the binomial theoremweb2.hunterspt-h.schools.nsw.edu.au/studentshared...binomial product: the...

Binomial product: The product of two or more expressions each consisting of two terms e.g. ( )a x n+

Choose notation (or combination): Describes the number of ways of choosing r items from n in no particular order

Coefficient: A constant multiplied by a pronumeral e.g. 3x2 has coefficient 3

TERMINOLOGY

The Binomial Theorem

10

467Chapter 10 The Binomial Theorem

DID YOU KNOW?

The binomial theorem was studied by mathematicians from very early times. For example, Euclid discovered the special expansion for ( )x y 2+ around 300 BC. Euclid is mainly famous for his work on geometry, but he also studied arithmetic and number theory. He founded the first school of mathematics, in Alexandria in Egypt.

Omar Khayyam, working around 1100 AD, discovered the expansions of ( ) ( )x y , x y4 5+ + and .( )x y 6+ He is mainly remembered as a poet (he wrote the Rubaiyat), but he was also an astronomer

and a mathematician. He wrote a book about algebra.Blaise Pascal (1623–62) is associated with the binomial theorem because of his famous ‘Pascal’s

triangle’, which is related to the coefficients of binomials. However, he approached the binomial theorem from the standpoint of probability rather than algebra.

Sir Isaac Newton (1642–1727) discovered the general rules for coefficients of binomial expansions, which Leonhard Euler (1707–83) later proved.

INTRODUCTION

You have alreadY studied binomial products. a special result of these products is .( )x y 2+ the binomial theorem looks at the expansion of the more general binomial ( )x y n+ .

the binomial theorem is associated with algebra and probability. in this chapter, you will study some properties of binomial products and their coefficients. in the next chapter you will see how the binomial theorem is used in probability theory.

Class Investigation

Can you complete the next 3 lines of Pascal’s triangle? is the triangle 1. always symmetrical?

1

1

3

1 1

3

1

4 4

1

1

1

1

2

6

Find the sum of each line.2. e.g. 1 3 3 1 8+ + + =Can you find a pattern for these sums? What would be the sum of the numbers in the nth line?

Combinations

468 Maths In Focus Mathematics Extension 1 HSC Course

in the Preliminary Course, you studied factorial notation and combinations. We revise these here.

Factorial notation:

! ( ) ( ) ( )… · ·n n n n n1 2 3 3 2 1= − − −

0! 1=

( )! !!C

n r rnn

r =−

n( )! !

!n r r

n=−ra k

Combinations:

We can also use choose notation:

ExamplEs

evaluate

1. 4!

Solution

4! 4 3 2 1

24

==

2. C107

Solution

( )! !!

! !!

C10 7 7

10

3 710

107 =

−

=

......

3 2 1 7 6 5 3 2 110 9 8 7 3 2 1

120

=

=

You can also evaluate these questions on a calculator.

´ ´ ´

´ ´ ´ ´ ´ ´´ ´ ´ ´ ´

´ ´ ´´ ´


3. 51a k

Solution

( )! !!

! !!

5 1 15

4 15

4 3 2 1 15 4 3 2 1

5

5 =−

=

=

=

1a k

4. C60

Solution

( )! !!

! !!

C6 0 0

6

6 06

6 5 4 3 2 1 16 5 4 3 2 1

1

60 =

−

=

=

=

5. 88a k

Solution

( )! !!

! !!

8 8 88

0 88

1 8 7 6 5 4 3 2 18 7 6 5 4 3 2 1

1

8 =−

=

=

=

8a k

Properties of coefficients

C 1n0 =

´ ´ ´ ´´ ´ ´ ´

´ ´ ´ ´ ´

´ ´ ´ ´ ´´ ´ ´ ´ ´2 ´

´ ´ ´ ´ ´ ´ ´ ´´ ´ ´ ´ ´ ´ ´=


Proof

( )

( )

a b C a C a b C a b C a b C ba b

C C C C C

C

1 0

1 0 1 1 0 1 0 1 0 0

1

Let and

n n n n n n n n n nn

n


n

n

0 11

22 2

33 3

0 11

22 2

33 3

0

f

f

+ = + + + + += =

+ = + + + + +=

− − −

− − −

C 1nn =

Proof

( )

( )

a b C a C a b C a b C a b C ba b

C C C C C

C

C C

0 1

0 1 0 0 1 0 1 0 1 1

1

1

Let and


n


n

nn

n nn

0 11

22 2

33 3

0 11

22 2

33 3

0`

f

f

+ = + + + + += =

+ = + + + + +== =

− − −

− − −

in general,

C Cnk

nn k= −

C C Cnk

nk

nk

11

1= +−−

− for ≤ ≤k n1 1−

Proof

(1 )

(1 )

( ) ( )

( ) ( )

( )

( ) ( )

( )

x C C x C x C x

x x C x C x C x

C x C x

x x x

C C x C x C x

C x C x C x C x C x

C C C x C C x

C C x C x

x x

x

C C x C x C x C x C x

1 1

1 1

1

LHS

RHS

n n n n nn

n

n n n n

nn

n nn

n

n n

n n n nn

n

n n n nn

n nn

n

n n n n n

nn

nn

n nn

n

n

n

n n n n nk

k nn

n

1 10

11

12

2 11

1

1 10

11

2 12

3

12

1 11

1 1

10

11

12

2 11

1

10

11

2 12

3 12

1 11

10

11

10

12

11

2

11

12

1 11

1

0 1 22

33

`

`

f

f

f

f

f

f f

+ = + + + ++ = + + +

+ ++ + +

= + + + ++ + + + + +

= + + + + ++ + +

= + += += + + + + + + +

− − − − −−

−

− − − −

−−

− −−

− −

− − − −−

−

− − − −−

− −−

− − − − −

−−

−−

− −−

−


equating coefficients gives

,

, ,

C C C

C C C

C C C

n n n

n n n

nn

nn

nn

11

11

0

21

21

1

11

11

2

f

= += += +

− −

− −

−−

−−

−

in general, .C C Cnk

nk

nk

11

1= +−−

−

ExamplEs

1. evaluate .C99

Solution

C 1=99

2. show that .C C38

5=8

Solution

C8

( )! !!

! !!

( )! !!

! !!

C

C

C

8 3 38

5 38

8 5 58

3 58

So

83

85

38

5

=−

=

=−

=

=

3. show that 7 6 6= +4 3 4a a ak k k

Solution

( )! !!

! !!

( )! !!

( )! !!

! !!

! !!

