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Here are solutions, answers, hints, and references for solutions for some of the review problems.
(A1)
Prove the identity
∑
n
i
=
k
(
i
k
)
=
(
n
+1
k
+1
)
by induction. (Here
n
is any positive integer.)
Solution.
We prove this by induction on
n
(the instructions didn’t say whether to argue by induction
on
n
, or on
k
, so it’s up to us). If
n < k
then both sides are zero; there are then no terms in the
sum on the left, while
(
n
+1
k
+1
)
= 0 by deﬁnition since
n
+ 1
< k
+ 1. Basis step:
n
=
k
. Then the
lefthand side is
∑
n
i
=
n
(
i
n
)
=
(
n
n
)
= 1. The righthand side is
(
n
+1
n
+1
)
= 1, so they’re equal.
Inductive step: Assume true for
n
, prove for
n
+ 1:
n
+1
X
i
=
k
±
i
k
²
=
n
X
i
=
k
±
i
k
²
+
±
n
+ 1
k
²
=
±
n
+ 1
k
+ 1
²
+
±
n
+ 1
k
²
=
±
n
+ 1
k
+ 1
²
by Pascal’s identity.
(A2)
Identify each of the following sets as ﬁnite, countably inﬁnite, or uncountable. (No proofs are
required.)
(A2a)
The set
Z
of all integers.
(A2b)
N
(A2c)
The set
R
of all real numbers.
(A2d)
The set of all rational numbers.
(A2e)
The set of all subsets of
Z
.
(A2f)
The set of all inﬁnite
sequences (
a
1
, a
2
, a
3
,
···
), where each
a
k
∈ {
0
,
1
}
.
Answers.
Countably inﬁnite:
Z
, the set of all rationals, and the set of all inﬁnite sequences of the
type described. Uncountable:
R
(see Example 20 page 235 of Rosen) and the set of all subsets of
Z
(certainly not ﬁnite; and we showed earlier in the semester that the power set of a set
S
can never
be placed into onetoone correspondence with
S
, so the only remaining possibility is that the power
set of
Z
be uncountable).
(A3)
Suppose that
a
0
= 1 and that for every positive integer
n
,
a
n
=
na
n

1
+ (

1)
n
. Show that
a
n
=
n
!
∑
n
k
=0
(

1)
k
k
!
for every
n
≥
0.
Solution.
We prove this by induction on
n
.
Basis step
n
= 1: the recurrence relation gives
1
a
0
+ (

1)
1
= 1

1 = 0, while on the other hand 1!
∑
1
k
=0
(

1)
k
k
!
=
1
0!

1
1!
= 1

1 = 0. So the
formula holds for
n
= 1.
Inductive step: Assume true for
n
.
a
n
+1
= (
n
+ 1)
a
n
+ (

1)
n
+1
= (
n
+ 1)
n
!
n
X
k
=0
(

1)
k
k
!
+ (

1)
n
+1
= (
n
+ 1)!
n
X
k
=0
(

1)
k
k
!
+ (

1)
n
+1
= (
n
+ 1)!
n
+1
X
k
=0
(

1)
k
k
!

(
n
+ 1)!
(

1)
n
+1
(
n
+ 1)!
+ (

1)
n
+1
= (
n
+ 1)!
n
+1
X
k
=0
(

1)
k
k
!

(

1)
n
+1
+ (

1)
n
+1
= (
n
+ 1)!
n
+1
X
k
=0
(

1)
k
k
!
.
(A4)
A game between two players
A, B
has these rules: There is a pile of
n
tokens. Player
A
goes
ﬁrst, and removes some number of tokens from the pile. Player
B
then does the same, and they
take turns. At each turn, the player must remove at least one token, and no more than 4 tokens.
The last player to remove a token is the loser.
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