the binomial theorem 1 objectives: pascal’s triangle 1 1 2 1 1 3 3 1 coefficient of (x + y) n when...

The binomial theorem 1 Objectives: Pascal’s triangle 1 1 1 2 1 1 3 3 1 Coefficient of (x + y) n when n is large Notatio n: ) ( r n n c r

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The binomial theorem 1

Objectives:

Pascal’s triangle

1 1

1 2 1 1 3 3

Coefficient of (x + y)n when n is large

Notation: )( rn ncr

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

(x + y)2 = x(x + y) + y(x + y) = x2 + 2xy + y2

(x + y)3 = (x + y)(x + y)2 = = x(x2 + 2xy + y2) + y(x2 + 2xy + y2)

(x + y)(x2 + 2xy + y2)

= x3 + 2x2y + xy2

+ x2y + 2xy2 + y3

= x3 + 3x2y + 3xy2 + y3

(x + y)4 = (x + y)(x + y)3 =

= x(x3 + 3x2y + 3xy2 + y3) + y(x3 + 3x2y + 3xy2 + y3)

= x4 + 3x3y + 3x2y2 + xy3

= x4 + 4x3y + 6x2y2 + 4xy3 + y4

x3 + 3x2y + 3xy2 + y3

+ x3y + 3x3y2 + 3x2y3 + y4

Page 3: 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

Expansion of (x + y)n

(x + y)2 = 1x2 + 2xy + 1y2

(x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

(x + y)1 = 1x + 1y

(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5

(1 + y)5 = 1(1)5 + 5(1)4y + 10(1)3y2 + 10(1)2y3 + 5(1)y4 + 1y5

= 1 + 5y + 10y2 + 10y3 + 5y4 + y5

Page 4: 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

Examples: Write down the expansions:

(x + y)2 = 1x2 + 2xy + 1y2

(x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3

(x + 3y)2 =

(x + 3y)2 = 1x2 + 2x(3y) + 1(3y)2 = x2 + 6xy + 9y2

(4 + y)3 =

(4 + y)3 = 1(4)3 + 3(4)2y + 3(4)y2 + 1y3 = 64 + 48y + 12y2 + y3

(x + y)3 = 1x3 + 3x2y + 3xy2 + 1y3

(2a + 3b)3 =

(2a + 3b)3 = 1(2a)3 + 3(2a)23b + 3(2a)(3b)2 + 1(3b)3

= 8a3 + 36a2b + 54ab2 + 27b3

Page 5: 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

Examples:

Write the coefficient of x3 in the expansion of (2x- 3)4.

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

x3 is the 2nd term : 4(2x)3(-3) = 4(8x3)(-3) = -96x3

The required coefficient is - 96

In the expansion of (1 + bx)4, the coefficient of x3 is 1372. Find the constant b.

y3 is the 4th term : 4(1)(bx)3 = 4b3x3 = 1372

4b3 = 1372

(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4

b3 = 343

b = 7

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