11x1 t14 11 some different types (2011)
TRANSCRIPT
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.