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1. Binom
ial
The
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
!

  



 
Cn r  
!  !
r n r
n
r

– 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
Blaise Pascal calculating.
(1623-1662)
 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
4! 3!
7!
7 3  
4! 3!
7!
(7 3)! 3!
• • •

C 
(7 • 6 • 5 • 4) • (3 • 2 •
1)
7 • 6 • 5 •
4
 35
(4 • 3 • 2 • 1) • (3 • 2 •
1)
 
4 • 3 • 2 •
1

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…
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
This triangle is called Pascal’s
Triangle (named after mathematician
Blaise Pascal).
Notice that row 5 comes from adding up row
4’s adjacent numbers.
(The first row is named row 0).
Row 0
Row 1
Row 2
Row 3
Row 4
Row 5
Row 6 1 6 15 20 15 6 1
This pattern will help us find the coefficients when we expand binomials...

6.  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:
 Note also that when r = 1 and r = (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)

nC0 
n!
n  0!0!

n!
n!0!
 1

nCn 
n!
n  n!n!

n!
0!n!
 1
Finding coefficient
nC1 
n!
n 1!1!

nn 1!
n 1!1!
 n

nCn 1 
n!
n  n 1!n 1!

nn 1!
1!n 1!
 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 2C1 2C2
n=3 3C0 3C1 3C2 3C3
n=4 4C0 4C1 4C2 4C3 4C4
n=5 5C0 5C1 5C2 5C3 5C4 5C5
n=6 6C0 6C1 6C2 6C3 6C4 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=0, nCr=1 0C0
1C0 1C1
2C0 2C2
3C0 3C3
4C0 4C4
5C0 5C5
6C0 6C6
r=n, nCr=1
n=0 1
n=1 1 1 C1
11
n=2 1 1 CCC1
221 1 22
n=3 1 1 CCCCC1
331 1 332 2 33
n=4 1 1 CCCCCCC1
441 1 442 2 443 3 44
n=5 1 1 CCCCCCCCC1
551 1 552 2 553 3 554 4 55
n=6 1 CCCCCC61 62 63 64 65 66
1 6C1 6C2 6C3 6C4 6C5 1
r=1, nCr=n
1 2 1
1 3 32 1
1 4 4C2 4C3 1
1 5 52 53 54 1
1 6 6C2 6C3 6C4 6C5 1
r=(n-1), nCr=n
1 3 3 1
1 4 4 1
1 5 5 1
1 6 6 1

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
4C0 x4
30
4C1 x3
31
4C2 x2
32
4C3 x1
33
4C4 x0
34
1 4 6 4 1
1x4 12x3  54x2 108x  81

 1x4
1 4x3
3 6x2
9 4x1
271x0
81

10. The Binomial Theorem
1 1 ( )n n n n r r n n
n r x y x nx y C x y nxy y       L  L  
n
C
 The general idea of the Binomial Theorem is that:
– The term that contains ar in the expansion (a + b)n is
or
n !

 ! !
– 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.)

n
n  r





arbn r
 
r n r a b
n r r
!
with
n r
( n r )! r
! 

Example: Use the Binomial Theorem to expand (x4 + 2)3.
3 0 C 3 1 C 3 2 C 3 3   C 4 3 (x 2)  4 3 (x ) ( ) (2)  4 2 x  4 2 (x )(2) 3 (2)
1  4 3 ) (x  3 ) 2 ( ) ( 4 2 x 3  4 2 ) 2 )( (x 1 3 (2)
6 12 8 12 8 4  x  x  x 

11. Find the eighth term in the expansion of (x + y)13 .
The eighth term is 13C7 x6y7.
13 7 C  
Therefore,
(13 • 12 • 11 • 10 • 9 • 8) •
7!
6! 7!
13!
6! 7!
•
•
1716
13 • 12 • 11 • 10 • 9 •
8
 
6 • 5 • 4 • 3 • 2 •
1
the eighth term of (x + y)13 is 1716 x6y7.
Example:
 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.

12. Proof of Binomial Theorem
 Binomial theorem for any positive integer n,
n n n n n n n n a b  c a  c a b c a b   c b   ........ 2 2
  n
n
2
1
0 1
Proof
The proof is obtained by applying principle of mathematical
induction.
Step: 1
Let the given statement be
  n
n n n n n n n n f n ab  c a  c a b c a b   c b   ( ) : ........ 2 2
Check the result for n = 1 we have
n
2
1
0 1
f a b  c a  c a b  a b  (1) : 1 1 1
1
1 1
0
1 1
Thus Result is true for n =1
Step: 2 Let us assume that result is true for n = k
k k k k k k k k f k ab  c a  c a b c a b   c b   ( ) : ........ 2 2
  k
k
2
1
0 1

13. Step: 3 We shall prove that f (k + 1) is also true,
k k k k k k k k f k a b c a c a b c a b c b
(  1) :   1   1        ........   k

  1
1
1 2 1
2
1
1
1 1
0

k
Now,
  k k a b (a b)(a b) 1     
   k 
k k k k k k k  a  b c a  c a b c a b   c b   ........ 2 2
k
2
1
0 1
From Step 2
 


 
c a  c a b  c a b  ........

c ab
 





   




........
1
1
1 2
0 1
1 2
1 2
1
0
k
k
k k
k
k k k k k
k
k
k k k k k k k
c a b c a b c ab c b
   


1 2
k k k k k k k k
c a c c a b c c a b
    
  





  



 
1
1
1 0 2 1
1
0
. ..
.....
k
k
k k
k
k
k
k
c c ab c b
k k c c c c c c
1 by using 1, , and 1 
1
      
1
0 
1
k
k
k
k
r
k
r
k
r

14. k k k k k k k c a c a b c a b c ab c b
             k
1 ........ 
 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
1
1 2 1 1
2
1
1
1 1
0

k
k k
k

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