binomial theorem 3 unit maths - school of · pdf filepolynomial can be expanded using the...

ONLINE: 3 UNIT MATHS 1

HSC MATHEMATICS: MATHEMATICS EXTENSION 1

3 UNIT MATHEMATICS

TOPIC 17 BINOMIAL THEOREM

n – factorial n!

For positive integers n the product of the first n natural numbers is called

n – factorial and is written

! (1)(2)(3) ( 1)( )n n n

0! 1 1! 1 2! (1)(2) 2 7! (1)(2)(3)(4)(5)(6)(7) 5040

A binomial is a polynomial of two terms such as 2 5x y or 32 ( )a b .

The Binomial Theorem is a quick way of expanding a binomial expression that

has been raised to some power n , for example, 12

5 6x .

http://www.physics.usyd.edu.au/teach_res/hsp/m3u/math3unit.htm




Consider the polynomial ( )n

x y where n is an integer 1, 2, 3,n This

polynomial can be expanded using the Binomial Theorem in terms of the

binomial coefficients 1,2, 3, 0n

k

nC n k n

k

n

k

is read as ‘n over k’

0

0 0

4 4 4 4

0 1 3 2

! !0

! ! ! !

01 1 1

0 0

4 4 4(4)(3)(2)(1) (4)(3)(2)(1)1 4

0 1 2(1)(3)(2)(1) (1)(2)(2)(1)

n n n

k k n k

n n

n

n n nn nC k n C C

k k n kk n k k n k

n nC C C

n

C C C C

4 4 4

3 1 4

6

4 4(4)(3)(2)(1)4 1

3 4(1)(2)(3)(1)C C C



Pascal’s triangle

An easy way to calculate the binomial coefficients is to use Pascal’s Triangle

where the binomial coefficients are arranged in a triangle. Each interior number

is the sum of the nearest two numbers in the previous row.



6 (6)(5)(4)(3)(2)(1)15

4 (4)(3)(2)(1)(1)(2

6 (6)(5)(4)(3)(

6 (6)(5)(4)(3)(2)(1)6

5 (5)(4)(3)(2)(1)(

2)(1)20

3 (3)(2)(1)(1)(2

)

6 (6)(5)(4)(3)(2)(1)1

2 (2)(1)(1)(2)(3)(4)(3

1)6

)

61

)

6 (6)(5)(4)(3)(2)(1)6

1 (1)(1)(2)(3)(

61

0(5)

5

4)

The construction of a Pascal Triangle is based upon the relationship between

binomial coefficients

1 1

11 1

n n n

k k kC C C k n

This is known as Pascal’s Triangle Identity.

Example 6 5 5

2 1 215 5 10 15LHS C RHS C C LHS RHS



Proof 1 1

11 1

n n n

k k kC C C k n

1 1

1 0 1

!1

! !

1 !!1 1 1 1

1! 1 ! 1! 2 !

n

k

n n n

nC k n

k n k

nnk C n C C n n

n n

QED

1 1

1

2

1 ! 1 !

1 ! ! ! !

1 ! 1 !1 1

1 ! ! 1 ! ! ( )( )

!

! !

n n

k k

n

k

k

n nC C

k n k k n k

n n

QED

k n k

k n k n k k k n k k n k

nC

k n k

The coefficients of the variables x and y in the expansion of n

x y are called the binomial

coefficients. The (k+1)th binomial coefficient of order n ( n a positive integer) is

!

! !

n

k

n nC

k k n k

(k+1)th binomial coefficient

n

k

nC

k

gives the number of combinations of n things k at a time.



Binomial Theorem

1 2 2 1

0 0

( )

( )

1 2 1

n n n n n n

k n k nn n k k n n k k

k

k k

x y x

x y

n n nx y x y x y y

n

nx y C x y

k

Example Expand the binomial expression 6( )a b

Construct a Pascal Triangle to find the binomial coefficients

6 6 5 4 2 3 3 2 4 5 6( ) 6 15 20 15 6a b a a b a b a b a b ab b

The sum of the binomial coefficients is equal to 2n and is obtained by setting

1x y

6 6

0

6 (1 1) 2 1 6 15 20 15 6 1 64k n

k

nn

k



The binomial theorem is proved by induction. That is, it is shown to hold for n = 1, and further

shown that if it holds for any given value of n then it also holds for the next higher value of n.

