2. Binomials
An expression in the form a + b is called a binomial,
because it is made of of two unlike terms.
We could use the FOIL method repeatedly to evaluate
expressions like (a + b)2, (a + b)3, or (a + b)4.
– (a + b)2 = a2 + 2ab + b2
– (a + b)3 = a3 + 3a2b + 3ab2 + b3
– (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
But to evaluate to higher powers of (a + b)n would be a
difficult and tedious process.
For a binomial expansion of (a + b)n, look at the
expansions below:
– (a + b)2 = a2 + 2ab + b2
– (a + b)3 = a3 + 3a2b + 3ab2 + b3
– (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
• Some simple patterns emerge by looking at these
examples:
– There are n + 1 terms, the first one is an and the last is bn.
– The exponent of a decreases by 1 for each term and the
exponents of b increase by 1.
– The sum of the exponents in each term is n.
3. For bigger exponents
To evaluate (a + b)8, we will find a way to calculate the
value of each coefficient.
(a + b)8= a8 + __a7b + __a6b2 + __a5b3 + __a4b4 + __a3b5 + __a2b6 + __ab7 + b8
– Pascal’s Triangle will allow us to figure out what the coefficients
of each term will be.
– The basic premise of Pascal’s Triangle is that every entry (other
than a 1) is the sum of the two entries diagonally above it.
The Factorial
In any of the examples we had done already, notice that
the coefficient of an and bn were each 1.
– Also, notice that the coefficient of an-1 and a were each n.
These values can be calculated by using factorials.
– n factorial is written as n! and calculated by multiplying the
positive whole numbers less than or equal to n.
Formula: For n≥1, n! = n • (n-1) • (n-2)• . . . • 3 • 2 • 1.
Example: 4! = 4 3 2 1 = 24
– Special cases: 0! = 1 and 1! = 1, to avoid division by zero in the
next formula.
4. The Binomial Coefficient
To find the coefficient of any term of (a +
b)n, we can apply factorials, using the
formula:
n
n!
n Cr
r
r! n r !
Blaise Pascal
(1623-1662)
– where n is the power of the binomial
expansion, (a + b)n, and
– r is the exponent of b for the specific term we are
calculating.
So, for the second term of (a + b)8, we would have n = 8
and r = 1 (because the second term is ___a7b).
– This procedure could be repeated for any term we choose, or all of
the terms, one after another.
– However, there is an easier way to calculate these coefficients.
Example :
7 C3
7!
7!
7
(7 3)! • 3! 4! • 3! 4! • 3!
(7 • 6 • 5 • 4) • (3 • 2 • 1)
(4 • 3 • 2 • 1) • (3 • 2 • 1)
7•6•5• 4
4 • 3 • 2 •1
35
5. Recall that a binomial has two terms...
(x + y)
The Binomial Theorem gives us a quick method to expand
binomials raised to powers such as…
(x + y)0
(x + y)1
(x + y)2
(x + y)3
Study the following…
Row
Row
Row
Row
Row
Row
Row
0
1
This triangle is called Pascal’s
1
Triangle (named after mathematician
1 1
Blaise Pascal).
2
1 2 1
3
1 3 3 1
Notice that row 5 comes from adding up
4
1 4 6 4 1 row 4’s adjacent numbers.
(The first row is named row 0).
5
1 5 10 10 5 1
6 1 6 15 20 15 6 1
This pattern will help us find the coefficients when we expand binomials...
6. Finding coefficient
What we will notice is that when r=0 and when r=n, then
nCr=1, no matter how big n becomes. This is because:
n C0
n!
n 0 ! 0!
n!
1
n! 0!
n Cn
n!
n n ! n!
n!
1
0! n!
Note also that when r = 1 and r = (n-1):
n
C1
n!
n 1 ! 1!
n n 1!
n 1 ! 1!
n
n Cn
1
n
n!
n 1 ! n 1!
n n 1!
1! n 1 !
So, the coefficients of the first and last terms will always be
one.
– The second coefficient and next-to-last coefficient will be n.
(because the denominators of their formulas are equal)
n
7. Constructing Pascal’s Triangle
Continue evaluating nCr for n=2 and n=3.
