21127, Concepts of Mathematics, Review Selected Solutions
1. Prove the following using either weak or strong induction:
(a) Let
P
(
n
) = 2
n
≥
n
+ 1
Base Case:
P
(0) = 2
0
= 1
≥
0 + 1
Inductive Hypothesis: Assume
P
(
k
) is true for some
k
∈
N
Inductive Step: We begin by citing the inductive hypothesis to conclude 2
k
≥
k
+ 1. We next
multiply both sides by two to obtain: 2
k
+1
≥
2
k
+ 2 =
k
+
k
+ 2. We next note that by definition
k
≥
0, and therefore 2
k
+1
≥
k
+
k
+ 2
≥
k
+ 2 as required.
(b) Go to office hours
(c) Let
P
(
n
) =
∑
n
k
=1
k
3
= (
∑
n
k
=1
k
)
2
Base Case:
P
(1) = 1
3
= 1 = 1
2
Inductive Hypothesis: Assume
P
(
r
) is true for some
r
∈
N
Inductive Step: We begin by considering the quantity (
∑
r
+1
k
=1
k
)
2
= (
∑
r
k
=1
k
+ (
r
+ 1))
2
. If we
let
α
=
∑
r
k
=1
k
then we have (
∑
r
+1
k
=1
k
)
2
= (
α
+ (
r
+ 1))
2
=
α
2
+ 2
α
(
r
+ 1) + (
r
+ 1)
2
. By the
inductive hypothesis we know that
α
2
=
∑
r
k
=1
k
3
.
Looking at the other two terms, we notice
that
α
=
∑
r
k
=1
k
=
(
r
+1)(
r
)
2
and therefore 2
α
(
r
+ 1) =
r
(
r
+ 1)
2
. We may finally conclude that
(
∑
r
+1
k
=1
k
)
2
= (
r
+ 1)
2
+
r
(
r
+ 1)
2
+
∑
r
k
=1
k
3
= (
r
+ 1)
3
+
∑
r
k
=1
k
3
=
∑
r
+1
k
=1
k
3
as desired.
(d) Go to office hours
(e) Before we begin, it is worth noting that the trick in this problem is inducting upon a+b. While
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 Spring '07
 GHEORGHICIUC
 Math, Inductive Reasoning, base case, inductive hypothesis

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