1. The binomial theorem 1 Objectives: Pascal’s triangle 1 1 2 1 1 3 3 1 Coefficient of (x + y) n when n is large Notation: ncrncr

2. Expansion of (x + y) n for n = 2, 3 and 4 (x + y) 2 =x(x + y) + y(x + y) =x 2 + 2xy + y 2 (x + y) 3 =(x + y)(x + y) 2 = = x(x 2 + 2xy + y 2 ) + y(x 2 + 2xy + y 2 ) (x + y)(x 2 + 2xy + y 2 ) = x 3 + 2x 2 y + xy 2 + x 2 y + 2xy 2 + y 3 = x 3 + 3x 2 y + 3xy 2 + y 3 (x + y) 4 =(x + y)(x + y) 3 = = x(x 3 + 3x 2 y + 3xy 2 + y 3 ) + y(x 3 + 3x 2 y + 3xy 2 + y 3 ) = x 4 + 3x 3 y + 3x 2 y 2 + xy 3 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 x 3 + 3x 2 y + 3xy 2 + y 3 + x 3 y + 3x 3 y 2 + 3x 2 y 3 + y 4

3. Expansion of (x + y) n (x + y) 2 =1x 2 + 2xy + 1y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 + 1y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4 (x + y) 1 =1x + 1y (x + y) 5 = 1x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5xy 4 + 1y 5 (1 + y) 5 = 1(1) 5 + 5(1) 4 y + 10(1) 3 y 2 + 10(1) 2 y 3 + 5(1)y 4 + 1y 5 = 1 + 5y + 10y 2 + 10y 3 + 5y 4 + y 5

4. Examples: Write down the expansions: (x + y) 2 =1x 2 + 2xy + 1y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 + 1y 3 (x + 3y) 2 = 1x 2 + 2x(3y) + 1(3y) 2 =x 2 + 6xy + 9y 2 (4 + y) 3 = 1(4) 3 + 3(4) 2 y + 3(4)y 2 + 1y 3 =64 + 48y + 12y 2 + y 3 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 + 1y 3 (2a + 3b) 3 = 1(2a) 3 + 3(2a) 2 3b + 3(2a)(3b) 2 + 1(3b) 3 = 8a 3 + 36a 2 b + 54ab 2 + 27b 3

5. Examples: Write the coefficient of x 3 in the expansion of (2x- 3) 4. (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4 x 3 is the 2 nd term :4(2x) 3 (-3) =4(8x 3 )(-3) =-96x 3 The required coefficient is - 96 In the expansion of (1 + bx) 4, the coefficient of x 3 is 1372. Find the constant b. y 3 is the 4 th term :4(1)(bx) 3 =4b 3 x 3 = 1372  4b 3 = 1372 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4  b 3 = 343  b = 7