1. How many different landscapes could be created?
Myriorama cards were invented in France around 1823 by Jean-Pierre Brès and further developed in England by John Clark. Early myrioramas were decorated with people, buildings, and scenery that could be laid out in any order to create a variety of landscapes. One 24-card set is sold as “The Endless Landscape.”
How many different landscapes could be created?
Math 30-1

2. 11.3 The Binomial Theorem Student Activity Race Expand (x + y)0
Math 30-1

3. Pascal’s Triangle and the Binomial Theorem
(x + y)n, n ∈ N
(x + y)0
= 1
(x + y)1
= 1x + 1y
(x + y)2
= 1x2 + 2xy + 1y2
(x + y)3
= 1x3 + 3x2y + 3xy2 +1 y3
(x + y)4
= 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x + y)5
= 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
(x + y)6
= 1x6 + 6x5y1 + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6
What is the relationship between the value of n and the number of terms in the expansion?
Math 30-1

4. Pascal’s Triangle and the Binomial Theorem
The numerical coefficients in a binomial expansion
can be found in Pascal’s triangle.
Pascal’s
Triangle
Pascal’s Triangle
using Combinatorics
n = 0
(a + b)0
1st Row
0C0 1
n = 1
(a + b)1
2nd Row
1C0
1C1 1 1
3rd Row
n = 2
(a + b)2 1 2 1
2C0
2C1
2C2
n = 3
(a + b)3
4th Row 1 3 3 1
3C0
3C1
3C2
3C3
n = 4
(a + b)4
5th Row 1 4 6 4 1
4C0
4C1
4C2
4C3
4C4 1 5 1
n = 5
(a + b)5 5 10 10
5C0
5C1
5C2
5C3
5C4
5C5
6th Row
What patterns do you notice in Pascal’s Triangle?
Math 30-1

5. The Binomial Theorem
The Binomial Theorem is a formula used for expanding
powers of binomials.
(a + b)3 = (a + b)(a + b)(a + b)
= a3 + 3a2b + 3ab2 + b3
The first term has no b. It is like choosing no b from three b’s.
The combination 3C0 is the coefficient of the first term.
The second term has one b. It is like choosing one b from three b’s.
The combination 3C1 is the coefficient of the second term.
The third term has two b’s. It is like choosing two b’s from three
b’s. The combination 3C2 is the coefficient of the third term.
The fourth term has three b’s. It is like choosing three b’s from
three b’s. The combination 3C3 is the coefficient of the fourth term.
(a + b)3
= 3C0 a3b0 + 3C1 a2b1 + 3C2 ab2 + 3C3 b3
Math 30-1

6. Binomial Expansion - the General Term
(a + b)3 = a3 + 3a2b + 3ab2 + b3
The degree of each term is 3.
For the variable a, the degree descends from 3 to 0.
For the variable b, the degree ascends from 0 to 3.
(a + b)3 = 3C0 a3 - 0b0 + 3C1 a3 - 1b1 + 3C2 a3 - 2b2 + 3C3 a3 - 3b3
(a + b)n = nC0 an - 0b0 + nC1 an - 1b1 + nC2 an - 2b2 + …..+ nCk an - kbk
The general term is the (k + 1)th term:
tk + 1 = nCk xn - k yk
The number of terms in the expansion of (a + b)n is
n + 1
Math 30-1

7. How many terms are in the expansion of each binomial?
If there are 17 terms in the expansion of the given binomial, what is the value of k?
Math 30-1

8. Binomial Expansion - Practice
x = 3x
y = 2
Expand the following.
a) (3x + 2)4
= 4C0
(3x)4
(2)0
+ 4C1
(3x)3
(2)1
+ 4C2
(3x)2
(2)2
+ 4C3
(3x)1
(2)3
+ 4C4
(3x)0
(2)4
= 1
(81x4)
+ 4
(27x3)
(2)
+ 6
(9x2)
(4)
+ 4
(3x)
(8)
+ 1
(16)
= 81x4
+ 216x3
+ 216x2
+ 96x
+ 16
n = 4
x = 2x
y = -3y
b) (2x - 3y)3
= 3C0(2x)3(-3y)0
+ 3C1(2x)2(-3y)1
+ 3C2(2x)1(-3y)2
+ 3C3(2x)0(-3y)3
= 1(8x3)(1)
+ 3(4x2)(-3y)
+ 3(2x)(9y2)
+ 1(1)(-27y3)
= 8x3
- 36x2y
+ 24x1y2
- 2y3
Math 30-1

