THE BINOMIAL THEOREM Robert Yen Hurlstone Agricultural High School This presentation is accompanied by workshop notes

THE BINOMIAL THEOREMTHE BINOMIAL THEOREM

Robert YenRobert YenHurlstone Agricultural High SchoolHurlstone Agricultural High School

This presentation is accompanied by workshop notesThis presentation is accompanied by workshop notes

INTRODUCTIONINTRODUCTION

Expanding (a + x)n

Difficult topic: high-level algebra

Targeted at better Extension 1 students

Master this topic to get ahead in the HSC exam

No shortcuts for this topic

BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLEBINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE

Coefficients

(a + x)1 = a + x 1 1

(a + x)2 = a2 + 2ax + x2 1 2 1

(a + x)3 = a3 + 3a2x + 3ax2 + x3 1 3 3 1

(a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x4 1 4 6 4 1

(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5 1 5 10 10 5 1

nnCCkk, A FORMULA FOR PASCAL’S TRIANGLE, A FORMULA FOR PASCAL’S TRIANGLE

1 0C0

1 1 1C0 1C1

1 2 1 2C0 2C1

2C2

1 3 3 1 3C0 3C1

3C2 3C3

1 4 6 4 1 4C0 4C1

4C2 4C3

4C4

1 5 10 10 5 1 5C0 5C1

5C2 5C3

5C4 5C5

1 6 15 20 15 6 1 6C0 6C1

6C2 6C3

6C4 6C5

6C6

1 7 21 35 35 21 7 1 7C0 7C1

7C2 7C3

7C4 7C5

7C6

7C7

1 8 28 56 70 56 28 8 1 8C0 8C1

8C2 8C3

8C4 8C5

8C6 8C7

8C8

nCk gives the value of row n, term k,

if we start numbering the rows and terms from 0

nnCCkk, A FORMULA FOR PASCAL’S TRIANGLE, A FORMULA FOR PASCAL’S TRIANGLE

n

kCk

n

CALCULATINGCALCULATING

MentallyMentally

102

20

123

3453

5

C

35C

1026

120

!2!3

!53

5

C


MentallyMentally

FormulaFormula

35C

102

20

123

3453

5

C

!!

!

knk

nCk

n

1026

120

!2!3

!53

5

C


MentallyMentally

FormulaFormula

because ...because ...

102

20

123

3453

5

C

!2!3

!5

12

12

123

345

123

3453

5

C

!!

!

knk

nCk

n

35C

THE BINOMIAL THEOREMTHE BINOMIAL THEOREM

(a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + nC3 an-3 x3

+ nC4 an-4 x4 + ... + nCn xn

=

kknn

kk

n xaC

0

Don’t worry too much about writing in notation: just have a good idea of the general term

The sum of terms from k = 0 to n

PROPERTIES OF nCk

1. nC0 = nCn = 1 1st and last

2. nC1 = nCn-1 = n 2nd and 2nd-last

3. nCk = nCn-k Symmetry

4. n+1Ck = nCk-1 + nCk Pascal’s triangle result: each

coefficient is the sum of the two

coefficients in the row above it

n+1Ck = nCk-1 + nCk Pascal’s triangle result

1 0C0

1 1 1C0 1C1

1 2 1 2C0 2C1

2C2

1 3 3 1 3C0 3C1

3C2 3C3

1 4 6 4 1 4C0 4C1

4C2 4C3

4C4

1 5 10 10 5 1 5C0 5C1

5C2 5C3

5C4 5C5

1 6 15 20 15 6 1 6C0 6C1

6C2 6C3

6C4 6C5

6C6

1 7 21 35 35 21 7 1 7C0 7C1

7C2 7C3

7C4 7C5

7C6 7C7

1 8 28 56 70 56 28 8 1 8C0 8C1

8C2 8C3

8C4 8C5

8C6 8C7

8C8

15 = 10 + 56C4 = 5C3 + 5C4

Example 1Example 1

(a) (a + 3)5 =

Example 1Example 1

(a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34

+ 5C5 35

= a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243

= a5 + 15a4 + 90a3 + 270a2 + 405a + 243

(b) (2x – y)4 =

Example 1Example 1

(a) (a + 3)5 = 5C0 a5 + 5C1 a4 31 + 5C2 a3 32 + 5C3 a2 33 + 5C4 a1 34

+ 5C5 35

= a5 + 5a4.3 + 10a3.9 + 10a2.27 + 5a.81 + 243

= a5 + 15a4 + 90a3 + 270a2 + 405a + 243

(b) (2x – y)4 = 4C0 (2x)4 + 4C1 (2x)3(-y)1 + 4C2 (2x)2(-y)2

+ 4C3 (2x)1(-y)3 + 4C4 (-y)4

= 16x4 + 4(8x3)(-y) + 6(4x2)(y2) + 4(2x)(-y3) + y4

= 16x4 – 32x3y + 24x2y2 – 8xy3 + y4

Example 2 (2008 HSC, Question 1(d), 2 marks)Example 2 (2008 HSC, Question 1(d), 2 marks)

(2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ...

