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SOLUTIONS TO HOMEWORK # 1
Problem 1.1.
We have to prove that the statement
P
n
: 1
2
+ 2
2
+
...
+
n
2
=
1
6
n
(
n
+ 1)(2
n
+ 1)
holds for
all
n
= 1
,
2
,...
.
Step 1
(”base of induction”).
P
1
: 1
2
=
1
6
·
1(1 + 1)(2
·
1 + 1) is true.
Step 2
. Assume
P
n
holds for some
n
=
k
∈
N
, that is
1
2
+ 2
2
+
...
+
k
2
=
1
6
k
(
k
+ 1)(2
k
+ 1)
(1)
We will prove that then that it holds for
n
=
k
+ 1 that is
P
k
+1
: 1
2
+ 2
2
+
...
+ (
k
+ 1)
2
=
1
6
(
k
+ 1)[(
k
+ 1) + 1][2(
k
+ 1) + 1]
holds. If (1) holds then
1
2
+ 2
2
+
...
+
k
2
+ (
k
+ 1)
2
=
1
6
k
(
k
+ 1)(2
k
+ 1) + (
k
+ 1)
2
=
=
k
+ 1
6
[
k
(2
k
+ 1) + 6(
k
+ 1)] =
=
1
6
(
k
+ 1)[2
k
2
+
k
+ 6
k
+ 6] =
1
6
(
k
+ 1)(
k
+ 2)(2
k
+ 3) =
=
1
6
(
k
+ 1)[(
k
+ 1) + 1][2(
k
+ 1) + 1]
.
So if
P
n
holds for
n
=
k
, then it holds for
n
=
k
+ 1.
By the principle of mathematical induction,
P
n
is true for
all
n
∈
N
.
Problem 1.6.
P
n
: (11)
n

4
n
is divisible by 7. This means (11)
n

4
n
= 7
s
for
some
s
∈
N
.
Step 1
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 Spring '08
 OLSAVSKYGREGORM

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