The Binomial Theorem (a+b)2 (a+b)(a+b) a2 +2ab + b2

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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

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

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

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

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= 12 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

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