7 4 47

3 47

6 3 36

6 4 46

3 36

2 46

7

6 6

LHS

RHS

=

=−

=

= +

=−

+−

= +

4

3 4

a

a a

k

k k

CONTINUED


! !!

! !!

! !!

! !!

! !( !) ( !)

! !( !)

! !!

LHS

3 3 46 4

2 4 36 3

4 36 4

4 36 3

4 34 6 3 6

4 37 6

4 37

7 6 6So

= +

= +

=+

=

=

=

= +4 3 4a a ak k k

1. evaluate7!(a) 8!(b) 3!(c) 5!(d) 0!(e)

(f) ! !10 2−3(4!)(g) 5(6!)(h)

(i) 59!

(j) 4!8!

2. evaluate(a) C4

6

(b) C102

(c) 96a k

(d) C811

(e) C130

(f) 55a k

(g) C14

(h) 125a k

(i) C12

(j) 108a k

3. show that

(a) 9 9=5 4a ak k

(b) C C=72 5

7

(c) 12 12=5 7a ak k

(d) C C=113 8

11

(e) C C=101 9

10

(f) 9 8 8= +7 6 7a a ak k k

(g) 11 10 10= +6 5 6a a ak k k

(h) C C C= +75 4 5

6 6

(i) 10 9 9= +6 5 6a a ak k k

(j) 7 6 6= +3 2 3a a ak k k

4. show that nn =k n k−a ak k.

5. evaluate x if C .C =x 277

6. if ,12 12=y3

a bk l evaluate y.

7. Find the value of a if C C= .a11

811

8. solve n 10 10= +6 5 6a a ak k k.

9. solve .CC +C = k7 720 19 19

10.1 exercises

´´

´´

´ ´


10. Factorise(a) ! ( ) !k k 1+ +(b) ! ( ) !r r 1+−(c) ( ) ! !n n1+ +(d) ( ) ! ( ) !k k1 1+ +−(e) ( ) ! ( ) !p p1 1+ − −(f) ! ( ) ! ( ) !t t t1 1+ + − −

11. simplify

(a) !

( ) !k

k 1−

(b) ( ) !( ) !kk

11+

−

(c) ( ) !( ) !nn

21

−−

(d) ( 1) !

!m

m+

(e) ( 2) !( 1) !kk

+−

12. simplify and leave your answer in factorial form.

(a) C

!6

83

(b) C

!85

9

(c) C

!54

6

(d) C

!74

9

(e) C10

!87

13. simplify

(a) 7

7

3

4

a

a

k

k

(b) C

C

5

116

11

(c) C

C12

3

412

(d) C

C9

6

97

(e) 14 148 7a ak k

(f) C

C6

63

4

(g) C

C8

5

86

14. simplify C

Cn

.nk

k

1−

15. show that .n nk

n11

1= +−−

−k ka b bk l l

Class Discussion

Combinations and Pascal’s triangle are closely related as Pascal’s triangle can be written in terms of combinations.

0C0

3C3

2C2

1C11C0

2C12C0

4C0

3C23C0

3C1

4C14C4

4C34C2

Check these values to show this is true. Can you write the next 1. 3 lines?Can you see from Pascal’s triangle that 2. ?C Cn n

k n k= −

Can you see from Pascal’s triangle that 3. ?CC +C =nk

nk

nk

11

1−−

−

¸


Investigation

expand these binomial products.1. (a) 0( )x1 +(b) 1( )x1 +(c) ( )x1 2+(d) ( )x1 3+(e) ( )x1 4+By looking at these expansions and Pascal’s triangle, could you 2. predict the expansion of ?( )x1 5+ What about ?( )x1 6+For each expansion, find the sum of the coefficients. What do you 3. notice? Could you predict the sum of coefficients of ( )x1 5+ or

?6( )x1 +

the binomial expansions and Pascal’s triangle are related.

ExamplEs

1.

expand (a) .( )x1 2+Find the sum of coefficients of the expansion.(b) how many terms are in the expansion?(c)

Solution

(a) .( )x x x1 1 22 2+ = + +sum of the coefficients:(b)

1 2 1 4

22

+ + ==

Notice that these coefficients form a row in Pascal’s triangle.there are 3 terms in the expansion.(c)

2.

expand (a) .( )x1 3+Find the sum of coefficients of the expansion.(b) how many terms are in the expansion?(c)

We can use the expansion .(a b) a 2ab b22+ + +=2

Binomial Theorem

You have already studied binomial products in the form of ( ) ( )x a y b+ + as well as special products such as perfect squares.

Expansion of (1 )x n+

The coefficient is the constant in each term.


Solution

(a) 2( ) ( ) ( )

( ) ( )

x x x

x x x

x x x x x

x x x

1 1 1

1 1 2

1 2 2

1 3 3

3

2

2 2 3

2 3

+ = + += + + += + + + + += + + +

(b) sum of the coefficients:

1 3 3 1 8

23

+ + + =

= Notice that the coefficients form a row in Pascal’s triangle.

there are 4 terms in the expansion.(c)

in general, the binomial expansion of ( )x1 n+ has coefficients that form the rows of Pascal’s triangle.

the sum of the coefficients of ( )x1 n+ is 2n

there are n 1+ terms in the binomial expansion of n( )x1 +

since Pascal’s triangle can also be written using the combinations Cnk, we can

write the binomial expansion as

xC C C x C x C x C xC xn n n(1 )x

x x x x xn n n n nk

nn0 1 2 3

n n nk

k nn

n nn

n

k n

0 1 22

33

11

2 3

f f

f f

+ = + + + + + + + +

= + + + + + + +

−−

a a a a a ak k k k k k

We can also write this binomial expansion in sigma notation.

C xn( )x

xnk

1 nk

k

k

n

k

k

n0

0

+ =

=

=

=a k

/

/

the ( )k 1 th+ term of the binomial expansion is Cnk x

k


ExamplEs

1.

expand the binomial product (a) .6( )x1 +Find the sum of its coefficients.(b) Write the binomial expansion in sigma notation.(c)

Solution

(a) C C x C x C x C x C x C x6( )x

x x x x x x

1

1 6 15 20 15 6

6 61

62

2 63

3 64

4 65

5 66

6

2 3 4 5 6

+ = + + + + + += + + + + + +

0

sum of coefficients of (b) ( )x1 n+ is 2n

so the sum of coefficients of 6( )x1 + is 26 or 64.

