Pascals Triangle Binomial Theorem

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Pascals Triangle & Binomial Theorem

Mundeep Gill [email protected]

IntroductionPascals Triangle and the Binomial Theorem are methods that can be used to

expand out expressions of the form

(a + b)n

Where aand bare either mathematical expressions or numerical values and n

is a given number (positive or negative).

Both Pascals Triangle and the Binomial Theorem can be used when n is

positive, however the Binomial Theorem is used when n is negative.

Pascals Triangle

0 11 1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

5 1 5 10 10 5 1

6 1 6 15 20 15 6 1

The numbers in Pascals Triangle indicate the coefficients that are required for

each term when expanding algebraic expressions.

For example, if expanding (a + b)4, Pascals Triangle can be used in the

following way:

The coefficients of terms when raised to the power of 4 are

1 4 6 4 1

Using this:

(a + b)4 = 1(a)4(b)0 + 4(a)3(b)1 + 6(a)2(b)2 + 4(a)1(b)3 + 1(a)0(b)4

= a4 + 4a3b + 6a2b2 + 4ab3 + b4

(taken from row 5 of

Pascals Triangle)


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For a general expansion, i.e. (a + b)n, you decrease the power of a from n to 0

and increase the power of b from 0 to n.

(NB: Within each term the powers should add up to n. For example, in the

above expansion: 1(a)4(b)0 + 4(a)3(b)1 + 6(a)2(b)2 + 4(a)1(b)3 + 1(a)0(b)4 within

each term the sum of all powers equals 4).

Example 1:Use Pascals Triangle to expand (x - 2)3.

Solution: The coefficients of an expansion involving the power of 3 are:

1 3 3 1

Therefore,

(x 2)3 = 1(x)3(-2)0 + 3(x)2(-2)1 + 3(x)1(-2)2 + 1(x)0(-2)3

= 1x3 + 3x2(-2) + 3x(4) + 1(-8)

= x3 6x2 + 12x 8

Example 2: Use Pascals Triangle to expand (3a + b)5.

Solution:The coefficients of an expansion involving the power of 5 are:

1 5 10 10 5 1

Therefore,

(3a + b)5 = 1(3a)5(b)0 + 5(3a)4(b)1 + 10(3a)3(b)2 + 10(3a)2(b)3

+ 5(3a)1(b)4 + 1(3a)0(b)5

= 243a5 + 5(81a4)(b)+ 10(27a3)(b2) + 10(9a2)(b3) + 5(3a)(b4) + b5

= 243a5 + 405a4b+ 270a3b2 + 90a2b3 + 15ab4 + b5


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Example 3: Use Pascals Triangle to expand

5

x

1x

+ .

Solution:5

x

1x

+ = ( )

0

5

x

1x1

+ ( )

1

4

x

1x5

+ ( )

2

3

x

1x10

+ ( )

3

2

x

1x10

+ ( )

4

1

x

1x5

+ ( )5

0

x

1x1

= 5x +x

5x 4+

2

3

x

10x+

3

2

x

10x+

4x

5x+

5x

1

= 5x + 35x + 10x +x

10 + 3x5 + 5x

1

Example 4: Expand (x2 + 1)3

Solution:

(x2 + 1)3 = 1(x2)3(1)0 + 3(x2)2(1)1 + 3(x2)1(1)2 + 1(x2)0(1)3

= x6 + 3x4 + 3x2 + 1

Questions

Use Pascals Triangle to expand the following expressions:

1. (x + 1)6

2. (3x 1)4

3. (2x + y)5

4. (x3 + 1)4

5.

3

x

12x

+


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Solutions:

1. (x + 1)6 = 1(x)6(1)0 + 6(x)5(1)1 + 15(x)4(1)2 + 20(x)3(1)3

+ 15(x)2(1)4 + 6(x)1(1)5 + 1(x)0(1)6

= x6 + 6x5 + 15x4 +20x3 + 15x2 + 6x + 1

2. (3x 1)4 = 1(3x)4(-1)0 + 4(3x)3(-1)1 + 6(3x)2(-1)2 + 4(3x)1(-1)3

+ 1(3x)0(-1)4

= 81x4 + 4(27x3)(-1) + 6(9x2)(1) + 4(3x)(-1) + 1(1)

= 81x4 108x3 + 54x2 12x + 1

3. (2x + y)5 = 1(2x)5(y)0 + 5(2x)4(y)1+ 10(2x)3(y)2 + 10(2x)2(y)3 + 5(2x)(y)4

+ 1(2x)0(y)5

= 32x5 + 5(16x4)(y)+ 10(8x3)(y2)+ 10(4x2)(y3)+ 10xy4 + y5

= 32x5 + 80x4y+ 80x3y2 + 40x2y3 + 10xy4 + y5

4. (x3 + 1)4 = 1(x3)4(1)0 + 4(x3)3(1)1 + 6(x3)2(1)2 + 4(x3)1(1)3 + 1(x3)0(1)4

= x12 + 4x9 + 6x6 + 4x3 + 1

5.

3

x

12x

+ = 1(2x)3

0

x

1

+ 3(2x)2

1

x

1

+ 3(2x)1

2

x

1

+ 1(2x)0

3

x

1

= 8x3 + 3(4x2)

x

1+ 6x

2x

1+

3x

1

= 8x3 + 12x +x

6+

3x

1


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The Binomial Theorem

The Binomial Theorem is an alternative method to expanding algebraicexpressions and is useful when dealing with large powers (where generatinglarge numbers of rows in Pascals Triangle would not be ideal). The Binomial

Theorem also has to be used when n is negative, since Pascals Triangle onlydeals with positive integers.