0

( )

k nn n k k

k

x yn

x yk

Suppose 0

( )

k nn n k k

k

x yn

x yk

is true. Now consider

1

0

1 1

0 0

11 1 1 1

1 0

1 1 1 1

0

Let 1 1

n n

nn k k

k

n nn k k n j j

k j

n nn n k k n n j j

k j

n n n n k k

k

x y x y x y

nx y x y

k

n nx y x y

k j

n n n nx x y y x y

k n j

j k k j

nx y x y x y

k

1

1 1

1 1 1

1

11 1 1 1 1

1 0

1

1

1Using Pascal's Identity

1

1 1

n nn k k

k

nn n n k k

k

n nn n n n k k n k k

k k

nx y

k

n nx y x y

k k

n n n

k k k

n nx y x y x Qy

k kEy x

D



Example (a) Use the binomial theorem to expand 5

2 3x .

(b) What is the 4th term in the expansion for increasing powers of x?

(c) Find the largest coefficient in the expansion.

(d) Expand 5

2 3x

(e) Use the binomial theorem to differentiate the function 5

2 3y x

(f) Use the binomial theorem to integrate 5

2 3y x from x = 0 to x = 1.

(a)

Always write down the binomial theorem to start answering the question

5

5

0

( )5k

n k k

k

x y x yk

5

5 5

0

52 3 2 3

kk k k

k

x xk

Use Pascal’s triangle to give the binomial coefficients 5

1 5 10 10 5 1k

5 5 4 3 2 2 2 3 3 4 4 5 5

2 3 4 5

2 3 (1)(2 )(1) (5)(2 )(3) (10)(2 )(3 ) (10)(2 )(3 ) (5)(2)(3 ) (1)(3 )

32 240 720 1080 810 243

x x x x x x

x x x x x



(b) increasing powers of x: 30 1 2 4 5xx x x x x 4th term 3k

4th term 5 3 3 3 3

4

52 3 1080

3t x x

in agreement with part (a)

(c) Let the coefficients be represented by k

a . Since x and y are positive, we can check whether

the coefficients are increasing or decreasing by considering the ratio 1

/ 0k k

a a k n

If 1 2 1 1 2 1

/ 1 / 1k k k k k k k k

a a a a a a a a

is the largest coefficient

5 1 1

1

5

52 3

51 3/ 1 2.6 2 1 3

5 1 22 3

k k

k k

k k

kka a k k k

k

k

Therefore 3

1080a is the largest coefficient in agreement with part (a).

(d) 5

5

0

( )5k

n k k

k

x y x yk

5

5 5

0

52 3 2 ( 1) 3

kk kk k

k

x xk

5 5 4 3 2 2 2 3 3 4 4 5 5

2 3 4 5

2 3 (1)(2 )(1) (5)(2 )(3) (10)(2 )(3 ) (10)(2 )(3 ) (5)(2)(3 ) (1)(3 )

32 240 720 1080 810 243

x x x x x x

x x x x x



(e)

5 4

44

0

4 3 2 2 3 4

2 3 4

5 2

( )

2 3 / (5)(3)(2 3 )

4Pascal's triangle 4 1 4 6 4 1

/ (15) 2 (4)(2) (3 ) (6)(2) (3 ) (4)(2)(3 ) (3 )

/ 240 1440 3240 3240 1215

2 3 32 240 720 1

kn k k

k

x y

y x dy dx x

x y nk

dy dx x x x x

dy dx x x x x

y x x x

3 4 5

2 3 4

080 810 243

/ 240 1440 3240 3240 1215

x x x

dy dx x x Ex Dx Q

(f)

15 5

0

16 6 6

0

5 2 3 4 5

12 3 4 5

0

12 3 4 5 6

0

2 3 2 3

1 12 3 5 2 864.5

18 18

2 3 32 240 720 1080 810 243

32 240 720 1080 810 243

32 120 240 270 162 40.5

32 120 240 270

y x I x dx

I x

y x x x x x x

I x x x x x dx

I x x x x x x

I

162 40.5 864.5 QED



Example Find the coefficients of x4 and x3 in the expansion of 6

12x

x

Check your answer by expanding the expression using the binomial theorem.

