11x1 t10 11 some different types

nnnn

βαi

tan1tancot1tantan1

that show totantan1tantantanfact that the Use

2006 Extension 1 HSC Q4d)

Some Different Looking Induction Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1tantantanfact that the Use

nnnnnn

tan1tan1tan1tan1tan

2006 Extension 1 HSC Q4d)

Some Different Looking Induction Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1tantantanfact that the Use

nnnnnn

tan1tan1tan1tan1tan

2006 Extension 1 HSC Q4d)

nnnn

tan1tan1tan1tantan

Some Different Looking Induction Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1tantantanfact that the Use

nnnnnn

tan1tan1tan1tan1tan

2006 Extension 1 HSC Q4d)

nnnn

tan1tan1tan1tantan

nnnn tan1tantantan1tan1

Some Different Looking Induction Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1tantantanfact that the Use

nnnnnn

tan1tan1tan1tan1tan

2006 Extension 1 HSC Q4d)

nnnn

tan1tan1tan1tantan

nnnn tan1tantantan1tan1

tan

tan1tantan1tan1 nnnn

Some Different Looking Induction Problems

nnnn

βαi

tan1tancot1tantan1

that show totantan1tantantanfact that the Use

nnnnnn

tan1tan1tan1tan1tan

2006 Extension 1 HSC Q4d)

nnnn

tan1tan1tan1tantan

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

1integersallfor that provetoinduction almathematic Use

nnnn

nii

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tancot2RHS

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS 2tancot2RHS

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS 2tancot2RHS12tantan1

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS 2tancot2RHS12tantan1

1tan2tancot

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS 2tancot2RHS12tantan1

1tan2tancot 112tancot

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS 2tancot2RHS12tantan1

1tan2tancot 112tancot

22tancot

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS

Hence the result is true for n

= 1

2tancot2RHS12tantan1

1tan2tancot 112tancot

22tancot

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS

Hence the result is true for n

= 1Step 2: Assume the result is true for n

= k, where k

is a positive

integer

2tancot2RHS12tantan1

1tan2tancot 112tancot

22tancot

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS

Hence the result is true for n

= 1Step 2: Assume the result is true for n

= k, where k

is a positive

integer

1tancot1 1tantan3tan2tan2tantan ..

kkkkei

2tancot2RHS12tantan1

1tan2tancot 112tancot

22tancot

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS

Hence the result is true for n

= 1Step 2: Assume the result is true for n

= k, where k

is a positive

integer

1tancot1 1tantan3tan2tan2tantan ..

kkkkei

Step 3: Prove the result is true for n

= k

+ 1

2tancot2RHS12tantan1

1tan2tancot 112tancot

22tancot

1tancot1 1tantan3tan2tan2tantan

1integersallfor that provetoinduction almathematic Use

nnnn

nii

Step 1: Prove the result is true for n

= 1

2tantanLHS

Hence the result is true for n

= 1Step 2: Assume the result is true for n

= k, where k

is a positive

integer

1tancot1 1tantan3tan2tan2tantan ..

kkkkei

Step 3: Prove the result is true for n

= k

+ 1

2tancot2RHS12tantan1

1tan2tancot 112tancot

22tancot

2tancot2

2tan1tan3tan2tan2tantan ..

kk

kkei

Proof:

2tan1tan1tantan3tan2tan2tantan2tan1tan3tan2tan2tantan

kkkkkk

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.

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

2004 Extension 1 HSC Q4a)

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n

= 3

2004 Extension 1 HSC Q4a)

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n

= 3

31

321

LHS

2004 Extension 1 HSC Q4a)

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n

= 3

31

321

LHS

31

232

RHS

2004 Extension 1 HSC Q4a)

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n

= 3

31

321

LHS

31

232

RHS

RHSLHS

2004 Extension 1 HSC Q4a)

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n

= 3

31

321

LHS

31

232

RHS

RHSLHS Hence the result is true for n

= 3

2004 Extension 1 HSC Q4a)

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n

= 3

31

321

LHS

31

232

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)

1221

521

421

32-1

;3integersallfor that provetoinduction almathematic Use

nnn

n

Step 1: Prove the result is true for n

= 3

31

321

LHS

31

232

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)

1221

521

421

32-1 ..

kkkei

Step 3: Prove the result is true for n

= k

+ 1

Step 3: Prove the result is true for n

= k

+ 1

kkkei

12

121

521

421

32-1 Prove ..

Proof:

Step 3: Prove the result is true for n

= k

+ 1

kkkei

12

121

521

421

32-1 Prove ..

12121

521

421

32-1

121

521

421

32-1

kk

k

Proof:

Step 3: Prove the result is true for n

= k

+ 1

kkkei

12

121

521

421

32-1 Prove ..

12121

521

421

32-1

121

521

421

32-1

kk

k

121

12

kkk

Proof:

Step 3: Prove the result is true for n

= k

+ 1

kkkei

12

121

521

421

32-1 Prove ..

12121

521

421

32-1

121

521

421

32-1

kk

k

121

12

kkk

121

12

k

kkk

Proof:

Step 3: Prove the result is true for n

= k

+ 1

kkkei

12

121

521

421

32-1 Prove ..

12121

521

421

32-1

121

521

421

32-1

kk

k

121

12

kkk

121

12

k

kkk

1

11

2

kk

kk

Proof:

Step 3: Prove the result is true for n

= k

+ 1

kkkei

12

121

521

421

32-1 Prove ..

12121

521

421

32-1

121

521

421

32-1

kk

k

121

12

kkk

121

12

k

kkk

1

11

2

kk

kk

12

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

12

121

521

421

32-1 Prove ..

12121

521

421

32-1

121

521

421

32-1

kk

k

121

12

kkk

121

12

k

kkk

1

11

2

kk

kk

12

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

12

121

521

421

32-1 Prove ..

12121

521

421

32-1

121

521

421

32-1

kk

k

121

12

kkk

121

12

k

kkk

1

11

2

kk

kk

12

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.