The general Binomial Theorem is:

(a + b)n = an + nan-1b +( )2!

1nn an-2b2 +

( )( )3!

2n1nn an-3b3

+( )( )( )

4!

3n2n1nn an-4b4 + + bn

A simpler form of the theorem is quoted by taking the special case in whicha = 1 and b = x.

(1 + x)n = 1 + nx +( )2!

1nn x2 +

( )( )3!

2n1nn x3 +

( )( )( )4!

3n2n1nn x4 + + xn

To illustrate how to apply the above theorem lets look at a simple example(where Pascals Triangle would probably be more ideal to use).

Use Pascals Triangle to expand (1 + x)4.The simpler form of the theorem can be used, which is:

(1 + x)n = 1 + nx + ( )2!

1nn x2 + ( )( )3!

2n1nn x3 + + xn

Where n in this case is 4.

(1 + x)4 = 1 + 4x +( )2!

144 x2 +

( )( )3!

24144 x3 + x4

= 1 + 4x +2

12x2 +

6

24x3 + x4

= 1 + 4x + 6x2 +4x3 + x4


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Example 2:

Use the Binomial Theorem to find the first four terms of

10

x3

11

+

Solution:The first four terms can be found by using

(1 + x)n = 1 + nx +( )2!

1nn x2 +

( )( )3!

2n1nn x3

Where x is replaced by x3

1and n = 10, hence

10

x3

11

+ = 1 + 10

x

3

1+

( )2

x3

1

2!

11010

+

( )( )3

x3

1

3!

21011010

= 1 + x3

10+

( )

2x9

1

2

910+

( )( )

3x27

1

6

8910

= 1 + x3

10+ 5x2 + 3x

9

40+

Example 3:

Use the Binomial Theorem to find the first four terms of (2 3z)11.

Solution:

Since the first term is not 1, we have to use the original form of the Theorem:

(a + b)n = an + nan-1b +( )2!

1nn an-2b2 +

( )( )3!

2n1nn an-3b3

Where a = 2, b = -3z and n = 11,

(2 3z)11 = (2)11 + 11(2)10(-3z) +( )2!

11111 (2)9(-3z)2

+( )( )

3!

21111111 (2)8(-3z)3

= 2048 + 11(1024)(-3z) +( )2

1011(512)(9z2) +

( )( )6

91011(256)(-27z3)

= 2048 33792z + 253440z2 1140480z3


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Example 4:

Use the Binomial Theorem to find the first four terms of ( )23

x1+ .

Solution:

The first four terms can be found by using

(1 + x)n = 1 + nx +( )2!

1nn x2 +

( )( )3!

2n1nn x3

Replacing n by2

3

( )23

x1+ = 1 +2

3x +

2!

12

3

2

3

x2 +3!

22

31

2

3

2

3

x3 +

= 1 +2

3x +

2

2

1

2

3

x2 +6

2

1

2

1

2

3

x3 +

= 1 +2

3x +

8

3x2

16

1x3 +


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Questions

Use the Binomial Theorem to find the first four terms of the following

expressions:

1. (1 + x)8

2. (1 4x)9

3. (2 + 5x)12

4. ( )21

x1+

5.23

x2

11

Solutions:

1. (1 + x)8

Using (1 + x)n = 1 + nx +( )2!

1nn x2 +

( )( )3!

2n1nn x3

(1 + x)8 = 1 + 8x +( )

2!

188 x2 +

( )( )

3!

28188 x3

= 1 + 8x +

( )

2

78

x

2

+

( )( )

6

678

x

3

+

= 1 + 8x + 28x2 + 56x3 +

2. (1 4x)9

Using (1 + x)n = 1 + nx +( )2!

1nn x2 +

( )( )3!

2n1nn x3

Where x is replaced by -4x and n = 9

(1 4x)9 = 1 + 9(-4x) + ( )2!

199 (-4x)2 + ( )( )3!

29199 (-4x)3 +

= 1 36x +( )2

89(16x2) +

( )( )6

789(-64x3) +

= 1 36x + 576x2 5376x3 +


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3. (2 + 5x)12

Using the original form of the Theorem:

(a + b)n = an + nan-1b +( )2!

1nn an-2b2 +

( )( )3!

2n1nn an-3b3

Where a = 2, b = 5x and n = 12

(2 + 5x)12 = 212 + 12(2)11(5x) +( )

2!

11212 (2)10(5x)2

+( )( )

3!

21211212 (2)9(5x)3 +

= 4096 + 12(2048)(5x) + 66(1024)(25x2)

+( )( )

6

101112(512)(125x3)

= 4096 + 122880x + 1689600x2 + 14080000x3

4. ( )21

x1+

Using (1 + x)n = 1 + nx +( )2!

1nn x2 +

( )( )3!

2n1nn x3 +

( )21

x1+ = 1 +2

1x +

2!

12

1

2

1

x2 +3!

22

11

2

1

2

1

x3 +

= 1 +2

1x +

2

2

1

2

1

x2 +6

2

3

2

1

2

1

x3 +

= 1 +2

1x

8

1x2 +

16

1x3 +


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5.23

x2

11

Using (1 + x)n = 1 + nx +( )2!

1nn x2 +

( )( )3!

2n1nn x3

Where x = x2

1 and n =

2

3

23

x2

11

= 1 +

2

3

x

2

1+

( )

2!

123

23

2

x2

1

+( )( )

3!

21 23

23

23

3

x2

1

+

= 1 x4

3+

( )

221

23

2x4

1+

( )( )

621

21

23

3x8

1+

= 1 x4

3+

32

3x2 +

128

1x3 +

Pascals Triangle Binomial Theorem

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