1(1)5 + 5(1)4(2x) + 10(1)3(2x)2 + 10(1)2(2x)3 + 5(1)(2x)4 + 1(2x)5 Binomial theorem (x+y)1 = 1x + 1y (x+y)2 1x2 + 2xy + 1y2 (x+y)3 1x3 + 3x2y + 3xy2 + 1y3 (x+y)4 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x+y)5 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 (1 + 2x)5 = 1(1)5 + 5(1)4(2x) + 10(1)3(2x)2 + 10(1)2(2x)3 + 5(1)(2x)4 + 1(2x)5 = 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5
(x+y)1 = 1x + 1y (x+y)2 1x2 + 2xy + 1y2 (x+y)3 1x3 + 3x2y + 3xy2 + 1y3 (x+y)4 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x+y)5 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 (1 - 2x)5 = 1(1)5 + 5(1)4(-2x) + 10(1)3(-2x)2 + 10(1)2(-2x)3 + 5(1)(-2x)4 + 1(-2x)5 = 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5
= 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2)2 + 4(1) (x+x2)3 + 1(x+x2)4 Expand (1 + x + x2)4 Treat as (1 + (x + x2))4 = 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2)2 + 4(1) (x+x2)3 + 1(x+x2)4 (x+x2)2 = x2 +2x3+x4 (x+x2)3 = x3 +3x2(x2)+3x(x2)2+(x2)3 = x3 + 3x4 + 3x5 + x6 Similarly (x+x2)4 = x4 +4x3(x2) +6x2(x2)2 +4x(x2)3 +(x2)4 = x4 +4x5 +6x6 +4x7 +x8
= 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2)2 + 4(1) (x+x2)3 + 1(x+x2)4 Expand (1 + x + x2)4 = (1 + (x + x2))4 = 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2)2 + 4(1) (x+x2)3 + 1(x+x2)4 = 1 + 4x +4x2 +6(x2+2x3+x4) +4( x3 + 3x4 + 3x5 + x6) + x4 +4x5 +6x6 +4x7 +x8 = 1+ 4x + 10x2 + 16x3 + 19x4 + 16x5 + 10x6 + 4x7 + x8
(1 - 2x)5 = 1(1)5 + 5(1)4(-2x) + 10(1)3(-2x)2 + 10(1)2(-2x)3 + 5(1)(-2x)4 + 1(-2x)5 = 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5 Remember nCr = n! (n-r)!r! 5C1 = 5 5C2 = 10 5C4 = 5 5C3 = 10 (1 - 2x)5 = 5C0(1)5 + 5C1(1)4(-2x) + 5C2(1)3(-2x)2 + 5C3(1)2(-2x)3 + 5C4(1)(-2x)4 + 5C5(-2x)5
Find the coefficient of x5y6 in the expansion of (x+y)11 (2y)6 = 26y6 So coefficient is 462 x 26 = 29568
Find the coefficient of x3 in the expansion of (x2+1/x)6 You get x3 by (x2)3 * (1/x)3 So you need 6C3 6C3 = 20 So coefficient is x3 = 20
1(1)5 + 5(1)4(-2x) + 10(1)3(-2x)2 + 10(1)2(-2x)3 + 5(1)(-2x)4 + 1(-2x)5 Find 0.985 to 3 sf Let x = 0.01 in the expression 0.985 = 1 – 10(0.01) + 40(0.01)2 – 80(0.01)3 + 80(0.01)4 – 32(0.01)5 0.985 = 1 – 0.1 + 0.004 – 0.00008 +…… 0.985 = 0.90392 = 0.904 (3sf)