66 6

0

6 6 66 66 6 2

0 0

( )6

Pascal's triangle 6 1 6 15 20 15 6 1

6 612 2 1 1 2

kk k

k

k kk k k kk k k

k k

x y x y nk

x x x xk kx

For the term in 4x we require 6 2 4 1k k , therefore, the coefficient

1ka

is

5

1

62 1 (6)(32) 192

1k

a

For the term in 3x we require 6 2 3 1.5k k , k is not an integer, there is no term in 3

x

4

6

6 2 2 4 612 64 240 160 60 / 12 /19 2 1 /x x x x x x

xx



Example Find the term in the expansion of 20

3 5 x with the greatest coefficient.


20

20

0

( )20k

n k k

k

x y x yk

20

20 20

0

203 5 3 5

kk k k

k

x xk

In the expansion of n

x y where x and y are both positive, then successive coefficients in the

expansion get larger and then smaller. Therefore, the ratio R of the (k+1)th coefficient to the kth

coefficient will exceed one or be equal to one until the largest term is reached.

If 1 2 1 1 2 1

/ 1 / 1k k k k k k k k

a a a a a a a a

is the largest coefficient

20 1 1

20

20!3 5

( 1)! 20 1 ! 5 120 1

20! 3 13 5

! 20 !

100 5 3 3 8 97 12.125 12 1 13

k k

k k

k kR k

k

k k

k k k k k k

The largest term is term is 7 13 1320

3 513

x



Example

(a) Expand 1n

x substitute 1x

(b) Expand 1n

x substitute 1x hence show that

1 3 5 0 2 4

n n n n n nC C C C C C

(c) Show that 1

1 3 5 0 2 42

n n n n n n nC C C C C C

(d) Prove 1

1

n n n

k k kC C C

which gives the Pascal’s triangle identity

(e) Show that

2

0

2 n

k

n n

n k


0

( )

k nn n k k

k

x yn

x yk

(a) 0

( 1)

k nn n k

k

xn

xk

0 1 2

0

1 2

k nn n n n n

n

k

xn

C C C Ck

(b) 0

( 1) 1k n

kn n k

k

xn

xk



0 1 2 3

0

1 3 5 0 2 4

1 0 1 1k n

k nn n n n n

n

k

n n n n n n

xn

C C C C Ck

C C C C C C

(c) Add the results from parts (a) and (b)

0 2 4

1

0 2 4 1 3 5

2 2

2

n n n n

n n n n n n n

C C C

C C C C C C

(d) 1

1 1

0

11

nn n k

k

nx x

k

coefficient for 1n k

x is

1n

k

1 1

0 0 0

1 ( 1) 1 ( 1)n n n

n n n k n k n k

k k k

n n nx x x x x x x

k k k

coefficient for 1n kx

is 1

n n

k k

Therefore, 1

1

n n n

k k kC C C

n

k

nC

k



(e) 2

2 2

0

( 1)2k n

n n k

k

xn

xk

We are interested in the term 2

nn

xn

when k n .

Also, 00 0

( 1) ( 1) ( 1)

k nn k

k

k n k nn n k n n n k

k k

x x xn n

x xk k

nx

k

The term in nx from the expansion expressed as a product is

0

kn

n k

k

nxx

n k

n

k

nx is formed by taking the product of terms in n k

x from the first expression and k

x from

the second expression in the product.

2

0 0

n nn k n

k

k

k

n n n nx x

k n k kx

k

n

n k

Hence, equating the coefficients of the terms for nx

2

0

2 n

k

n n

n k


binomial theorem 3 unit maths - school of · pdf filepolynomial can be expanded using the...

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