Transcript IB Math SL

PreCalculus
Section 8-5
The Binomial Theorem
Objectives
• Use the Binomial Theorem to calculate
binomial coefficients.
• Use binomial coefficients to write binomial
expansions.
• Use Pascal’s Triangle to calculate binomial
coefficients.
Binomial Coefficients
Binomial Coefficients
• We know that a binomial is a polynomial
that has two terms. In this section, you will
study a formula that provides a quick
method of raising a binomial to a power.
• To begin, look at the expansion of (x + y) n.
The Binomial Expansion
The binomial expansions of (x + y) n
( x  y )0  1
( x  y )1  x  y
( x  y ) 2  x 2  2 xy  y 2
( x  y )3  x 3  3 x 2 y  3 xy 2  y 3
( x  y ) 4  x 4  4 x 3 y  6 x 2 y 2  4 xy 3  y 4
( x  y )5  x 5  5 x 4 y  10 x 3 y 2  10 x 2 y 3  5 xy 4  y 5
reveal several patterns.
Binomial Coefficients
( x  y )0  1
( x  y )1  x  y
1.
The sum of the powers of each term is n. For instance, ( x  y)
in the expansion of (x + y)5 the sum of the powers of (( xx  yy))
each term is 5.
( x  y)
4+1=5
2
 x 2  2 xy  y 2
3
 x 3  3 x 2 y  3 xy 2  y 3
4
 x 4  4 x 3 y  6 x 2 y 2  4 xy 3  y 4
5
 x 5  5 x 4 y  10 x 3 y 2  10 x 2 y 3  5 xy 4  y 5
3+2=5
(x + y)5 = x5 + 5x4y1 + 10x3y2 + 10x2y3 + 5x1y4 + y5
2. The coefficients increase and then decrease in a symmetric pattern. The
coefficients of a binomial expansion are called binomial coefficients.
3. The expansion of (x + y)n begins with x n and ends with y n .
4. The variables in the terms after x n follow the pattern x n-1y , x n-2y2 , x n-3y3 and
so on to y n . With each term the exponent on x decreases by 1 and the
exponent on y increases by 1.
Binomial Coefficients
• To find them, you can use the Binomial
Theorem.
!
The Factorial Symbol
!
0! = 1 1! = 1
n! = n(n-1) · . . . · 3 · 2 · 1
n must be an integer greater than or equal to 2
What this says is if you have a positive integer followed by the factorial symbol
you multiply the integer by each integer less than it until you get down to 1.
6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
!
Your calculator can compute factorials. The ! symbol is under the
“catalog" key.
!
Your Turn:
Evaluate (a) 5! (b) 7!
Check using your calculator.
Solution (a) 5!  5  4  3  2 1  120
(b) 7!  7  6  5  4  3  2 1  5040
Binomial Coefficients
Binomial Coefficient
For nonnegative integers n and r, with r < n,
 n
n!
n Cr    
 r  r !(n  r )!
The symbols n Cr
 n
and   for the binomial
r
coefficients are read “n choose r”
Your calculator can compute these as well. It is also under the “catalog" key and nCr.
Example: Evaluating Binomial Coefficients
 6
Evaluate (a)  
 2
Solution
(a)
(b)
8
(b)  
 0
 n
n!
n Cr    
 r  r !(n  r )!
6!
6!
6 ∙ 5 ∙ 4! 6 ∙ 5 30
6
=
=
=
=
=
= 15
2
2! 6 − 2 ! 2! ∙ 4!
2! ∙ 4!
2!
2
8!
8!
1 1
8
=
=
= = =1
0
0! 8 − 0 ! 0! ∙ 8! 0! 1
Your Turn: Finding Binomial
Coefficients
• Find each binomial coefficient. Check using
your calculator.
 n
n!
n Cr    
 r  r !(n  r )!
BINOMIAL EXPANSION
Expanding Binomials
The Binomial Theorem
 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
 xy  y
1
 2
 3
r
 n  1
The x's start out to the nth power and decrease by 1 in power each
term. The y's start out to the 0 power and increase by 1 in power
each term. The binomial coefficients are found by computing the
combinations. And the sum of the powers for x and y is n.
Example: Expanding Binomials
 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
xy

