1. If (1+x)2 = (1+x)(1+x)
= 1+2x+x2
BINOMIAL EXPANSION How to expand if
(2 + 3x)9 = ?????
DEFINITION OF BINOMIAL EXPANSION
A sequence have 2 term or more is named as binomial sequence or binomial expression.The example
are :
(a + x), (1 - x), (2x + 3y)
Definition 1:
“ FACTORIAL” for a number is a multiplication / product of a sequence number.
Symbol: !
Example : a product of first 7 positive integer are ;
7 x 6 x 5 x 4 x 3 x 2 x 1 read as 7 factorial therefore 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
General form is given by the definition below.
For any integer positive n, we definite “ FACTORIAL n ” which sign by n! as
n! = n(n-1)(n-2)( n-3)………3x2x1 where, 0! = 1
Example 1.
Find the value of ,
a) 9! 18!
d)
b) 6! 15!5!
9!
c)
6!
Definition 2:
Co-eficient of Binomial ;
n n n n n!
Symbol @ Cr is define as = Cr =
r r n r !r!

2. We can use the button on our calculator to find these values
Example 2:
Find the value of ,
5 20
a) c)
3 17
n 6 5
b) d)
0 4 3
EXERCISE 1 EXERCISE 2 ANSWER :
Find the value of ; Without using the
calculator, find:
a) 8! EX. 1:
b) 4!7! 5
7! a) a) 40320
c) 2
0! b) 120960
18 c) 5040
d) 6! – 3! b)
8! 3 d) 714
e) e) 56
6!
14! f) 1001
f) g) 1344
10!4!
h) 30240
4!8!
g) EX. 2:
6!
a) 10
h) 8! – ( 2!7!)
b) 816

3. BINOMIAL THEOREM
Expansion in general form (a + b ) n where a, b is any number and n is integer positive is given as
n n 0 n n 1 1 n n 2 2 n n r r n 0 n
(a + b ) n = a b + a b + a b +……….+ a b +……+ a b
0 1 2 r m
or
n!
a 0b n
n! n! n! n! n!(n n)!
(a + b ) n = a nb0 + a n 1b1 + a n 2 b 2 +….+ a n r b r +..........+
0!(n 0)! 1!(n 1)! 2!(n 2)! r!(n r )!
Or
(a + b)n = an + nC1an – 1b + nC2an – 2b2 + nC3an – 3b3 + ... + nCran – rbr + ... + xn
Or we simplify generally as : (a + b ) n = n
Cran – rbr …………[ general form]
Note:
a) Remember !! power of a is desending,power of b is ascending
b) First term , T1= an
c) Final term T(n)= bn
n n r r
d) Term of ( r+1) , T( r+1) = a b
r
Example 3 :
By using the binomial theorem,expand all the following;
a) ( 3+ x) 5 x
5
b) ( 4-x) 6 d) 2
2
c) ( 1- 3x) 5

4. PASCAL'S TRIANGLE
Another way to expand the ( a + b )n , where n is more than 2
For small values the easiest way to determine the value of several consecutive binomial
coefficients is with Pascal's Triangle:
How Pascal's Triangle is Constructed
1. We start to generate Pascal’s triangle by writing down the number 1. Then we write a new row
with the number 1 twice:
1
1 1
2. We then generate new rows to build a triangle of numbers. Each new row must begin and end
with 1:
1
1 1
1 * 1
1 * * 1
3. The remaining numbers in each row are calculated by adding together the two numbers in the
row above which lie above-left and above-right. So, adding the two 1’s in the second row gives 2,
and this number goes in the vacant space in the third row:
1
1 1
1 2 1
1 * * 1
4. The two vacant spaces in the fourth row are each found by adding together the two numbers in
the third row which lie above-left and above-right: 1 + 2 = 3, and 2 + 1 = 3. This gives:
1
1 1
1 2 1
1 3 3 1
We can continue to build up the triangle in this way to write down as many rows as we wish.
Pascal’s triangle
n=0 1
n=1 1 1
n=2 1 2 1
n=3 1 3 3 1
n=4 1 4 6 4 1
n=5 1 5 10 10 5 1
Exercise 1
1. Generate the seventh, eighth, and ninth rows of Pascal’s triangle.

5. USING PASCAL’S TRIANGLE TO THE BINOMIAL EXPRESSION.
Example:
1. By using the pascal’s triangle, expand ( 2+3x)4 .
STEP:
- First create the pascal’s triangle….where n=4
n=0 1
n=1 1 1
n=2 1 2 1
Take the value in n=3 1 3 3 1
this row as a co- n=4 1 4 6 4
eficient 1
- Recognize expression of a and b ….
So, for ( 2+3x)4 …… 2 is a and 3x is b
- Then put it in the normal way as binomial….where the power of a is descrease
( menurun) and power of b is increase ( menaik)…..
Result:
( 2+3x)4 = 1 (2)4 (3x)0 + 4 (2)3 (3x)1 + 6 (2)2 (3x)2+ 4 (2)1 (3x)3+ 1 (2)0 (3x)4
See..all the co-efiecient is from the pascal’s triangle..which is 1,4,6,4,1
Example:
By using the binomial theorem,expand all the following;
a) ( 3+ x) 5
b) ( 4-x) 6
c) ( 1- 3x) 5
d) 5
x
2
2