11x1 t14 11 some different types (2011)

nnnn

βαi

tan1tancot1tantan1

that show totantan1

tantantanfact that the Use

2006 Extension 1 HSC Q4d)

Some Different Looking Induction

Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1

tantantanfact that the Use

nn

nnnn

tan1tan1

tan1tan1tan

2006 Extension 1 HSC Q4d)

Some Different Looking Induction

Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1

tantantanfact that the Use

nn

nnnn

tan1tan1

tan1tan1tan

2006 Extension 1 HSC Q4d)

nn

nn

tan1tan1

tan1tantan

Some Different Looking Induction

Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1

tantantanfact that the Use

nn

nnnn

tan1tan1

tan1tan1tan

2006 Extension 1 HSC Q4d)

nn

nn

tan1tan1

tan1tantan

nnnn tan1tantantan1tan1

Some Different Looking Induction

Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1

tantantanfact that the Use

nn

nnnn

tan1tan1

tan1tan1tan

2006 Extension 1 HSC Q4d)

nn

nn

tan1tan1

tan1tantan

nnnn tan1tantantan1tan1

tan

tan1tantan1tan1

nnnn

Some Different Looking Induction

Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1

tantantanfact that the Use

nn

nnnn

tan1tan1

tan1tan1tan

2006 Extension 1 HSC Q4d)

nn

nn

tan1tan1

tan1tantan

nnnn tan1tantantan1tan1

tan

tan1tantan1tan1

nnnn

nnnn tan1tancottan1tan1

Some Different Looking Induction

Problems

2006 Extension 1 Examiners Comments;

(d) (i) This part was not done well, with many candidates

presenting circular arguments.

2006 Extension 1 Examiners Comments;

(d) (i) This part was not done well, with many candidates

presenting circular arguments.

However, a number of candidates successfully rearranged

the given fact and easily saw the substitution required.

2006 Extension 1 Examiners Comments;

(d) (i) This part was not done well, with many candidates

presenting circular arguments.

However, a number of candidates successfully rearranged

the given fact and easily saw the substitution required.

Many candidates spent considerable time on this part,

which was worth only one mark.

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tancot2RHS

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS 2tancot2RHS

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS 2tancot2RHS

12tantan1

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS 2tancot2RHS

12tantan1

1tan2tancot

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS 2tancot2RHS

12tantan1

1tan2tancot

112tancot

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS 2tancot2RHS

12tantan1

1tan2tancot

112tancot

22tancot

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS

Hence the result is true for n = 1

2tancot2RHS

12tantan1

1tan2tancot

112tancot

22tancot

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS

Hence the result is true for n = 1

Step 2: Assume the result is true for n = k, where k is a positive

integer

2tancot2RHS

12tantan1

1tan2tancot

112tancot

22tancot

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS

Hence the result is true for n = 1

Step 2: Assume the result is true for n = k, where k is a positive

integer

1tancot1

1tantan3tan2tan2tantan ..

kk

kkei

2tancot2RHS

12tantan1

1tan2tancot

112tancot

22tancot

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS

Hence the result is true for n = 1

Step 2: Assume the result is true for n = k, where k is a positive

integer

1tancot1

1tantan3tan2tan2tantan ..

kk

kkei

Step 3: Prove the result is true for n = k + 1

2tancot2RHS

12tantan1

1tan2tancot

112tancot

22tancot

1tancot1

1tantan3tan2tan2tantan

1 integers allfor that prove toinduction almathematic Use

nn

nn

nii

Step 1: Prove the result is true for n = 1

2tantanLHS

Hence the result is true for n = 1

Step 2: Assume the result is true for n = k, where k is a positive

integer

1tancot1

1tantan3tan2tan2tantan ..

kk

kkei

Step 3: Prove the result is true for n = k + 1

2tancot2RHS

12tantan1

1tan2tancot

112tancot

22tancot

2tancot2

2tan1tan3tan2tan2tantan ..

kk

kkei

Proof:

2tan1tan1tantan3tan2tan2tantan

2tan1tan3tan2tan2tantan

kkkk

kk

Proof:

2tan1tan1tancot1 kkkk

2tan1tan1tantan3tan2tan2tantan

2tan1tan3tan2tan2tantan

kkkk

kk

Proof:

2tan1tan1tancot1 kkkk

2tan1tan1tantan3tan2tan2tantan

2tan1tan3tan2tan2tantan

kkkk

kk

11tan2tancot1tancot1 kkkk

Proof:

2tan1tan1tancot1 kkkk

2tan1tan1tantan3tan2tan2tantan

2tan1tan3tan2tan2tantan

kkkk

kk

11tan2tancot1tancot1 kkkk

1tancot2tancot1tancot2 kkkk

Proof:

2tan1tan1tancot1 kkkk

2tan1tan1tantan3tan2tan2tantan

2tan1tan3tan2tan2tantan

kkkk

kk

11tan2tancot1tancot1 kkkk

1tancot2tancot1tancot2 kkkk

2tancot2 kk

Proof:

2tan1tan1tancot1 kkkk

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

2tan1tan1tantan3tan2tan2tantan

2tan1tan3tan2tan2tantan

kkkk

kk

11tan2tancot1tancot1 kkkk

1tancot2tancot1tancot2 kkkk

2tancot2 kk

Proof:

2tan1tan1tancot1 kkkk

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

Step 4: Since the result is true for n= 1, then the result is true for all

positive integral values of n, by induction

2tan1tan1tantan3tan2tan2tantan

2tan1tan3tan2tan2tantan

kkkk

kk

11tan2tancot1tancot1 kkkk

1tancot2tancot1tancot2 kkkk

2tancot2 kk

2006 Extension 1 Examiners Comments;

(ii) Most candidates attempted this part. However, many

only earned the mark for stating the inductive step.