(c) C xn( )x1 nk

k

k

n

0+ =

=/

so C x6( )x1 kk

k

6

0

6

+ ==/

2. evaluate the coefficient of x9 in the binomial expansion of .( )x1 12+

Solution

the general term of a binomial expansion is nCkxk.

so in the expansion of ( )x1 12+ the term involving x9 is 12C9x9.

the coefficient is .C 220=129

3. how many terms are in the expansion of ?( )x1 20+

Solution

there are n 1+ terms in the expansion of .( )x1 n+ so there are 20 1+ or 21 terms in the expansion of .20( )x1 +

4. expand .x5 k

k 0

5

= ka k/

Solution

x x x x x x

x x x x x1 5 10 10 5

5 5 5 5 5 5 5k

k 0

52 3 4 5

2 3 4 5

= + + + + +

= + + + + += k 0 1 2 3 4 5a a a a a a ak k k k k k k/

You could simply add the coefficients from part (a).

This is the 10th term since

the 1st term is C x0

012.


5. Find the 4th term of the binomial product .( )x1 + 9

Solution

the ( )k 1 th+ term of ( )x1 n+ is Cnkx

k.For the 4th term:k

k

1 4

3

+ ==

so the 4th term of ( )x1 9+ is .C x x843

3 3=9

1. (i) expand each binomial product.

Find the sum of coefficients.(ii) Find the number of terms.(iii) Write each product in sigma (iv)

notation.(a) ( )x1 3+(b) ( )x1 4+(c) ( )x1 7+(d) ( )x1 6+(e) ( )x1 5+

2. Find the sum of coefficients of each (i)

binomial product in index formthe number of terms(ii) (a) ( )x1 25+(b) ( )x1 34+(c) ( )x1 17+(d) ( )x1 63+(e) 40( )x1 +

3. (i) Write as a binomial product and

write out the expansion of (ii)

(a) C xkk

k

4

0

4

=/

(b) x7 k

k 0

7

= ka k/

(c) C x3k

k

k 0

3

=/

(d) C xkk

k

6

0

6

=/

(e) x8 k

k 0

8

= ka k/

4. Find the coefficient of x4 in the expansion of(a) ( )x1 7+(b) 9( )x1 +

(c) C xkk

k

6

0

6

=/

(d) ( )x1 8+(e) ( )x1 5+

5. Find the coefficient of x3 in the expansion of(a) ( )x1 9+

(b) C x5k

k

k 0

5

=/

(c) ( )x1 11+(d) ( )x1 10+

(e) x4 k

k 0

4

= ka k/

6. Find the coefficient of x2 in the expansion of (a) 3( )x1 +

(b) C xkk

k

6

0

6

=/

(c) ( )x1 10+(d) ( )x1 7+(e) ( )x1 14+

10.2 exercises


7. Find the coefficient of x5 in the expansion of(a) ( )x1 10+(b) ( )x1 12+(c) ( )x1 8+(d) ( )x1 15+

(e) x7 k

k 0

7

= ka k/

8. Find the coefficient of the 5th term of(a) ( )x1 5+(b) ( )x1 7+

(c) x8 k

k 0

8

= ka k/

(d) 9( )x1 +(e) ( )x1 11+

9. Find the 3rd term of(a) ( )x1 8+(b) ( )x1 3+(c) ( )x1 4+(d) ( )x1 7+

(e) x6 k

k 0

6

= ka k/

10. Find the kth term of(a) ( )x1 n 1+ +

(b) ( )x1 n2+

(c) xn 1 k

k

n

0

1 −=

−

kb l/

(d) ( )x1 n2 1+ +

(e) ( )x1 n3 1+ −

Remember the general term is the (k 1)th+ term so the 5th term is .C xn

44

Expansion of ( )a x n+

Investigation

expand these binomial products.1. (a) ( )a x 0+(b) ( )a x 1+(c) ( )a x 2+(d) ( )a x 3+(e) ( )a x 4+By looking at these expansions and Pascal’s triangle, could you 2. predict the expansion of ?( )a x 5+ What about ?( )a x 6+For each expansion, find the sum of the coefficients. What do you 3. notice? Could you predict the sum of coefficients of 5( )a x+ or

?( )a x 6+

the binomial expansion of ( )a x n+ and Pascal’s triangle are related in the same way as ( )x1 n+ .

the sum of the coefficients Cnk of ( )a x 2isn n+

there are n 1+ terms in the binomial expansion of ( )a x n+


We can write the binomial expansion as

C a C a x C a x C a x

C a x C x

n n( )a x

a a x a x a x a x xn n n n nk

nn0 1 2 3

n n n n n n n

nk

n k k nn

n

n n n n n k k n

0 11

22 2

33 3

1 2 2 3 3

f

f

f f

+ = + + + ++ + +

= + + + + + + +

− − −

−

− − − −a a a a a ak k k k k k

We can also write this binomial expansion in sigma notation.

C a xn n( )a x

a xnk

kn k k

k

n

n k k

k

n0

0

+ =

=

−

=

−

=a k

/

/

Notice that this is the same as n( )a x a xnk

k n k

k

n

0+ = −

=a k/ since n

k n k= −

na ak k. Can you

see why?

the ( )k 1 th+ term of the binomial expansion is C a xnk

n k k−

ExamplEs

1. expand .( )x4 5+

Solution

C C x C x C x C x C x4 4 4 4 45( )

( ) ( ) ( ) ( ) ( ) ( )

x

x x x x x

x x x x x

4

1 1024 5 256 10 64 10 16 5 4 1

1024 1280 640 160 20

50

5 51

4 52

3 2 53

2 3 54

4 55

5

2 3 4 5

2 3 4 5

+ = + + + + += + + + + += + + + + +

2. expand .( )x3 2 4+

Solution

C( ) ( ) ( ) ( ) ( ) ( )( ) ( )C x C x C x x C3 3 2 3 2 3 2 22 34( )

( ) ( )( ) ( )( ) ( )( ) ( )

x

x x x x

x x x x

3 2

1 81 4 27 2 6 9 4 4 3 8 1 16

81 216 216 96 16

4 40

4 41

3 42

23

44

4

4 3 2

4 3 2

+ = + + + += + + + += + + + +

CONTINUED


3. expand .( ) ( )C x y2 −3k

k k

k

3

0

3−

=/

Solution

( ) ( )

( ) ( ) ( ) ( )( ) ( )

C x y

C x C x y C x y C y

2

2 2 2

−

− − −23

3

2 33

( )

( ) ( ) ( ) ( )( ) ( )

x y

x x y x y y

x x y xy y

2

1 8 3 4 3 2 1

8 12 6

kk k

k

3 3

0

33

30

31 2 3

3

3 2 2

3 2 2 3

= −

= + + += + − + − + −= − + −

−

=/

4. if ( ) a b1 3 34+ = + , evaluate a and b.