y

1
 2
 3
r
 n  1
Expand: (x + 2)5
5 4 1 5
x (2) +
1
2
x3(2)2 +
5 2 3 5
x (2) +
x(2)4 + (2)5
3
4
=
x5 +
=
x5 + 5(x4)(2) + 10(x3)(4) + 10(x2)(8) + 5(x)(16) + 32
= x5 + 10x4 + 40x3 + 80x2 + 80x + 32
Your Turn:
 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
 xy  y
1
 2
 3
r
 n  1
Write the expansion of the expression (x + 1)3.
Solution:
Therefore, the expansion is as follows.
 3 2
 3 1 2 3
 x  1  x    x  1    x  1  1
1
 2
 3
 3
The binomial coefficients are    3,    3 .
1
 2
3
3
(x + 1)3 = x3 + (3)x2(1) + (3)x(12) + (13)
= x3 + 3x2 + 3x + 1
Your Turn:
 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
 xy  y
1
 2
 3
r
 n  1
Write the binomial expansion of ( x  y)9 .
Solution:
𝑥+𝑦
𝑥+𝑦
9
9
= 𝑥9 +
9 6 3
9 4 5
9 3 6
9 2 7
9
9 8
9 7 2
9 5 4
𝑥 𝑦+
𝑥 𝑦 +
𝑥 𝑦 +
𝑥 𝑦 +
𝑥 𝑦 +
𝑥 𝑦 +
𝑥 𝑦 +
𝑥𝑦 8 + 𝑦 9
1
2
3
4
5
6
7
8
= 𝑥 9 + 9𝑥 8 𝑦 + 36𝑥 7 𝑦 2 + 84𝑥 6 𝑦 3 + 126𝑥 5 𝑦 4 + 126𝑥 4 𝑦 5 + 84𝑥 3 𝑦 6 + 36𝑥 2 𝑦 7 + 9𝑥𝑦 8 + 𝑦 9
Example:
Expand: (3x – 2y)4
 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
 xy  y
1
 2
 3
r
 n  1
=
(3x)4 +
4
4
4
(3x)3(-2y)1 +
(3x)2(-2y)2 +
(3x)1(-2y)3+ (-2y)4
1
2
3
= (81x4) + 4(27x3)(-2y) + 6(9x2)(4y2) + 4(3x)(-8y3) + (16y4)
=
81x4 – 216x3y + 216x2y2 – 96xy3 + 16y4
Your Turn:
5
b

Expand  a   .
2

 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
 xy  y
1
 2
 3
r
 n  1
Solution
b
xa , y
2
𝑏
𝑎−
2
5
𝑏
𝑎−
2
5
𝑏
𝑏
5 4
5 3
= 𝑎5 +
𝑎 −
+
𝑎 −
1
2
2
2
= 𝑎5 + 5 𝑎4
𝑏
𝑎−
2
5
2
𝑏
5 2
+
𝑎 −
3
2
3
𝑏
5
+
𝑎 −
4
2
4
𝑏
+ −
2
2
3
4
5
𝑏
𝑏
𝑏
𝑏
𝑏
−
+ 10 𝑎3
+ 10 𝑎2 −
+ 5 𝑎
+ −
2
4
8
16
32
5 4
5 3 2 5 2 3
5
1 5
4
= 𝑎 − 𝑎 𝑏 + 𝑎 𝑏 − 𝑎 𝑏 + 𝑎𝑏 − 𝑏
2
2
4
16
32
5
5
Your Turn:
Find the first three terms of the following expansion (x2 – 1)20 .
 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
 xy  y
1
 2
 3
r
 n  1
𝑥2 − 1
20
= 𝑥2
20
+
20
1
𝑥2
19
−1
1
+
20
2
= x40 + 20(x38)(-1) + 190(x36)(1) + . . .
= x40 – 20x38 + 190x36 – . . .
𝑥2
18
−1 2 +. . .
Binomial Expansions
• Sometimes you will need to find a specific
term in a binomial expansion.
• Instead of writing out the entire expansion,
you can use the fact that, from the Binomial
Theorem,
𝑥+𝑦
𝑛
= 𝑥𝑛 +
𝑛 𝑛−1
𝑛 𝑛−2 2
𝑛 𝑛−3 3
𝑛 𝑛−𝑟 𝑟
𝑛
𝑥
𝑦+
𝑥
𝑦 +
𝑥
𝑦 +. . . +
𝑥
𝑦 +. . . +
𝑥𝑦 𝑛−1 + 𝑦 𝑛
3
𝑛−1
1
2
𝑟
the (r +
1)th
𝑛 𝑛−𝑟 𝑟
term is
𝑥
𝑦 or nCr x n–ry r .
𝑟
The Binomial Theorem –General Term
General term in the expansion is;
C
n r
n-r
r
x y
Or
𝒏 𝒏−𝒓 𝒓
𝒙 𝒚
𝒓
Which is the (r + 1)th term
Example:
Find the sixth term of (a + 2b)8.
• Solution:
Because the formula is for the (r + 1)th term, r is one
less than the number of the term you need. So, to find
the sixth term in this binomial expansion, use r = 5,
n = 8, x = a and y = 2b.
nCr
xn – ryr = 8C5a8 – 5(2b)5
= 56  a3  (2b)5
= 56(25)a3b5
= 1792a3b5
Your Turn:
10
(
a

2
b
)
Find the fourth term of
.
Remember the
General term
𝒏 𝒏−𝒓 𝒓
𝒙 𝒚
𝒓
Solution Using n = 10, r = 3, x = a, y = 2b in the
formula, we find the fourth term is
10  7
3
7
3
7 3
a
(2
b
)

120
a
8
b

960
a
b .
 
