11x1 t10 11 some different types
TRANSCRIPT
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.