MJC JC2 H3 Paper 9820 Prelim Solutions final .pdf - 2017...

MJC/2017 JC2 Prelim Marking Scheme/H3 Math (9820/01)/Math Dept Page 1 of 13
2017 H3 M
ATH
(9820/01) JC 2 P
RELIM
– M
ARKING
S
CHEME
Qn Solution Mark Scheme
1 Mathematical Induction 9
(i) Let S( n ) be the statement a player can always find a correct starting
token to win the game when there are n green tokens, n red tokens,
n ∈ ! + .
S(1): 1 green token, 1 red token.
Choose the green token first.
Therefore, S(1) is true.
Assume S( k ) is true for some k ∈ ! + . i.e. a player can always find a
correct starting token to win the game when there are k green tokens
and k red tokens.
S( 1 k + ): 1 k + green tokens, 1 k + red tokens.
Given any order of
1 k +
green tokens and
1 k +
red tokens, there must
exist an ordered pair of green and red tokens (in that clockwise order
where G first, then R). (In the worst case scenario, after a series of
1 k +
consecutive green tokens, the next token must be red.)
Remove the pair of green and red tokens. There are k green tokens and
k red tokens remaining. From the inductive step, the player can find a
starting point to win the game. The ordered pair of green and red
tokens can be reinserted in the same spot where they were taken from.
From the same starting spot, the player can still pick all 2 2 k + tokens
while fulfilling the condition that at least as many green tokens as red
tokens are picked up at all times.
Therefore, S( k ) is true implies S( 1 k + ) is true.
Since S(1) is true and S( k ) is true implies S( 1 k + ) is true, by
mathematical induction, S( n ) is true for n ∈ !
+
. A player can always
find a correct starting token to win the game when there are n green
tokens and n red tokens.
B1 – statement
B2 – correct
consideration of
S(1)
B1 – S( k )
assumption
B2 – some
explanation for
inductive step
(must mention G
first then R)
B1 - conclusion
(ii) No, the player might not be able to win the game for certain
arrangements of tokens.
A counter example for the case when 2 n = would be if the tokens
were arranged as follows.
To win the game, the player must start with a green token and then
picks up 2 red tokens consecutively, causing the player to not be able
to win the game.
B1
B1 -
counterexample
R R
G G
R R

MJC/2017 JC2 Prelim Marking Scheme/H3 Math (9820/01)/Math Dept Page 2 of 13
Qn Solution Mark
Scheme
2 Recurrence Relations & Series and Sequences 10 Marks
(i)
( )
( )
( )
1
2 1
2
2 1 2 1
2
1 2 1
2
2 1 1 2
1
1 2
e
e
e
e
e Shown
n
n k
n n n
n
n
k
n n n n n
n
k
n n n
n
k
n n n n n
n
n n
k
n n
a
a a a
a
a a a a a
a a a
a a a a a
a a
a a
−
− −
−
− − − −
− − −
− − − −
−
− −
⎛ ⎞
⎛ ⎞ ⎛ ⎞
= +
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
⎛ ⎞
⎛ ⎞
= +
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
⎛ ⎞
= +
⎜ ⎟
⎝ ⎠
⎛ ⎞
− =
⎜ ⎟
⎝ ⎠
− =
1
3 3
1 2
3 2
3 2
1
2 1
e
e 1 e
1 e
r
n n
r r
k
r r
r r
n
k k
k
a a
a a
a a
a a
−
= =
− −
−
⎛ ⎞
− =
⎜ ⎟
⎝ ⎠
⎛ ⎞
−
⎜ ⎟
⎝ ⎠
− =
−
∑ ∑
3 4
3 2
1 2
2 3
1
1 2
...
n n
n n
n n
n n
a a
a a
a a
a a
a a
a a
− −
− −
−
− −
+ −
+
+ −
+ −
3 2
2
1
1 1
3 2
3
1 1
1
3
2
1
1
1
e 1 e
1 e
e 1 e
e
1 e e 1
e
e
e 1
e
e
e 1
n
k k
n
n
k
n
k k
k
n
n
k k
n
k
k
k
n
k
k
k
a a
a a
a
a
−
−
−
−
−
⎛ ⎞
−
⎜ ⎟
⎝ ⎠
− =
−
⎛ ⎞
−
⎜ ⎟
⎝ ⎠
= +
− −
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
−
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
−
( )
1
1
e
f
e 1
k
k
k ∴ =
−
.