General term Tk = 12Ck (2x)12-k(3y)k = 1 mark

For coefficient of x8y4, substitute k = ?

Example 2 (2008 HSC, Question 1(d), 2 marks)Example 2 (2008 HSC, Question 1(d), 2 marks)

(2x + 3y)12 = 12C0 (2x)12 + 12C1 (2x)11(3y)1 + 12C2 (2x)10(3y)2 + ...

General term Tk = 12Ck (2x)12-k(3y)k = 1 mark

For coefficient of x8y4, substitute k = 4:

T4 = 12C4 (2x)8(3y)4

= 12C4 (28)(34) x8y4

Coefficient is 12C4 (28)(34) or 10 264 320.

It’s OK to leave the coefficient unevaluated, especially if the

question asks for ‘an expression’.

TTkk is not the is not the kkthth term term

Tk is the term that contains xk

Simpler to write out the first few terms rather than memorise the notation

Better to avoid referring to ‘the kth term’: too confusing

HSC questions ask for ‘the term that contains x8’ rather than ‘the 9th term’

Example 3 (2005 HSC, Question 2(b), 3 marks)Example 3 (2005 HSC, Question 2(b), 3 marks)

General term Tk = 12Ck (2x)12-k

= 12Ck 212-k x12-k (-x-2)k

= 12Ck 212-k x12-k (-1)k x-2k

= 12Ck 212-k x12-3k (-1)k

For term independent of x:

...1

21

221

22

210

212

1

211

11212

012

12

2

xxC

xxCxC

xx

k

x

2

1


General term Tk = 12Ck (2x)12-k

= 12Ck 212-k x12-k (-x-2)k

= 12Ck 212-k x12-k (-1)k x-2k

= 12Ck 212-k x12-3k (-1)k

For term independent of x:12 – 3k = 0 k = 4

T4 = 12C4 28 x0 (-1)4

= 126 720

...1

21

221

22

210

212

1

211

11212

012

12

2

xxC

xxCxC

xx

k

x

2

1

FINDING THE GREATEST COEFFICIENT

(1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4

+ 1792x5 + 1792x6 + 1024x7 + 256x8

Example 4Example 4

(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...

General term Tk = 8Ck 18-k (2x)k

= 8Ck 1 (2k) xk

= 8Ck 2k xk

tk = 8Ck 2k Leave out xk as we are only interested in the coefficient

Example 4Example 4

(a) (1 + 2x)8 = 8C0 18 + 8C1 17 (2x)1 + 8C2 16 (2x)2 + 8C3 15 (2x)3 +...

General term Tk = 8Ck 18-k (2x)k

= 8Ck 1 (2k) xk

= 8Ck 2k xk

tk = 8Ck 2k

(b)

Leave out xk as we are only interested in the coefficient

kk

kk

k

k

C

C

t

t

2

28

11

81

Example 4Example 4

(b)

Ratio of consecutive factorials

kk

kk

k

k

C

C

t

t

2

28

11

81

!

!1

n

n

Example 4Example 4

(b)

Ratio of consecutive factorials

kk

kk

k

k

C

C

t

t

2

28

11

81

1

82

2.1

8.

1

1

2.!7

!8.

!1

!

2!8!

!8

!18!1

!8 1

k

k

k

k

k

k

k

k

kkkk

81...567

1...5678

7!

8! eg

1!

!1

nn

n

Example 4Example 4

(c) For the greatest coefficient tk+1, we want:

tk+1 > tk

16 – 2k > k + 1

-3k > -15

k < 5

k = 4 k must be a whole number

1

1

82

11

k

k

t

t

k

k

for the largest possible integer

value of k

Example 4Example 4

(c) Greatest coefficient tk+1 = t5

= 8C5 25

= 56 32

= 1792

tk = 8Ck 2k

THE BINOMIAL THEOREM FOR (1 + THE BINOMIAL THEOREM FOR (1 + xx))nn

(1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 +

... + nCn xn

=

kn

kk

n xC0

Example 5Example 5

(1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn

Sub x = ?

