1. The Binomial Theorem (a+b)2 (a+b)(a+b) a2 +2ab + b2
(a+b)3 (a+b)(a2+2ab+b2) a3+3a2b2+3ab2+b3
Where have we seen these leading coefficients before?
Pascal’s Triangle

2. The Binomial Theorem Tr+1 =( )an-rbr
(a+b)n = an+ [nC1]a(n-1)b+….+ [nCr]a(n-r)br+…+bn
(a+b)n = an+ ( )a(n-1)b+….+ ( )a(n-r)br+…+bn n n r 1
Finding General Term…
Tr+1 =( )an-rbr n r

3. Example 3
Write down the first 3 and last 2 terms of the expansion of (2x+ )12
(2x+ )12=(2x)12 + ( )(2x)11( ) + ( )(2x)10( )2 +
( )(2x)( )11 + ( )12 12 12 2 1 12 11

4. Example 4 Find the 7th term of (3x- )14. a= (3x), b= (- ) and n=14
So, as Tr+1= ( )an-rbr, we let r = 6
T7= ( )(3x)8(- )6

5. Example 5 a) The coefficient of x6 b) The constant term
In the expansion of (x2+ )12, find:
a) The coefficient of x6
b) The constant term
a= (x2), b= ( ) and n= Tr+1 = ( )(x2)12-r( )r
= ( )x24-2r
= ( )4rx24-3r
If 24-3r=0 Then 3r = 24 R = 8 T9 = ( )48x0 The constant term is… ( )48 or 32,440,320
If 24-3r=6
Then 3r=18
R=6
T7 = ( )46x6
The coefficient of x6 is
( )46 or 3,784,704

6. Example 6 Find the coefficient of x5 in the expansion of (x+3)(2x-1)6.
=(x+3)[(2x)6+( )(2x)5(-1)+( )(2x)4(-1)2+…]
=(x+3)(26x6-( )25x5+( )24x4-…)
So, the terms containing x5 are ( )24x5 from (1)
and -3( )25x5 from (2)
The Coefficient of x5 is ( )24-3( )25=-336 1 2