1. The Binomial
Theorem
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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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