When we include all the possible values of r such that
0≤r≤n, we get the figure below:
n=0
0C0
n=1
1C0 1C1
n=2
2C0
n=3
n=4
3C0
4C0
n=5
5C0
n=6
6C0 6C1
3C1
4C1
5C1
2C1
6C2
3C2
4C2
5C2
2C2
4C3
5C3
6C3
3C3
4C4
5C4
6C4
5C5
6C5
6C6
8. Knowing what we know about nCr and its values when
r=0, 1, (n-1), and n, we can fill out the outside values
of the Triangle:
r=n, nCr=1
r=1, nCr=n
r=(n-1), nCr=n
n=0
1
0C0
n=1
r=0, nCr=1
1 0 1C
1C1 1C1 1
1
n=2
1 1 2 2C 11 C
1
2C01 C2 1 2C2 2
2
n=3
1 0 3C33 33C 111C
C2 3
3
31
3C111 C1 3C2 2 3C3 3
n=4
1 0 4CC 44C 44C 111C
C3 4
4
4 14 C2
4C111 4 1 4C2 2 4C3 3 4C4 4
n=5
1 0 5C55 55C 55C 55C 111C
54 5
51
2
3
5C111 C1 5C2 2 5C3 3 5C4 4 5C5 5
n=6
1 0 6CC 66C 66C 66C 66C 111C
C3 C4 C5 6
6
6 16 C2
6C111 6 1 6C2 2 6C3 3 6C4 4 6C5 5 6C6 6
9. Using Pascal’s Triangle
We can also use Pascal’s Triangle to expand
binomials, such as (x - 3)4.
The numbers in Pascal’s Triangle can be used to find
the coefficients in a binomial expansion.
For example, the coefficients in (x - 3)4 are represented
by the row of Pascal’s Triangle for n = 4.
x
3
4
4 C0 x
1x
4
4
1
3
0
4 x
4 C1 x
3
3
3
4
6
4
1
3
6 x
1
2
4 C2 x
9
2
4 x
3
1
2
1
4 C3 x
27
1x 4 12x 3 54x 2 108x 81
1x
1
0
3
81
3
4 C4 x
0
3
4
10. The Binomial Theorem
( x y)n
with nCr
x n nx n 1 y nCr x n r y r nxy n 1 y n
n!
(n r )!r !
The general idea of the Binomial Theorem is that:
– The term that contains ar in the expansion (a + b)n is
n
n
r n r
r
ab
or
n!
arbn
n r ! r!
r
– It helps to remember that the sum of the exponents of each term
of the expansion is n. (In our formula, note that r + (n - r) = n.)
Example: Use the Binomial Theorem to expand (x4 + 2)3.
(x 4
2)3
4 3
C0(x )
3
4 3
1 (x )
4 2
C1( x ) (2)
3
4
2
C2(x )( 2)
3
3 ( x 4 ) 2 (2) 3 (x 4 )( 2) 2
x12 6 x8 12 x 4 8
1 (2)
(2)
3 C3
3
3
11. Example:
Find the eighth term in the expansion of (x + y)13 .
Think of the first term of the expansion as x13y 0 .
The power of y is 1 less than the number of the term in
the expansion.
The eighth term is 13C7 x 6 y7.
13
C7
13!
6! • 7!
(13 • 12 • 11 • 10 • 9 • 8) • 7!
6! • 7!
13 • 12 • 11 • 10 • 9 • 8
1716
6 • 5 • 4 • 3 • 2 •1
Therefore,
the eighth term of (x + y)13 is 1716 x 6 y7.
12. Proof of Binomial Theorem
Binomial theorem for any positive integer n,
a b
n
n
c0an
n
c1a n 1b nc2an 2b2 ........ ncnbn
Proof
The proof is obtained by applying principle of mathematical
induction.
Step: 1
Let the given statement be
f (n) : a b
n
n
c0an
n
c1an 1b nc2an 2b2 ........ ncnbn
Check the result for n = 1 we have
f (1) : a b
1
1
c0a1 1c1a1 1b1 a b
Thus Result is true for n =1
Step: 2
Let us assume that result is true for n = k
f (k ) : a b
k
k
c0ak
k
c1ak 1b k c2ak 2b2 ........ k ck bk
13. Step: 3
We shall prove that f (k + 1) is also true,
k 1
f (k 1) : a b
k 1
c0ak
1
k 1
c1ak b
k 1
c2ak 1b2 ........ k 1ck 1bk
Now,
a b
k 1
(a b)( a b) k
k
a b
c0 a k
k
c1a k 1b k c2 a k 2b 2 ........
k
ck b k
From Step 2
k
c0 a k
1
1
k
c1a k b k c2 a k 1b 2 ........ k ck ab k
k
k
c0 a k
c0 a k b k c1a k 1b 2 ........ k ck 1ab k
k
c1
k
c0 a k b
k
c2
...
by using
k 1
c0
1, k cr
k
cr
k
1
k
k
ck b k
1
c1 a k 1b 2 .....
k
ck
k
ck 1 ab k
cr , and k ck
1
k
ck b k
1
k 1
ck
1
1
14. k 1
c0 a k
1
k 1
c1a k b
k 1
c2 a k 1b 2 ........
k 1
ck ab k
k 1
ck 1b k
Thus it has been proved that f(k+1) is true when ever
f(k) is true,
Therefore, by Principle of mathematical induction f(n) is
true for every Positive integer n.
1
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