[Aiming to prove: [Aiming to prove: ] ]n

k

n

k

nC 20

Example 5Example 5


Sub x = 1:

(1 + 1)n = nC0 + nC1 (1) + nC2 (12) + ... + nCn (1n)

2n = nC0 + nC1 + nC2 + ... + nCn

This will make the x’s disappear and make the LHS become 2n

k

n

k

nn C

0

2

Example 6Example 6

(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n

Term with xn = 2nCn xn

Coefficient of xn = 2nCn

[Aiming to prove ][Aiming to prove ] n

nn

kk

n CC 2

0

2

Example 6Example 6

(1 + x)2n = 2nC0 + 2nC1 x + 2nC2 x2 + ... + 2nC2n x2n

Term with xn = 2nCn xn

Coefficient of xn = 2nCn


(1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCn xn)

(nC0 + nC1 x + nC2 x2 + ... + nCn xn)

If we expanded the RHS, there would be many terms

[Aiming to prove ][Aiming to prove ]

nn

n

kk

n CC 2

0

2

Example 6Example 6

(1 + x)n.(1 + x)n = (nC0 + nC1 x + nC2 x2 + ... + nCn xn)

(nC0 + nC1 x + nC2 x2 + ... + nCn xn)

Terms with xn

= nC0(nCn xn) + nC1 x (nCn-1 xn-1) + nC2 x2 (nCn-2 xn-2) + ...

+ nCn xn (nC0)

Coefficient of xn

= nC0 nCn + nC1

nCn-1 + nC2 nCn-2 + ... + nCn

nC0

= (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2

by symmetry of Pascal’s triangle

[Aiming to prove ][Aiming to prove ]

nCk = nCn-k

nn

n

kk

n CC 2

0

2

Example 6Example 6

By equating coefficients of xn on both sides of

(1 + x)2n = (1 + x)n.(1 + x)n

2nCn = (nC0)2 + (nC1)2 + (nC2)2 + ... + (nCn)2

n

kk

nn

n CC0

22


(i) (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn

Differentiating both sides:

n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1

= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1

(ii) Substitute x = ? to prove result:

[Aiming to prove: n 3n-1 = nC1 + ... + r nCr 2r-1 + ... + n nCn 2n-1 ]

The general term




n(1 + x)n-1 = 0 + nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1

= nC1 + 2 nC2 x + ... + r nCr xr-1 + ... + n nCn xn-1

(ii) Substitute x = 2 to prove result:

n(1 + 2)n-1 = nC1 + 2 nC2 2 + ... + r nCr 2r-1 + ... + n nCn 2n-1

n 3n-1 = nC1 + 4 nC2 + ... + r nCr 2r-1 + ... + n nCn 2n-1

Example 8 (2008 HSC, Question 6(c), 5 marks)Example 8 (2008 HSC, Question 6(c), 5 marks)

(i) (1 + x)p+q = p+qC0 + p+qC1 x + p+qC2 x2 + ... + p+qCp+q xp+q

The term of independent of x will be

q

qp

x

x 1




qqp

q

qq

qp

Cx

xC

is,that ,

q

qp

x

x 1




(ii) (1 + x)p = pC0 + pC1 x + pC2 x2 + ... + pCp xp

qqp

q

qq

qp

Cx

xC

is,that ,

q

qp

x

x 1


(ii)


Terms independent of x

=


(ii)



= pC0 qC0 + pC1 x qC1 + pC2 x2 qC2 + ... + pCp xp qCp

= 1 + pC1 qC1 + pC2 qC2 + ... + pCp

qCp

px

1

x

1

2

1

x

p ≤ q


(ii)


= pC0 qC0 + pC1 x qC1 + pC2 x2 qC2 + ... + pCp xp qCp

= 1 + pC1 qC1 + pC2 qC2 + ... + pCp

qCp

By equating the terms independent of x on both sides of

p+qCq = 1 + pC1 qC1 + pC2 qC2 + ... + pCp

qCp

1 + pC1 qC1 + pC2 qC2 + ... + pCp

qCp = p+qCq

px

1

x

1

2

1

x

q

pq

qp

xx

x

x

111

1

From (i)

And now ...And now ...

A fairly hard identity to prove:A fairly hard identity to prove:

Example 9 from 2002 HSCExample 9 from 2002 HSC

Question 7(b), 6 marksQuestion 7(b), 6 marks



= c0 + c1 x + c2 x2 + ... + cn xn [1]

Identity to be proved involves (n + 2) 2n-1 so try ...