Solution

expand ( )1 3 4+

( )C( ) ( ) ( )C C C C1 1 1 1 +4 2 4( )

( ) ( ) ( )

1 3 3 3 3 3

1 4 3 6 3 4 3 3

1 4 3 6 9 4 27 81

1 4 3 6 3 4 9 3 9

1 4 3 18 4 3 3 9

28 16 3

40

4 41

3 42

2 43

1 3 44

2 3 4

+ = + + += + + + += + + + += + + + += + + + += +

so a 28= and b 16= .

5. By writing 0.999 as . ,1 0 001− evaluate 0.9995 correct to 3 decimal places.

Solution

5

( . ) ( . ) ( . )

1( 0.001) ( 0.001)

C C C C

C C

1 1 0 001 1 0 001 1 0 001+ − − −− −

2

4

3

. ( . )

( . ) ( . ) ( . ) ( . )

( . ).

0 999 1 0 001

1 5 0 001 10 0 001 10 0 001 5 0 001

0 0010 995

5 5

50

5 51

4 52

3 53

2 3

54

55

2 4

5

= −= + +

+ += + − + − + − + −

+ −=

6. Find the 3rd term of the expansion of 6( )x2 3− .

Solution

the ( )k 1 th+ term of ( )a x n+ is C a x .nk

n k k−

For the 3rd term: k

k

1 3

2

+ ==

´ ´´


so the 3rd term of 6( )x2 3− is( ) ( )C x2 3− 26

26 2−

( ) ( )

( )

x

x

x

15 2 9

135 16

2160

4

4

4

===

1. Write each product in sigma notation.(a) ( )x6 5 12+(b) ( )a b 18−(c) ( )y3 2 24+

16( )x y2−(d) ( )x y5 7 12+(e)

2. expand each binomial product.(a) ( )a x 4+(b) ( )a x 6+(c) ( )a x 5+(d) ( )a2 1 3+(e) ( )x 2 7−(f) ( )x4 32 4+(g) ( )x3 2 6−(h) ( )a b4 5 3−(i) ( )m2 3 5+(j) ( )x1 2 8−

3. (i) Write as a binomial product and

find the expansion of(ii)

(a) C a xkk k

k

3 3

0

3−

=/

(b) ( )2−( )x35 k k

k

5

0

5−

= ka k/

(c) ( )C x y2 kk

k

k

6 6

0

6−

=/

(d) ( )b5( )C a2 kk

k

k

4 4

0

4−

=/

(e) 7 x yk k

k

7

0

7 −

= ka k/

(f) ( )q3−( )C p4 k3 − kk

k

3

0

3

=/

(g) ( )n2( )C 3 kk

k

k

5 5

0

5−

=/

(h) ( )b−( )C a2 kk

k

k

6 6

0

6−

=/

(i) ( )c4−( )ab34 k k

k

4

0

4−

= ka k/

(j) ( )x−( )27 k k

k

7

0

7−

= ka k/

4. Write as a binomial product without expanding.

(a) ( )7( )C a k10k

k

k

10

0

10−

=/

(b) )y(−( )69 k k

k

9

0

9−

= ka k/

(c) ( )b4( )C a3kk k

k

7 7

0

7−

=/

(d) ( )3( )C xkk k

k

8 2 8

0

8−

=/

(e) ( )q−( )C p5kk k

k

11 11

0

11−

=/

(f) ( )9−( )xn 4 n k k

k

n

0

−

= ka k/

(g) ( )b( )C a3nk

n k k

k

n2

0

−

=/

(h) ( )2−k

( )xn 31 n k k

k

n1

0

1 + + −

=

+

b l/

(i) ( )b6( )C an 1−k

n k k

k

n2 1

0

1− −

=

−

/

(j) ( )y7( )C x8nk

n k k

k

n2 2 2

0

2−

=/

10.3 exercises


5. expand(a) ( )2 1 5+(b) ( )3 1 6−

(c) k3 −( ) ( )2 2 33k

k

0

3

−= ka k/

(d) ( )3 5 4+

(e) 5 ( ) ( )C 3 2kk

k k

0

55 −

=

−/

(f) x32

4

+b l

(g) x x1 5

+b l

(h) x12

3

−b l

(i) x x6 2k

k

k

6

0

6−

= ka bk l/

(j) a b3 2

3

−c m

6. evaluate a and b if ( ) a b2 3 23+ = + .

7. if ( ) a b2 5 54− = + , evaluate a and b.

8. if ( ) ( ) a b3 2 2 26k

k k

0

66 − = +

=

−

ka k/ ,

evaluate a and b.

9. evaluate x and y if ( ) x y3 1 5− = + .

10. if ( ) ( ) p q4 24k

k k

0

44 = +

=

−

ka k/ ,

evaluate p and q.

11. if ( ) a b2 3 2 33+ = + , evaluate a and b.

12. By writing 0.99 as . ,1 0 01− find to 4 decimal places.

0.99(a) 3

0.99(b) 4

0.99(c) 5

0.99(d) 6

0.99(e) 7

13. By writing 0.98 as . ,1 0 02− evaluate to 4 decimal places.

0.98(a) 2

0.98(b) 5

0.98(c) 3

0.98(d) 4

0.98(e) 6

14. By writing 1.01 as . ,1 0 01+ evaluate to 4 decimal places.

1.01(a) 3

1.01(b) 5

1.01(c) 4

1.01(d) 7

1.01(e) 6

15. Find the coefficient of the 3rd term in the expansion of(a) ( )x3 6+

(b) C x 2kk k

k

5 5

0

5−

=/

(c) ( )a5 7−(d) ( )n3 2 9+

(e) ( ) ( )C t2 3−kk k

k

10 10

0

10−

=/

16. Find the coefficient of the 5th term in the expansion of(a) ( )x y2 8+

(b) ( ) ( )a2 17 k k

k

7

0

7

−−

= ka k/

(c) ( ) ( )a b7 45 k k

k

5

0

5

−−

= ka k/

(d) ( )n2 3 9−

(e) ( ) ( )C x 3kk k

k

6 2 6

0

6−

=/

17. Find the coefficient of the 6th term (in coefficient and index form) in the expansion of(a) ( )a6 15+

(b) ( ) ( )C x5 4kk k

k

10

0

10

−−

=

10/(c) ( )a b2 3 14+(d) ( )x3 5 9−

(e) ( ) ( )t q20 7 5k k

k

20

0

20

−−

= ka k/

18. Find the ( 1)k th+ term in the expansion of(a) ( )x 5 8+(b) ( )a2 3 12+(c) ( )a b5 6−(d) ( )x y4 15−(e) ( )a b3 2 21−