3
Example:
Find the 8th term of the expansion of (2x – 3)12
Remember the
General term
𝒏 𝒏−𝒓 𝒓
𝒙 𝒚
𝒓
8th term = 12 (2x)5(-3)7
7
= 792(32x5)(-2187)
= 1,647,360 x7y8
Your Turn:
In the expansion of (x + 2y)15 , find the term containing y8.
Remember the
General term
𝒏 𝒏−𝒓 𝒓
𝒙 𝒚
𝒓
the term containing y8 =
15
8
𝑥
7
2𝑦
8
= 6435(x7)(256y8)
= 1,647,360 x7y8
Pascal’s Triangle
Pascal
June 19, 1623-August 19,1662
Born in Clermont-Ferrand,
France
Mathematician, physicist,
religious philosopher
Instrumental in development
of economics and social
science
“Contradiction is not a sign of falsity, nor the lack of
contradiction a sign of truth.”
Quoted in W H Auden and L Kronenberger, The Viking Book of Aphorisms (New York 1966).
History
Pascal’s Triangle has a long line of history behind it. It might be
understood that because it’s named after the mathematician Blaise Pascal,
that he was the one that invented it. This however is not entirely true.
There were many people that had worked with and even created a
Pascal’s Triangle long before Pascal was even thought of. Don’t believe
me? Click on the picture and learn more about the history of Pascal’s
Triangle…….
Or click here to
see some more
history…..
Blaise Pascal
Pascal’s Triangle
There is a convenient way to remember the pattern for
binomial coefficients. By arranging the coefficients in a
triangular pattern, you obtain the following array, which is
called Pascal’s Triangle. This triangle is named after the
famous French mathematician Blaise Pascal (1623–1662).
4 + 6 = 10
15 + 6 = 21
Pascal’s Triangle
The first and last number in each row of Pascal’s Triangle is
1. Every other number in each row is formed by adding the
two numbers immediately above the number. Pascal noticed
that the numbers in this triangle are precisely the same
numbers as the coefficients of binomial expansions, as
follows.
Pascal’s Triangle
• The top row of Pascal’s Triangle is called the
zeroth row because it corresponds to the binomial
expansion (x + y)0 = 1.
• Similarly, the next row is called the first row
because it corresponds to the binomial expansion
(x + y)1 = 1(x) + 1(y).
• In general, the nth row of Pascal’s Triangle gives
the coefficients of (x + y)n.
Example:
Using Pascal’s Triangle
Use the seventh row of Pascal’s Triangle to
find the binomial coefficients.
8C0 8C1 8C2 8C3 8C4 8C5 8C6 8C7 8C8
Solution:
Binomial Expansion using
Pascal’s Triangle
• Expand (x+y)n
• There are three parts to consider
(a) the coefficients from Pascal’s Triangle.
(b) the powers of x (decreasing from n in above
expansion to 0).
(c) the powers of y (increasing from 0 in the
above expansion).
• All elements in the expansion are in the form xn–ryr
i.e. powers always add up to n.
Example:
Give the expansion of (x + 1)6
(a) Pascal’s Triangle
We want the row starting 1 6 …….
1
1
5
6
10
15
10
20
5
15
1
6
1
Continued:
Give the expansion of (x + 1)6
(b) Powers of x x6, x5, x4, …….
(c) Powers of y 10, 11, 12, … (ALL 1)
So the full expansion is:
(x + 1)6 =
1x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1
Your Turn:
Give the expansion of (1 - 3x)5
(a) Pascal’s Triangle 1 5 10 10 5 1
(b) Powers of “x”
All 1
1, (-3x), (-3x)2, (-3x)3, (-3x)4, (-3x)5
(c) Powers of “y”
(Increasing)
So the full expansion of (1 - 3x)5
= 1+ 5∙(-3x) + 10∙(-3x)2 + 10∙(-3x)3 + 5∙(-3x)4 + (-3x)5
= 1- 15x + 90x2 - 270x3 + 405x4 - 243x5
Binomial Expansion with
Pascal’s Triangle
1
1
This is good for lower
powers but could get
very large. For larger
powers it is easier to
use combinations to
calculate the binomial
coefficients.
1
1
1
1
2
3
4
5
1
1
3
6
1
4
10 10
1
5
1
Pascal's Triangle will give us the coefficients for a binomial expansion
of any power if we extended it far enough.
Binomial Theorem
Summary
The Binomial Theorem
Points to be remmebered :
• Coefficients are arranged in a Pascal triangle.
• Summation of the indices of each term is equal to
the power (order) of the expansion.
• The first term of the expansion is arranged in
descending order after the expansion.
• The second term of the expansion is arranged in
ascending order order after the expansion.
• Number of terms in the expansion is equal to the
power of the expansion plus one.
The Binomial Theorem
 n
 n
 n
 n
 n  n1
n
( x  y)n  xn    xn1 y    x n2 y 2    x n3 y3  ...    x nr y r  ...  
 xy  y
1
 2
 3
r
 n  1
The General Term
𝑛 𝑛−𝑟 𝑟
𝑥
𝑦
𝑟
Homework
• Section 8.5 pg 624 – 626;
– Vocab. Check: #1 – 4 all
– Exercises: #1 – 43 odd, 49 – 71 odd, 77