12x1 t08 03 binomial theorem

Calculating Coefficients without Pascal’s Triangle

Calculating Coefficients without Pascal’s Triangle

!!!

knknCk

n

Calculating Coefficients without Pascal’s Triangle

Proof

!!!

knknCk

n

Calculating Coefficients without Pascal’s Triangle

Proof

!!!

knknCk

n

Step 1: Prove true for k = 0

Calculating Coefficients without Pascal’s Triangle

10

CLHS n

Proof

!!!

knknCk

n

Step 1: Prove true for k = 0

Calculating Coefficients without Pascal’s Triangle

10

CLHS n

Proof

1!!

!!0!

nn

nnRHS

!!!

knknCk

n

Step 1: Prove true for k = 0

Calculating Coefficients without Pascal’s Triangle

10

CLHS n

RHSLHS

Proof

1!!

!!0!

nn

nnRHS

!!!

knknCk

n

Step 1: Prove true for k = 0

Calculating Coefficients without Pascal’s Triangle

10

CLHS n

RHSLHS

Proof

1!!

!!0!

nn

nnRHS

!!!

knknCk

n

Step 1: Prove true for k = 0

Hence the result is true for k = 0

0integer an is wherefor trueisresult theAssume:2 Step rrk

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

!1!1!.. 1

rnrnCei r

n

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

!1!1!.. 1

rnrnCei r

n

Proof

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

!1!1!.. 1

rnrnCei r

n

Proof11

1

rn

rn

rn CCC

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

!1!1!.. 1

rnrnCei r

n

Proof11

1

rn

rn

rn CCC

rn

rn

rn CCC

1

11

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

!1!1!.. 1

rnrnCei r

n

Proof11

1

rn

rn

rn CCC

rn

rn

rn CCC

1

11

!!

!!!1

!1rnr

nrnr

n

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

!1!1!.. 1

rnrnCei r

n

Proof11

1

rn

rn

rn CCC

rn

rn

rn CCC

1

11

!!

!!!1

!1rnr

nrnr

n

!!1

1!!1rnr

rnn

0integer an is wherefor trueisresult theAssume:2 Step rrk

!!!..

rnrnCei r

n

1for trueisresult theProve:3 Step rk

!1!1!.. 1

rnrnCei r

n

Proof11

1

rn

rn

rn CCC

rn

rn

rn CCC

1

11

!!

!!!1

!1rnr

nrnr

n

!!1

1!!1rnr

rnn

!!1

11!rnr

rnn

!!1

11!rnr

rnn

!!1

11!rnr

rnn

!!1

!rnr

rnn

!!1

11!rnr

rnn

!!1

!rnr

rnn

!1!1!

rnrn

!!1

11!rnr

rnn

!!1

!rnr

rnn

!1!1!

rnrn

Hence the result is true for k = r + 1 if it is also true for k = r

!!1

11!rnr

rnn

!!1

!rnr

rnn

!1!1!

rnrn

Hence the result is true for k = r + 1 if it is also true for k = r

Step 4: Hence the result is true for all positive integral values of n by induction

The Binomial Theorem

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

This extends to;

n

k

kknk

nn baCba0

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

This extends to;

n

k

kknk

nn baCba0

411Evaluate .. Cge

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

This extends to;

n

k

kknk

nn baCba0

411Evaluate .. Cge

!7!4!11

411 C

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

This extends to;

n

k

kknk

nn baCba0

411Evaluate .. Cge

!7!4!11

411 C

1234891011

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

This extends to;

n

k

kknk

nn baCba0

411Evaluate .. Cge

!7!4!11

411 C

1234891011

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

This extends to;

n

k

kknk

nn baCba0

411Evaluate .. Cge

!7!4!11

411 C

1234891011

3

The Binomial Theorem n

nnk

knnnnn xCxCxCxCCx 2

2101

n

k

kk

n xC0

integer positive a is and

!!!where n

knknCk

n

NOTE: there are (n + 1) terms

This extends to;

n

k

kknk

nn baCba0

411Evaluate .. Cge

!7!4!11

411 C

1234891011

330

3

(ii) Find the value of n so that;

(ii) Find the value of n so that;

85 a) CC nn

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

1135 ofexpansion in the 5th term theFind

baiii

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

1135 ofexpansion in the 5th term theFind

baiii

k

kkk b

aCT

35 1111

1

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

1135 ofexpansion in the 5th term theFind

baiii

k

kkk b

aCT

35 1111

1

4

74

115

35

baCT

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

1135 ofexpansion in the 5th term theFind

baiii

k

kkk b

aCT

35 1111

1

4

74

115

35

baCT

4

7474

11 35b

aC

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

1135 ofexpansion in the 5th term theFind

baiii

k

kkk b

aCT

35 1111

1

4

74

115

35

baCT

4

7474

11 35b

aC unsimplified

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

1135 ofexpansion in the 5th term theFind

baiii

k

kkk b

aCT

35 1111

1

4

74

115

35

baCT

4

7474

11 35b

aC unsimplified

4

78178125330b

a

(ii) Find the value of n so that;

85 a) CC nn

knn

kn CC

1358

nn

820

87 b) CCC nn

kn

kn

kn CCC

1

11

19n

1135 ofexpansion in the 5th term theFind

baiii

k

kkk b

aCT

35 1111

1

4

74

115

35

baCT

4

7474

11 35b

aC unsimplified

4

78178125330b

a

4

72088281250b

a

9

2

213in oft independen termObtain the

xxxiv

9

2

213in oft independen termObtain the

xxxiv

k

kkk x

xCT

213 929

1

9

2

213in oft independen termObtain the

xxxiv

k

kkk x

xCT

213 929

1

0 with termmeans oft independen term xx

9

2

213in oft independen termObtain the

xxxiv

k

kkk x

xCT

213 929

1

0 with termmeans oft independen term xx

0192 xxx kk

9

2

213in oft independen termObtain the

xxxiv

k

kkk x

xCT

213 929

1

0 with termmeans oft independen term xx

0192 xxx kk

60318

0218

kk

xxx kk

9

2

213in oft independen termObtain the

xxxiv

k

kkk x

xCT

213 929

1

0 with termmeans oft independen term xx

0192 xxx kk

60318

0218

kk

xxx kk

6

326

97 2

13

x xCT

9

2

213in oft independen termObtain the

xxxiv

k

kkk x

xCT

213 929

1

0 with termmeans oft independen term xx

0192 xxx kk

60318

0218

kk

xxx kk

6

326

97 2

13

x xCT

6

36

9

23C

9

2

213in oft independen termObtain the

xxxiv

k

kkk x

xCT

213 929

1

0 with termmeans oft independen term xx

0192 xxx kk

60318

0218

kk

xxx kk

6

326

97 2

13

x xCT

6

36

9

23C

Exercise 5D; 2, 3, 5, 7, 9, 10ac, 12ac, 13,

14, 15ac, 19, 25