19. Find the kth term in the expansion of (a) ( )x2 9+

(b) ( ) ( )C x y5 2kk k

k

5 5

0

5−

=/

(c) ( ) ( )d3 28 k k

k

8

0

8

−−

= ka k/

(d) 13( )m n6−(e) ( )a b3 2 20−

20. Find the ( )k 1 th− term in the expansion of

(a) ( )x3 6+

(b) C a 2kk k

k

9 9

0

9−

=/

(c) ( )a5 3 6+(d) ( )x3 4 18−

(e) ( ) ( )C x y3 2−8k

k k

k

2 8

0

8−

=/

21. Write out the expansion of ( )sin x1 5+ .

22. simplify ( ) ( )2 1 2 14 4+ + − .

ExamplEs

1.

(a) Find the general ( )k 1 th+ term of the binomial product .( )x3 22 6−(b) Find the coefficient of x4 in the expansion of .( )x3 22 6−

Solution

the (a) ( )k 1 th+ term is ( ) ( )C x3 2− .kk k6 2 6 −

(b) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

C x

C x

C x

3 2

3 2

3 2

−−−

kk k

kk k k

kk k k

6 2 6

6 6 2 6

6 6 12 2

==

−

− −

− −

For the term involving x4:x xk12 2 4=−

so k

k

k

k

12 2 4

12 2 4

8 2

4

− == +==

When 4,k= we have the 5th term of the expansion ( 1)k + .

( ) ( ) ( )C x3 2−( ) ( ) ( )C x3 2−6 4

2

− 4T ( )5

64

12 2 4

64

4 4

==

−

the coefficient is ( ) ( )C 3 2 2160− =464

2 .

CONTINUED

Further Work with Coefficients

once you are familiar with binomial products and their expansions, you can do further work with coefficients.


2. Find the constant term of .xx22

2

8

+c m

Solution

the ( 1)k th+ term is ( )xx

8 2kk

2 82

−

ka ck m

xx

xx

x x

x

2

2 1

2

2

8

8

8

8

kk

k

k kk

k k k

k k

16 22

16 22

16 2 2

16 4

=

=

=

=

−

−

− −

−

k

k

k

k

a e

a e

a

a

k o

k o

k

k

the constant term does not involve x, so we look for the term with x0:x xk16 4 0=−

so k

k

k

16 4 0

16 4

4

− ===

When 4,k= we have the 5th term of the expansion ( 1)k +

T x

x

8 2

8 2

8 2

1120

( )5

4 16 4 4

4 0

4

=

=

=

=

−

4

4

4

a

a

a

k

k

k

3. Find the coefficient of x3 in the expansion of .xx

2 52

9

−c m

Solution

( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

T C a b

C xx

C xx

C x

C x

2 5

2 5

2 5

2 5

kn

kn k k

kk

k

kk k

k

k

kk k k k

kk k k

1

9 92

9 9 92

9 9 9 2

9 9 9 3

=

= −

= −

= −= −

+−

−

− −

− − −

− −

c

e

m

o

For the term involving ,x3

( ) ( ) ( )

( ) ( )

x xk

k

k

T C x

x

x

9 3 3

6 3

2

2 5

36 128 25

115 200

( )

k9 3 3

2 19

29 2 2 9 3 2

3

3

`

`

=− =

==

= −==

−

+− −

the coefficient of x3 is 115 200.


Greatest coefficient

You might have noticed that the coefficients often increase in size and then decrease. We can find the greatest coefficient without having to expand the binomial product.

in order to do this, we need to compare the (k + 1)th and kth terms.

in the expansion of ( ) ,a b n+ T

T

kn k

ab1

k

k 1 = − ++

Proof

b

and

[ ( )]! ( )!!

( )! !!

( )! ( )!!

( )! !!

( )! !!

!( )! ( )!

( ) ( ) . . . ( ) ( ) . . . .( ) ( ) . . . ( ) ( ) . . .

T C a b

T C a b

T

T

C a b

C a b

n k kn a b

n k kn a

n k kn a

n k kn b

n k kn

nn k k

ab

n k n k k k kn k n k k k

ab

kn k

ab

1 1

1 11 1

1 3 2 1 1 2 3 2 11 3 2 1 1 2 3 2 1

1

( )

( )

( )

kn

kn k k

kn

kn k k

k

k

nk

n k k

nk

n k k

n k k

n k k

1

11 1

1

11 1

1 1

$ $ $ $ $

$ $ $ $

==

=

=

− − −

−

=

− + −

−

=−

− + −

=− − − − −

− + − − −

= − +

+−

−− − −

+

−− − −

−

− − −

−

b l

For this to work, k 0.≠

ExamplEs

1. Find the greatest coefficient in the expansion of .( )x3 10+

Solution

k10 −T C x

T C x

C x

3

3

3

( )k k

k

k kk k

kk k

110

101

10 1 1

101

11 1

===

+

−− − −

−− −

CONTINUED

´ ´

´

´


Comparing coefficients:

·

( ) ! !!

![ ( )] ! ( ) !

3(10 )(10 1) . . . 3 2 1 ( 1)( 2) . . . 2 1(10 1)(10 )(10 1) . . . 3 2 1 ( 1)( 2) . . . 2 1

T

T

C

C

C

C

CC

k kk k

k k k k kk k k k k

kk

kk

3

3

3

31

1010

3 1010 1 1

310 1

311

k

k

kk

kk

k

k

kk

1

101

11

10 10

101

10

1010

1

$ $ $ $

$ $ $ $

=

=

=

=−

− − −

=− − − − −

− + − − − − −

= − +

= −

+

−−

−

−

−

For the coefficient of

the coefficient of

i.e.

T T

T

T

kk

1

311 1

>

>

>

k k

k

k

1

1

−

+

+

,

.

k k

k

k

11 3

11 4

2 75

>>>

−

so for 1, 2,k = the coefficient of .T T>k k1+

For , , , , , ... ,k 3 4 5 6 7= the coefficient of .T T<k k1+

the term with the greatest coefficient occurs when 2k =

( )

T C x

T C x

T x

x

3

3

45 3

295 245

k kk k

110 10

2 110

210 2 2

38 2

2

====

+−

+−

so the greatest coefficient is 295 245.

