the binomial expansion1

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The Binomial Expansion


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CURRICULUM CONTENT


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In this presentation we will develop a formula toenable us to find the terms of the expansion of

n b a )(

where n is any positive integer.

We call the expansion binomial as the originalexpression has 2 parts.

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222 b ab a

2)( ba ))(( baba

We know that

so the coefficients of the terms are 1, 2 and 1

We can write this as22

b ab a 1 2 1



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223ab b a a

3)( b a 2))(( b a b a

)2)(( 22

b a b a b a

322 b ab b a

3223b ab b a a

so the coefficients of the expansion of

are 1, 3, 3 and 1

3)( ba

1 2 1

1 2 1

331 1



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So, we now have

3)( b a

2)( b a

CoefficientsExpression

1 2 1

1 3 3 1

4)( b a 1 4 6 4 1



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So, we now have

3)( b a

2)( b a


1 2 1

1 3 3 1

4)( b a 1 4 6 4 1

Each number in a row can be found by adding the 2coefficients above it.



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So, we now have

3)( b a

2)( b a


1 2 1

1 3 3 1

4)( b a 1 4 6 4 1

The 1st and last numbers are always 1.

Each number in a row can be found by adding the 2coefficients above it.



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So, we now have

3)( b a

2)( b a


1 2 1

1 3 3 1

1

)( b a

1 1

0)( b a

4)( b a 1 4 6 4 1

To make a triangle of coefficients, we can fill inthe obvious ones at the top.

1



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The triangle of binomial coefficients is calledPascal’s triangle, after the French mathematician.

. . . but it’s easy to know which row we want as,for example,

3)( b a starts with 1 3 . . .

10)( b a will start 1 10 . . .

Notice that the 4th row gives the coefficients of

)( b a 3



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Exercise

Find the coefficients in the expansion of6

)( b a

Solution: We need 7 rows

1 2 1

1 3 3 1

1 1

1

1 4 6 4 11 5 10 110 5

1 6 15 120 15 6Coefficients



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We usually want to know the complete expansion not just the coefficients.

5)( b a e.g. Find the expansion of

Pascal’s triangle gives the coefficients Solution:1 5 10 110 5

The full expansion is

Tip: The powers in each term sum to 5

543223405babbabababa 1 5 10 10 5 1

1

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Example 1

1. Find the expansion of in ascendingpowers of x .

5

)21( x

Solution: The coefficients are

1 5 10 110 5

5432)2()2(5)2(10)2(10)2(51 x x x x x

543232808040101 x x x x x

So,

5)21( x

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EXERCISE 1



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20)( b a If we want the first few terms of the expansionof, for example, , Pascal’s triangle is not

helpful.

We will now develop a method of getting the

coefficients without needing the triangle.



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Specific Term of an ExpansionIf only a specific term of an expansion is required,the binomial theorem can be used to determine such aterm without computing all the rows of Pascal’s

triangle or all the preceding coefficients.

where r is any integer from 0 to n .

r r n

r

n

r baC U

1

Generalizations

The binomial expansion of in ascendingpowers is given by

n

b a )(

nnnnnnn

n

bbaC baC baC

ba

...

)(

22

2

1

1

0

0

Binomial Theorem

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e.g.2 Find the first 4 terms in the expansion ofin ascending powers of x .18)1( x

22

18

)( xC

Solution:

18)1( x

018

0

18

)()1( xC )(118

x C

...)( 3

3

18

x C

1 x 18 2153 x ...816 3 x



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EXERCISE 2



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e.g.4 Find the 5th term of the expansion ofin ascending powers of x .

12)2( x

48

41 )2(12

xC U r

Solution: The 5th term contains 4x

Powers of a + b

It is

48)2(495 x4126720 x

These numberswill always bethe same.

r r n

r

n

r baC U

1



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EXERCISE 3



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The binomial expansion of in ascendingpowers of x is given by

n b a )(

n n n n n n n

n

b b a C b a C a C

b a

...

)(

222

110

SUMMARY

The ( r + 1 ) th term is r r n r

n b a C

The expansion of isn x )1(

n n n n n x x C x C C x ...)1(

2210

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Exercise

1. Find the 1st

4 terms of the expansion ofin ascending powers of x .

8

)32( x

Solution:

353

8262

871

880

8)3(2)3(2)3(22 x C x C x C C

2. Find the 6th term of the expansion ofin ascending powers of x .

13)1( x

3248384161283072256 x x x

55

13)( x C Solution:

51287 x

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the binomial expansion1

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