SOLUTIONS FOR HOMEWORK 1
1.3.
Defne Proposition
P
n
:
∑
n
k
=1
k
3
=
n
2
(
n
+ 1)
2
/
4 We shall show,
by induction, that
P
n
For any
n
∈
N
. The base is easy to establish For
n
= 1. Indeed,
P
1
states that 1
3
= 1
2
(1 + 1)
2
/
2, which is true.
The induction step consists oF proving that
P
n
implies
P
n
+1
. To this
end, assume
∑
n
k
=1
k
3
=
n
2
(
n
+ 1)
2
/
4, and show that
∑
n
+1
k
=1
k
3
= (
n
+
1)
2
(
n
+ 2)
2
/
4. We have:
n
+1
s
k
=1
k
3
=
n
s
k
=1
k
3
+ (
n
+ 1)
3
=
n
2
(
n
+ 1)
2
4
+ (
n
+ 1)
3
=
(
n
+ 1)
2
4
p
n
2
+ 4(
n
+ 1)
P
=
(
n
+ 1)
2
(
n
+ 2)
2
4
.
1.5.
We prove the equality
∑
n
k
=0
2

k
= 2
−
2

n
by induction. The
base oF inductions is taken care oF by veriFying the above equality For
n
= 1: 1 + 2

1
= 2
−
2

1
. This is easily checked.
The induction step consists oF showing that, iF
∑
n
k
=0
2

k
= 2
−
2

n
,
then
∑
n
+1
k
=0
2

k
= 2
−
2

(
n
+1)
. Use the induction hypothesis to get
n
+1
s
k
=0
2

k
=
n
s
k
=0
2

k
+ 2

(
n
+1)
= 2
−
2

n
+ 2

(
n
+1)
= 2
−
2
·
2

(
n
+1)
+ 2

(
n
+1)
= 2
−
2

(
n
+1)
,
and we are done.
1.6.
Defne Proposition
P
n
: 7 divides 11
n
−
4
n
. We show by induction
that
P
n
is true For every
n
. The base oF induction is easy to veriFy:
when
n
= 1, 7 divides 11
1
−
4
1
= 7. The inductive step consists oF
showing that, iF 7 divides 11
n
−
4
n
, then 7 also divides 11
n
+1
−
4
n
+1
.
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 Winter '08
 staff
 Negative and nonnegative numbers, Natural number

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