2. Find the greatest coefficient in the expansion of .( )a3 2 7+

Solution

( )

( )

( )

T C a

C a

T C a

C a

C a

3 2

3 2

3 2

3 2

3 2

( )

k kk k

kk k k

k kk k

kk k

kk k k

17 7

7 7 7

71

7 1 1

71

8 1

71

8 1 8

=====

+−

− −

−− − −

−− −

−− − −

k 0.>

´

´´

\


Comparing coefficients:

C

C

C

C3 2

3 2

[ ( )] ! ( ) !!

( ) ! !!

( ) ! !( !)

![ ( )] ! ( ) !

( ) ! ![ ( )] ! ( ) !

3(7 )(7 ( 1)) . . . 3 2 1 ( 1)( 2) . . . 3 2 12(7 ( 1))(7 )(7 ( 1) . . . 3 2 1( 1)( 2)( 3) . . . 3 2 1

( ( ))

( )

T

T

k k

k k

k kk k

k kk k

k k k k kk k k k k k

kk

kk

kk

3

2

37 1 1

7

27

7

3 72 7

77 1 1

3 72 7 1 1

32 7 1

32 8

316 2

k

k

kk k

kk k

k

k

1

71

8 1

7 7

71

7

$ $ $ $ $

$ $ $ $

=

=

=

− − −

−

=−

− − −

=−

− − −

=− − + − −

− − − − + − − −

=− −

=−

= −

+

−− −

−

−

For the coefficient of T T>1k k+

the coefficient of 1T

T>1

k

k +

i.e. k

k

k k

k

k

316 2 1

16 2 3

16 5

351

>

>>

>

−

−

so for 1,k= 2 and 3, the coefficient of T T>1k k+ .For 4,k= 5, … the coefficient of T T<1k k+ .so the term with the greatest coefficient occurs when 3k= .this is the 4th term:

2

( )

( )

T C a

C a

C a

a

3 2

3

3 2

22 680

–4

73

7 3 3

73

4 3

73

4 3 4

4

====

so the greatest coefficient is 22 680.

k .0>

´


1. Find the 5th term of the

expansion of .x y2 3 27

+b l

2. Find the 4th term in the

expansion of .x x23

8

−b l

3. Find the coefficient of x2 in the

expansion of .( )xx

21841

r

rr

0

1818

=

−

ra ck m/

4. (a) Write out the expansion of

.xx3 3

5

+b l

(b) Find the greatest coefficient.

5. Find the term of x

x23

8

3−c m that is

independent of x.

6. Find the coefficient of x3 in each expansion.(a) ( )x 4 6−(b) ( )x3 8+

(c) )3C ( ) (x2 −kk

k k5

0

55

=

−/

(d) ( )x1 3 9−(e) ( )x 1 123 +

7. Find the constant term in each binomial product.

(a) xx23 62

+c m

(b) a

a10 13k

k k

0

10 10

2

3

=

−

ka c ek m o/

(c) ( )x6 2 7−

(d) x x4 12

−b l

(e) aa9 12k

k k

0

9 9 2

=

−

ka b ck l m/

8. Find the coefficient of a5 in the expansion of ( )a2 3 11− .

9. What is the term independent of

n in ( ) ?C n39

n21

kk

k

k9

0

9

2−

=c m/

10. Findthe coefficient of (a) y2 and the coefficient of (b) y5 in the

expansion of yy12

72

−f p .

11. Find the constant term of(a) ( )x4 8+(b) ( )x 6 3+

(c) x x8 2

r

rr

0

88

=

−

ra bk l/

(d) xx

3 2 9

2+c m

(e) yy12

10

3−e o

12. Find the coefficient of x3 in the expansion of

(a) C x 58

kk

k k

0

88

=

−/

(b) ( )x3 211r

r r

0

1111

=

−

ra k/

13. Find the coefficient of x in the

expansion of .x x3 7

−b l

14. Find the term independent of y

in the expansion of .yy

2 1 12

3+e o

15. the coefficients of x4 and x5 in the expansion of ( )x3 n− are equal in magnitude but opposite in sign. Find the value of n.

16. Find the values of a and b if the 5th term of ( )ax b 10+ is 13440x6 and the 8th term is -960x3.

17. in the expansion of ( ) ,ax b 6+ the coefficient of x4 is 2160 and the coefficient of x5 is 576− . Find a and b.

18. (a) simplify .C

C

k

k

81

8

−

(b) hence, or otherwise, find the greatest coefficient of the expansion of ( ) .x1 8+

10.4 exercises


19. (a) simplify C

C

k

k

61

6

−

.

(b) hence or otherwise, find the greatest coefficient in the expansion of ( )x2 6+ .

20. Find the greatest coefficient of(a) ( 3)x 9+(b) (2 1)a 10+(c) (4 3 )x 5+

(d) ( )x y28 k k

k

8

0

8−

= ka k/

(e) (3 1)n 11+(f) ( )a b2 6−(g) (3 4 )y 7+(h) ( )x y4 3 9−

(i) ( ) ( )t15 2 5k k

k

15

0

15−

= ka k/ (leave

your answer in coefficient and index form)(j) ( )n m4 7 12− (leave your answer in coefficient and index form)

21. Find the greatest coefficient of(a) ( )x 22 7+(b) ( )x2 1 5−(c) ( 3 )x y 9+

(d) x x22

8

+b l

(e) 2x x32

11

−b l

Further proofs

there is some further work you can do with coefficients and the expansion of the binomial product. some of these involve differentiation and integration.