2006 Extension 1 Examiners Comments;

(ii) Most candidates attempted this part. However, many

only earned the mark for stating the inductive step.

They could not prove the expression true for n=1 as they

were not able to use part (i), and consequently not able to

prove the inductive step.

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

2004 Extension 1 HSC Q4a)

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n = 3

2004 Extension 1 HSC Q4a)

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n = 3

3

1

3

21

LHS

2004 Extension 1 HSC Q4a)

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n = 3

3

1

3

21

LHS

3

1

23

2

RHS

2004 Extension 1 HSC Q4a)

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n = 3

3

1

3

21

LHS

3

1

23

2

RHS

RHSLHS

2004 Extension 1 HSC Q4a)

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n = 3

3

1

3

21

LHS

3

1

23

2

RHS

RHSLHS

Hence the result is true for n = 3

2004 Extension 1 HSC Q4a)

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n = 3

3

1

3

21

LHS

3

1

23

2

RHS

RHSLHS

Hence the result is true for n = 3

Step 2: Assume the result is true for n = k, where k is a positive

integer

2004 Extension 1 HSC Q4a)

1

221

5

21

4

21

3

2-1

;3 integers allfor that prove toinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n = 3

3

1

3

21

LHS

3

1

23

2

RHS

RHSLHS

Hence the result is true for n = 3

Step 2: Assume the result is true for n = k, where k is a positive

integer

2004 Extension 1 HSC Q4a)

122

15

21

4

21

3

2-1 ..

kkkei

Step 3: Prove the result is true for n = k + 1

Step 3: Prove the result is true for n = k + 1

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

Proof:

Step 3: Prove the result is true for n = k + 1

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

1

21

21

5

21

4

21

3

2-1

1

21

5

21

4

21

3

2-1

kk

k

Proof:

Step 3: Prove the result is true for n = k + 1

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

1

21

21

5

21

4

21

3

2-1

1

21

5

21

4

21

3

2-1

kk

k

1

21

1

2

kkk

Proof:

Step 3: Prove the result is true for n = k + 1

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

1

21

21

5

21

4

21

3

2-1

1

21

5

21

4

21

3

2-1

kk

k

1

21

1

2

kkk

1

21

1

2

k

k

kk

Proof:

Step 3: Prove the result is true for n = k + 1

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

1

21

21

5

21

4

21

3

2-1

1

21

5

21

4

21

3

2-1

kk

k

1

21

1

2

kkk

1

21

1

2

k

k

kk

1

1

1

2

k

k

kk

Proof:

Step 3: Prove the result is true for n = k + 1

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

1

21

21

5

21

4

21

3

2-1

1

21

5

21

4

21

3

2-1

kk

k

1

21

1

2

kkk

1

21

1

2

k

k

kk

1

1

1

2

k

k

kk

1

2

kk

Proof:

Step 3: Prove the result is true for n = k + 1

Step 4: Since the result is true for n = 3, then the result is true for

all positive integral values of n > 2 by induction.

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

1

21

21

5

21

4

21

3

2-1

1

21

5

21

4

21

3

2-1

kk

k

1

21

1

2

kkk

1

21

1

2

k

k

kk

1

1

1

2

k

k

kk

1

2

kk

Proof:

Hence the result is true for

n = k + 1 if it is also true

for n = k

Step 3: Prove the result is true for n = k + 1

Step 4: Since the result is true for n = 3, then the result is true for

all positive integral values of n > 2 by induction.

kkkei

1

2

1

21

5

21

4

21

3

2-1 Prove ..

1

21

21

5

21

4

21

3

2-1

1

21

5

21

4

21

3

2-1

kk

k

1

21

1

2

kkk

1

21

1

2

k

k

kk

1

1

1

2

k

k

kk

1

2

kk

2004 Extension 1 Examiners Comments;

(a) Many candidates submitted excellent responses

including all the necessary components of an

induction proof.

2004 Extension 1 Examiners Comments;

(a) Many candidates submitted excellent responses

including all the necessary components of an

induction proof.

Weaker responses were those that did not verify for n = 3,

had an incorrect statement of the assumption for n = k or

had an incorrect use of the assumption in the attempt to

prove the statement.

2004 Extension 1 Examiners Comments;

(a) Many candidates submitted excellent responses

including all the necessary components of an

induction proof.

Weaker responses were those that did not verify for n = 3,

had an incorrect statement of the assumption for n = k or

had an incorrect use of the assumption in the attempt to

prove the statement.

Poor algebraic skills made it difficult or even impossible for

some candidates to complete the proof.

2004 Extension 1 Examiners Comments;

(a) Many candidates submitted excellent responses

including all the necessary components of an

induction proof.

Weaker responses were those that did not verify for n = 3,

had an incorrect statement of the assumption for n = k or

had an incorrect use of the assumption in the attempt to

prove the statement.

Poor algebraic skills made it difficult or even impossible for

some candidates to complete the proof.

Rote learning of the formula S(k) +T(k+1) = S(k+1) led

many candidates to treat the expression as a sum rather

than a product, thus making completion of the proof

impossible.