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9.5
The Binomial Theorem
Let’s look at the expansion of (x + y)n
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 +2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
Expanding a binomial using Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Pascal’s Triangle
Write the
next row.
1 6 15 20 15 6 1
Expand (x + 3)4 From Pascal’s triangle write down
the 4th row.
1 4 6 4 1
These numbers are the same numbers that are the
coefficients of the binomial expansion.
The expansion of (a + b)4 is:
1a4b0 + 4a3b1 + 6a2b2 + 4a1b3 + 1a0b4
Notice that the exponents always add up to 4 with
the a’s going in descending order and the b’s in
ascending order.
Now substitute x in for a and 3 in for b.
1a4b0 + 4a3b1 + 6a2b2 + 4a1b3 + 1a0b4
x4 + 4x3(3)1 + 6x2(3)2 + 4x(3)3 + 34
This simplifies to x4 + 12x3 + 54x2 + 108x + 81
Expand (x – 2y)4 This time substitute x in for a
and -2y for b. Use ( ).
x4 + 4x3(-2y)1 + 6x2(-2y)2 + 4x(-2y)3 + (-2y)4
The final answer is:
x4 – 8x3y + 24x2y2 – 32xy3 + 16y4
The Binomial Theorem
In the expansion of (x + y)n
(x + y)n = xn + nxn-1y + … + nCmxn-mym + … +nxyn-1 + yn
the coefficient of xn-mym is given by
  !
!
!
m
m
n
n
Cm
n


Find the following binomial coefficients.
8C2 =
10C3 =
7C3 =
7C4 =

!
2
!
6
!
8



!
2
!
6
!
6
7
8
28

!
3
!
7
!
10





!
3
!
7
!
7
8
9
10
120





1
2
3
5
6
7
35







1
2
3
4
4
5
6
7
35
Find the 6th term in the expansion of (3a + 2b)12
Using the Binomial Theorem, let x = 3a and y = 2b
and note that in the 6th term, the exponent of y is
m = 5 and the exponent of x is n – m = 12 – 5 = 7.
Consequently, the 6th term of the expansion is:

5
7
5
12 y
x
C    5
7
2
3
!
5
!
7
!
7
8
9
10
11
12
b
a






= 55,427,328 a7b5

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Pc 9-5.ppt

  • 1. 9.5 The Binomial Theorem Let’s look at the expansion of (x + y)n (x + y)0 = 1 (x + y)1 = x + y (x + y)2 = x2 +2xy + y2 (x + y)3 = x3 + 3x2y + 3xy2 + y3 (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
  • 2. Expanding a binomial using Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 Pascal’s Triangle Write the next row. 1 6 15 20 15 6 1
  • 3. Expand (x + 3)4 From Pascal’s triangle write down the 4th row. 1 4 6 4 1 These numbers are the same numbers that are the coefficients of the binomial expansion. The expansion of (a + b)4 is: 1a4b0 + 4a3b1 + 6a2b2 + 4a1b3 + 1a0b4 Notice that the exponents always add up to 4 with the a’s going in descending order and the b’s in ascending order. Now substitute x in for a and 3 in for b.
  • 4. 1a4b0 + 4a3b1 + 6a2b2 + 4a1b3 + 1a0b4 x4 + 4x3(3)1 + 6x2(3)2 + 4x(3)3 + 34 This simplifies to x4 + 12x3 + 54x2 + 108x + 81 Expand (x – 2y)4 This time substitute x in for a and -2y for b. Use ( ). x4 + 4x3(-2y)1 + 6x2(-2y)2 + 4x(-2y)3 + (-2y)4 The final answer is: x4 – 8x3y + 24x2y2 – 32xy3 + 16y4
  • 5. The Binomial Theorem In the expansion of (x + y)n (x + y)n = xn + nxn-1y + … + nCmxn-mym + … +nxyn-1 + yn the coefficient of xn-mym is given by   ! ! ! m m n n Cm n  
  • 6. Find the following binomial coefficients. 8C2 = 10C3 = 7C3 = 7C4 =  ! 2 ! 6 ! 8    ! 2 ! 6 ! 6 7 8 28  ! 3 ! 7 ! 10      ! 3 ! 7 ! 7 8 9 10 120      1 2 3 5 6 7 35        1 2 3 4 4 5 6 7 35
  • 7. Find the 6th term in the expansion of (3a + 2b)12 Using the Binomial Theorem, let x = 3a and y = 2b and note that in the 6th term, the exponent of y is m = 5 and the exponent of x is n – m = 12 – 5 = 7. Consequently, the 6th term of the expansion is:  5 7 5 12 y x C    5 7 2 3 ! 5 ! 7 ! 7 8 9 10 11 12 b a       = 55,427,328 a7b5