Class Investigation

Find a pattern for the coefficient Cnk by using differentiation:

=

( )

( )

…

…

x C C x C x C x C x

x

C C C C C

C

10

1 0 0 0 0 0

1

Let

n n n n n nn

n

n n n n n nn

n

n

0 1 22

33

0 1 22

33

0`

+ = + + + + +=

+ + + + + +=

(1)

differentiating both sides of (1) gives

( )

( )

n x C C x C x n C xx

n C C C n C

n C

1 2 30

1 0 2 0 3 0 0

Let

n n n n nn

n

n n n n nn

n

n

11 2 3

2 1

11 2 3

2 1

1`

f

f

+ = + + + +=

+ = + + + +=

− −

− −

(2)

differentiating both sides of (2) gives

C C x

C C

C

0

+

+ +

( ) ( ) ( )

( ) ( ) ( )

( )( )

n n x n n C x

x

n n n n C

n n

n nC

1 1 2 3 2 10

1 1 0 2 3 2 1 0

1 2

21

Let

n n

nn

n nn

n

n

22 3

2

12 3

2

2

2

$

$

`

f

f

− + = + + −=

− + = + −− =−

=

− −

− −

n n

n n

n

Continue differentiating to find a general pattern for .Cnk


ExamplEs

1. Prove the sum of coefficients in ( )a b n+ is equal to .2n

Solution

( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

…

…

…

a b C a C a b C a b C b

a b

C C C C

C C C C

1

1 1 1 1 1 1 1 1

2

Let

n n n n n n n nn

n

n n n n n n n nn

n

n n n n nn

0 11

22 2

0 11

22 2

0 1 2`

+ = + + + += =

+ = + + + += + + + +

− −

− −

2. By comparing coefficients of x4 in both sides of (1 ) (1 ) (1 ) ,x x x8 4 4+ = + + show that

( )C Ckk

4 2 84

0

4

==/

Solution

Coefficient of x4 in ( )x1 8+ is .C84

terms of x4 in ( ) ( )x x1 14 4+ + are given byx· · · · · .C C x C x C C x C x C x C x C x C4

04

44 4

14

33 4

22 4

22 4

33 4

14

44 4

0+ + + +∴ coefficient in x4 is C C C C C C C C C C4

04

44

14

34

24

24

34

14

44

0+ + + +( ) ( ) ( ) ( ) ( )C C C C C4

02 4

12 4

22 4

32 4

42= + + + +

(since ,C C C C40

44

41

43= = )

( )Ck

k0

44 2=

=/

∴ )C C=(k

k4

0

42 8

4=/ (since coefficients of x4 are equal)

3. if ( )x xnr

1 n

r

nr

0+ =

=a k/

show that nr

3 4r

r

nn

0=

=a k/

Solution

.

.

( )

x

nr

nr

3

1 3 3

4 3

Let

Then

i.e.

n

r

nr

n r

r

n0

0

=

+ =

=

=

=

a

a

k

k

/

/


4. if n( ) ,x C x1 n

r

n

rr

0+ =

=/ show that .r C n2n

r

n

rn

1

1==

−/

Solution

( ) …x C C x C x C x C x1 n n n n n nn

n0 1 2

23

3+ = + + + + +differentiating both sides

( )

( ) ( ) ( ) ( )

n x C C x C x n C xx

n C C C n C

n C C C n C

r C

1 2 31

1 1 2 1 3 1 1

2 2 3

Let

n n n n nn

n

n n n n nn

n

n n n n nn

n

r

n

r

11 2 3

2 1

11 2 3

2 1

11 2 3

1

f

f

f

+ = + + + +=

+ = + + + += + + + +

=

− −

− −

−

=/

5. Given the expansion ( ) ,x x x xn n n nn

10 1 2

n n2 f+ = + + + +a a a ak k k k show

that … .( )

xn n x n x nn n

xnx

0 1 2 2 3 1 1

1 1n n2 3 1 1

+ + + + =+ +

+ −+ +

a a a ak k k k

Solution

integrating both sides:

( )

( )

x dx

nx

C

1

11

LHS n

n 1

1

= +

=+

++

+

#

x x x dx

x x xnx C

n n n nn

n n n nn2 3 1

0 1 2

1 2

RHS n

n

2

2 3 1

2

f

f

= + + + +

= + + + ++

++

0

a a a a

a a a a

k k k k

k k k k

#

n

n

1

1

+

+

( )

( )n

xC x x x

nx C

nx

C x x xnx

n n n nn

n n n nn

11

2 3 1

11

2 3 1

0 2

0 1 2

Son

n

1

2 3 1

2

3

2 3 1

f

f

++

+ = + + + ++

+

++

+ = + + + ++

+

+

1a a a a

a a a a

k k k k

k k k k

substitute :x 0=

( )n

Cn

nC

Cn

n n n nn1

1 00

20

30

10

11 0

11

0 1 2

n n1

3

2 3 1

3

3

f+

++ = + + + +

+

++ =

= −+

+ +

a a a ak k k k

( )

( )n

xn

x x n xnx

nx

x x xnx

n n nn

n n n nn

11

11

2 2 3 1

11 1

2 3 1

0 1

0 1 2

Son n

n n

1 2 3 1

1 2 3 1

f

f

++

−+

= + + + ++

++ −

= + + + ++

+ +

+ +

a a _ a

a a a a

k k i k

k k k k


ExamplE

Given that (1 ) ,x x x xn n n nn0 1 2

n n2 f+ = + + + +a a a ak k k k prove that n n n n

n0 1 22 2 2 3n n2 f+ + + + =a a a ak k k k .

Solution

( )x x x xn n n nn

10 1 2

n n2 f+ = + + + +a a a ak k k k

substitute :x 2=

+ +

23 + +

( ) ,n n n nn

n n n nn

1 2 2 2 20 1 2

0 1 2

n n

n n

2

2

f

f

+ = + +

2 2` = + +

a a a a

a a a a

k k k k

k k k k

1. Find the coefficient of x3 in the expansion of

(a) ( )31C xx

3kk

k

k

19 19

0

19

−−

=c m/

(b) ( ) ( )x x1 15+ +(c) ( ) ( )x x1 12 8+ +(d) ( ) ( )x x1 14 7+ +

2. By equating coefficients of ( )x1 6+ and ,( )x1 +( )x1 3 3+ show

that .26 3 3 3 3= +2 0 2 1 1a a a a ak k k k k

3. By equating coefficients of ( )x1 10+ and ( ) ( ) ,x x1 1 9+ + show

that .10 9 9= +4 4 3a a ak k k

4. By equating coefficients of ,xk use ( ) ( )x x1 1= + +( )x1 n n1+ + to prove that .C C Cn

kn

kn

k1

1= ++−

5. (a) Factorise .( )x1 6+ +( )x1 5+Find the coefficient (b)

of x3 in the expansion of .( )x1+ +( )x1 5 6+

6. Find the coefficient of x2 in

( )x xx 11 35+ +b l .

7. the coefficients of x4 and x5 are equal in the expansion of

.( )x1 n+ evaluate n.

10.5 exercises

in order to do questions like these examples, it is a good idea to look at what you are trying to prove first.

Notice that in example 4, the result to be proven included an .x –n 1

this means that we use differentiation since .( )x nxdxd n n 1= − also, when

differentiating, the constant term is zero, so the sum of terms starts from k 1= rather than .k 0=

in example 5, the result included .xn 1+ this means that we use integration

since .x dx Cnx

1n

n 1

= ++

+

# remember with integration that we need to evaluate C,

so we use a substitution for x, usually ,x 0= to evaluate it.

