Permutation
Probability
68
Multiple Choice
A group of 3 lawn tennis players S, T, U. A team consisting of 2 players is to be formed. In how many ways can we do so?
3
20
5
6
69
Multiple Choice
Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} having 3 elements.
720
120
100
60
70
Multiple Choice
uppose we have a set of 6 letters { A,B,C,D,E,F}. In how many ways can we select a group of 3 letters from this set?
3
6
30
60
Binomial Theorem

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