The document discusses the binomial theorem, which provides a formula for expanding binomial expressions of the form (a + b)^n. It explains that the theorem allows calculating terms of the expansion without using repeated FOIL multiplication. Pascal's triangle is introduced as a way to determine the coefficients of each term. The key points of the binomial theorem are defined, including that the sum of the exponents in each term equals n. An example expansion is shown. Proofs of properties like the coefficients when r=0, 1, n-1, n are provided.

1. Binom
ial
The
Theorem
By iTutor.com
T- 1-855-694-8886
Email- info@iTutor.com

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

15. Call us for
more
Information:
1-855-694-
8886
www.iTutor.com
Vis
it
The End