You can see from the examples, that we quite often use substitution of different values of x in these types of questions. sometimes this is all that is needed.


8. Given that n( ) ,x xnk

1 k

k

n

0+ =

=a k/

show that

nk

6 7k

k

nn

0=

=a k/(a)

(b) nk

2 3k

k

nn

0=

=a k/

nk

3 2k

k

nn

0

2==a k/(c)

9. if n( ) ,x xnk

1 k

k

n

0+ =

=a k/ use

differentiation to show that

.k nk

n2k

nn

1

1==

−a k/

10. Given ( ) ,a x a xnk

n n k k

k

n

0+ = −

=a k/

show that

(a) n 1=0a k

(b) n( )a an1 n k

k

n

0+ = −

= ka k/

(c) ) a1( ) (a n1 n k n k

k

n

0− = − −

= ka k/

(d) ( )n a x k a xnn

k

nn k k1

1

1+ =−

=

− −

ka k/

(e) nn =1a k

11. if n( )x1 +

…x x xn n n nn0 1 2

n2= + + + +a a a ak k k k

show that … ( )n n n n n

n2 4 8 2

0 1 2 3n− + − + + −a a a a ak k k k k

( ) .1 n= −

12. Given that ( )x1 n+

… ,x x xn n n nn0 1 2

n2= + + + +a a a ak k k k

show that

(a) …n n n nn

20 1 2

n+ + + + =a a a ak k k k

(b) … ( )n n n nn

1 00 1 2

n− + − + − =a a a ak k k k

13. Given ( ) ,x xn1 2k

nk2

0

2

+ ==

n

ka k/

(a) n 42showk

nn

0

2

== ka k/

(b) by differentiating both sides,

k nn 42show thatk

nn

1

2

== ka k/

14. if ( ) ,x xnr

1 n

r

nr

0+ =

=a k/

show by integration that

r( )

.r

nr n1

1

11

r

n

0=

+−

+=a k/

15. show that

…n n nn

nn0 2

11 3

12 1

1+ + + ++a a a ak k k k

n 1

2 1n 1

= −+

+ by integrating ( )x1 n+ .


1. Write the binomial expansion of x x3 12

+b l in sigma notation.

2. if ( ) ,x xnr

1r

nr n

0= +

=a k/ show that

.r nnr

2r

nn

1

1==

−a k/3. Find the coefficient of x5 in the

expansion of .( )x3 2 7−

4. By equating coefficients of ( )x1 8+ and ( ) ( ) ,x x1 13 5+ + show that

.83

33

50

32

51

31

52

30

53

= + + +b b b b b b b b bl l l l l l l l l

5. if n( ) ,x x x xn n n nn

10 1 2

n2 f+ = + + + +a a a ak k k k show that

(a) n 10

=a k

(b) )1 0=n(n n n n nn0 1 2 3

f− + − + −a a a a ak k k k k

6. Find the constant term of

.x x12 1

k

kk

0

1212

=

−kb al k/

7. simplify .( ) ( )3 1 3 15 5+ − −

8. use ( ) ( ) ( )x x x1 1 1n n 1+ = + + − to prove

that kn n1 1− − .n

k = + k 1−a a ak k k

9. expand .( )x y2 3 5+

10. Find the coefficient of x3 in the expansion of ( ) ( ) .x x1 14 5+ +

11. Find the term independent of x in

.( )xx 314 2 kk

0

14 14 −

kb al k/

12. Find the greatest coefficient of

.x210

r

r r

0

1010

=

−rb l/

13. if the coefficient of x3 is 4320 and the coefficient of x4 is −1620 in the expansion of ( ) ,ax b 5+ find a and b.

14. (a) simplify .C

C

k

k

101

10

−

hence, or otherwise, find the greatest (b) coefficient of .( )x3 10+

15. the coefficients of x2 and x3 in the expansion of ( )x2 n− are equal in magnitude but opposite in sign. evaluate n.

16. Find the coefficient of x4 in the expansion of .8( ) ( )x x1 1 7+ + +

17. Find the greatest coefficient in the expansion of .( )x3 2 9+

18. expand

(a) ( )pn 3 2n k

k

k

0

5−

= ka k/

(b) ( )5 2 6−

(c) 3xx2

4

−c m

(d) )C b( )a2 n k−n (kk

k

0

3

−=/

(e) ( )2 3 2 7−

19. if ( ) a b5 3 5 35− = + , evaluate a and b.

20. Given ( ) ,a x a xnk

n n k k

k

n

0+ = −

=a k/ show that

(a) ( )a ank

2 2n k n k

k

n

0+ = −

=a k/

(b) ( )n a k ank

1 –n

k

nn k1

1+ =

=

−a k/

(c) ( )

na x

na x an

k 1 1

n k k

k

n n n1

0

1 1

+=

++ −− +

=

+ +

a k/

Test Yourself 10


Challenge Exercise 10

1. By considering coefficients of x4 on both sides of ( ) ( ) ( ) ,x x x1 1 120 10 10+ = + + show that

.( ) ( )· ·C C C C C C2204

100

104

101

103

102

2= + +

2. (a) show that

.n2( ) 1 ( )x x xx1 11n nn

+ + = +a k

show that (b)

… .n n n nn

nn0 1 2

22 2 2 2

+ + + + =a a a a bk k k k l

3. By equating coefficients on both sides of ( ) ( ) ( ) ,x x x1 1 1n n n2+ − = − show

)1 0=2

( nkk

nk

0−

=a k/ if n is odd.

4. if ( )x C x3 2 kk

k15

0

15

+ ==/

show (a) ( )( )

C

C

kk

3 12 15

k

k 1 =+−+

find the greatest coefficient (b) Ck (leave your answer in index form)

5. show that k

.k n12

2 13 1k n

k

n 1 2 1

0

2

=+ +

−+ +

=

n2c m/

6. Prove that 1nr r n1

21

3

r

n r n

0

1 1

=+ +

−=

+ +a k/ by

finding the integrals

n( )x dx10

2+# and .x dxn

rr

nr

00

2=

=a k/#

7. Prove C· · …( )( )

( ) ( )…( )k k

n n n n k1 2 3 1

1 2 1n

k =−

− − − +

using mathematical induction.

the binomial theoremweb2.hunterspt-h.schools.nsw.edu.au/studentshared...binomial product: the...

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