[Aiming to prove: c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1 ]

To save time, this question asks us to abbreviate nCk to ck



= c0 + c1 x + c2 x2 + ... + cn xn [1]



n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1

Substituting x = ? to give 2n-1 on the LHS:




= c0 + c1 x + c2 x2 + ... + cn xn [1]



n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1

Substituting x = 1 to give 2n-1 on the LHS:

n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1

n 2n-1 = c1 + 2c2 + ... + ncn [2]

How do we make the LHS say (n + 2) 2n-1?




= c0 + c1 x + c2 x2 + ... + cn xn [1]Differentiating both sides:

n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1 Substituting x = 1:

n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1

n 2n-1 = c1 + 2c2 + ... + ncn [2]

Add 2 (2n-1) to both sides. But that’s 2n.




= c0 + c1 x + c2 x2 + ... + cn xn [1]


n(1 + x)n-1 = c1 + 2c2 x + ... + ncn xn-1

Substituting x = 1:

n(1 + 1)n-1 = c1 + 2c2 1 + ... + ncn 1n-1

n 2n-1 = c1 + 2c2 + ... + ncn [2]

To prove the 2n identity, sub x = 1 into [1] above:

(1 + 1)n = c0 + c1 1 + c2 12 + ... + cn 1n

2n = c0 + c1 + c2 + ... + cn [3]



(i) n 2n-1 = c1 + 2c2 + ... + ncn [2]

2n = c0 + c1 + c2 + ... + cn [3]

[2] + [3]:

n 2n-1 + 2n = c1 + 2c2 + ... + ncn + c0 + c1 + c2 + ... + cn

2n-1 (n + 2) = c0 + 2c1 + 3c2 + ... + (n + 1)cn

c0 + 2c1 + 3c2 + ... + (n + 1)cn = (n + 2) 2n-1


Each coefficient ck increases by 1 as

required


(ii) Answer involves dividing by (k + 1)(k + 2) and alternating

–/+ pattern so try integrating (1 + x)n from (i) twice and substituting x = -1.

(1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]

[Aiming to find ] 211

4.33.22.1210

nn

cccc nn


(ii) (1 + x)n = c0 + c1 x + c2 x2 + ... + cn xn [1]

Integrate both sides:

To find k, sub x = 0:

[Aiming to find ]

kxn

cx

cx

cxcx

nnnn

13221

01

1321

1

1

Don’t forget the constant of integration

1

1

1321

1

11

1

000011

1

132210

1

1

nx

n

cx

cx

cxcx

n

nk

kn

nnn

n

211

4.33.22.1210

nn

cccc nn


(ii)

Integrate again to get 1.2, 2.3, 3.4 denominators:

To find d, sub x = 0:

[Aiming to find ]

dn

x

nn

xcxcxcxcx

nn

nnn

1214.33.221

21

1 142

31

202

211

4.33.22.1210

nn

cccc nn


(ii)

Sub x = ? for –/+ pattern:


4.33.22.1210

nn

cccc nn


(ii)

Sub x = -1 for –/+ pattern:


4.33.22.1210

nn

cccc nn


(ii)


4.33.22.1210

nn

cccc nn

2

1

21

1

21

12

21

1

1

1

21)1(

4.33.22.1

21

1

1

1

21

1)1(

4.33.22.10

210

2210

n

nn

n

nn

n

nnnnn

cccc

nnnnn

cccc

nn

nn

Example 10 (2007 HSC, Question 4(a), 6 marks)Example 10 (2007 HSC, Question 4(a), 6 marks)

(i) P(both green) = 0.1 0.1

= 0.01

(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20

P(X = 2) =


(i) P(both green) = 0.1 0.1

= 0.01

(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20

P(X = 2) = 20C2 0.12 0.918

= 0.28517 ...

0.285


(i) P(both green) = 0.1 0.1 = 0.01

(ii) p = 0.1, q = 1 – 0.1 = 0.9, n = 20

P(X = 2) = 20C2 0.12 0.918

= 0.28517 ... 0.285

(iii) P(X > 2) = 1 – P(X = 0) – P(X = 1) – P(X = 2)

= 1 – 20C0 0.10 0.920 – 20C1 0.110.919 – 0.285 from (ii)

= 1 – 0.920 – 20(0.1)0.919 – 0.285 = 0.32325 ... 0.32

HOW TO STUDY FOR MATHS (P-R-A-C)

1. Practise your maths

2. Rewrite your maths

3. Attack your maths

4. Check your maths

WORK HARD AND BEST OF LUCK

FOR YOUR HSC EXAMS!

Documents

THE BINOMIAL THEOREM Robert Yen Hurlstone Agricultural High School This presentation is accompanied by workshop notes