9. Binomial Expansion - Practice
(2x - 3x -1)5
= 5C0(2x)5(-3x -1)0
+ 5C1(2x)4(-3x -1)1
+ 5C2(2x)3(-3x -1)2
n = 5
x = 2x
y = -3x -1
+ 5C3(2x)2(-3x -1)3
+ 5C4(2x)1(-3x -1)4
+ 5C5(2x)0(-3x -1)5
= 1(32x5)
+ 5(16x4)(-3x -1)
+ 10(8x3)(9x -2)
+ 10(4x2)(-27x -3)
+ 5(2x)(81x -4)
+ 1(-243x -5)
= 32x5
- 240x3
+ 720x1
- 1080x -1
+ 810x -3
- 243x -5
Math 30-1

10. Finding a Particular Term in a Binomial Expansion
a) Determine an expression for the eighth term in the expansion of (3x - 2)11.
tk + 1 = nCk xn - k yk
t8 = t7 + 1 so k = 7
n = 11
x = 3x
y = -2
k = 7
t7 + 1 = 11C7 (3x) (-2)7
t8 = 11C7 (3x)4 (-2)7
= 330(81x4)(-128)
= x4
Determine the coefficient of the term containing Ax2y5 in the
expansion of(x - 3y)7.
tk + 1 = nCk xn - k yk
n = 7
k = 5
Ax2y5 = 7C5 (x)2(-3y)5
= 21 (x2)(-243y5)
= -5103x2y5
Math 30-1

11. Determine an expression for the term containing x 3 in the expansion of (2x - 3)6.
tk + 1 = nCk xn - k yk
When the exponent on x is 3, and the value of n is 6, the value of k must be 3.
t3 + 1 = 6C3 (2x)6 - 3 (-3)3
Math 30-1

12. Determine the value of the constant term of the expansion of
tk + 1 = nCk xn - k yk
x18 - 3k
= x0
n = 18
x = 2x
y = -x -2
k = ?
tk + 1 = 18Ck (2x)18 - k (-x -2)k
18 - 3k = 0
-3k = -18
k = 6
tk + 1 = 18Ck k x18 - k (-1)k x -2k
tk + 1 = 18Ck k (-1)k x18 - k x -2k
tk + 1 = 18Ck k (-1)k x18 - 3k
Substitute k = 6:
t6 + 1 = 18C (-1)6 x18 - 3(6)
Therefore, the constant
term is
t7 = 18C6 212 (-1)6
t7 =
Math 30-1

13. Determine a Particular Term in the Binomial
One term in the expansion of (2x - m)7 is x4y3.
Determine an expression to represent m.
tk + 1 = nCk an - k bk
n = 7
a = 2x
b = -m
k = 3
tk + 1 = 7C3 (2x)4 (-m)3
tk + 1 = (35) (16x4) (-m)3
x4y3 = (560x4) (-m)3
Therefore, m is 3y.
3y = m
Math 30-1

14. Assignment
Worksheet
Math 30-1

15. Pascal’s triangle is an array of natural numbers. The sum of any two
adjacent numbers is equal to the number directly below them.
Sum of
each row
0 Row 1 1 20
1st Row 1 1 2 21
2nd Row 1 2 1 4 22
3rd Row 1 3 3 1 8 23
4th Row 1 4 6 4 1 16 24 1 32 25
5th Row 1 5 10 10 5 26 1 6 15 20 15 6 1 64
6th Row 1 7 21 35 35 21 7 1
128 27
7th Row
nth Row 2n
Math 30-1