1. The Binomial
Theorem

2. Pascal’s Triangle and the Binomial Theorem
(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
7.5.2

3. The Binomial Theorem
The Binomial Theorem is a formula used for expanding
powers of binomials.
Each term of the answer is the
product of three first-degree
factors. For each term of the
answer, an a and/or b is taken
from each first-degree factor.
(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 first 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 third term.
(a + b)3
= 3C0 a3 + 3C1 a2b + 3C2 ab2 + 3C3 b3
7.5.3

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
2nd Row
n = 1
(a + b)1 1
1C0
1C1 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
n = 5
(a + b)5 1 5 10 10 5 1
5C0
5C1
5C2
5C3
5C4
5C5
6th Row
7.5.4

5. 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 an - k bk
7.5.5

6. Binomial Expansion - Practice
a = 3x
b = 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)
= 81x x x2 + 96x +16
n = 4
a = 2x
b = -3y
b) (2x - 3y)4
= 4C0(2x)4(-3y)0
+ 4C1(2x)3(-3y)1
+ 4C2(2x)2(-3y)2
+ 4C3(2x)1(-3y)3
+ 4C4(2x)0(-3y)4
= 1(16x4)
+ 4(8x3)(-3y)
+ 6(4x2)(9y2)
+ 4(2x)(-27y3)
+ 81y4
= 16x4 - 96x3y + 216x2y xy3 + 81y4
7.5.6

7. 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
a = 2x
b = -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(-81x-5)
= 32x5
- 240x3
+ 720x1
- 1080x-1
+ 810x-3
- 81x-5
7.5.7

8. Finding a Particular Term in a Binomial Expansion
a) Find the eighth term in the expansion of (3x - 2)11.
tk + 1 = nCk an - k bk
n = 11
a = 3x
b = -2
k = 7
t7 + 1 = 11C7 (3x) (-2)7
t8 = 11C7 (3x)4 (-2)7
= 330(81x4)(-128)
= x4
b) Find the middle term of (a2 - 3b3)8.
n = 8, therefore, there are nine terms. The fifth term is the middle term.
tk + 1 = nCk an - k bk
n = 8
a = a2
b = -3b3
k = 4
t4 + 1 = 8C4 (a2) (-3b3)4
t5 = 8C4 (a2) 4 (-3b3)4
= 70a8(81b12)
= 5670a8b12
7.5.8

9. Finding a Particular Term in a Binomial Expansion
Find the constant term of the expansion of
tk + 1 = nCk an - k bk
x18 - 3k
= x0
n = 18
a = 2x
b = -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 =
7.5.9

10. Finding a Particular Term in a Binomial Expansion
Find the numerical coefficient of
the x11 term of the expansion of
tk + 1 = nCk an - k bk
x20 - 3k
= x11
n = 10
a = x2
b = -x-1
k = ?
tk + 1 = 10Ck (x2)10 - k (-x-1)k
20 - 3k = 11
-3k = -9
k = 3
tk + 1 = 10Ck x20 - 2k (-1)k x-k
tk + 1 = 10Ck(-1)k x20 - 2k x-k
tk + 1 = 10Ck (-1)k x20 - 3k
Substitute k = 3:
t3 + 1 = 10C3 (-1)3 x20 - 3(3)
Therefore, the numerical
coefficient of the x11 term
is -120.
t4 = 10C3 (-1)3x11
t4 = -120 x11
7.5.10

11. Finding a Particular Term of the Binomial
One term in the expansion of (2x - m)7 is x4y3. Find 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
7.5.11

12. THANK
YOU