MATH 55 2/26 DISCUSSION QUESTIONS
1.
Prove the following statements for all positive integers
n
by using induction.
(a) 1
2
+ 2
2
+
...
+
n
2
=
n
(
n
+1)(2
n
+1)
6
(b) 6 divides
n
3

n
(c) 1
2

2
2
+ 3
2

...
+ (

1)
n

1
n
2
= (

1)
n

1
n
(
n
+1)
2
Solution:
(a) I’ll leave the base case
n
= 1 to you. Suppose that
P
(
k
) is true. Then
1
2
+ 2
2
+
...
+
k
2
+ (
k
+ 1)
2
=
k
(
k
+ 1)(2
k
+ 1)
6
+ (
k
+ 1)
2
= (
k
+ 1)
k
(2
k
+ 1) + 6
k
+ 6
6
= (
k
+ 1)
2
k
2
+ 7
k
+ 6
6
=
(
k
+ 1)(
k
+ 2)(2
k
+ 3)
6
as we wanted to show.
(b) I’ll leave the base case
n
= 1 to you (notice that 6 divides 0). For the induction step, assume
P
(
k
), so
k
3

k
= 6
m
for some integer
m
. Then
(
k
+ 1)
3

(
k
+ 1) =
k
3
+ 3
k
2
+ 3
k
+ 1

k

1 = (
k
3

k
) + 3
k
2
+ 3
k
= 6
m
= (3
k
2
+ 3
k
)
,
so we need to show that 6 divides 3
k
2
+ 3
k
. Well,
k
2
+
k
is even since
k
and
k
2
have the same
parity (we showed this a long time ago), so
k
2
+
k
= 2
y
for some integer
y
. Then 3
k
2
+ 3
k
= 6
y
so it’s a multiple of 6 as needed.
(c) I’ll leave the base case
n
= 1 to you again. Assume
P
(
k
). Then
1
2

2
2
+ 3
2

...
+ (

1)
k

1
k
2
+ (

1)
k
(
k
+ 1)
2
= (

1)
k

1
k
(
k
+ 1)
2
+ (

1)
k
(
k
+ 1)
2
= (

1)
k
(
k
+ 1)
2

k
(
k
+ 1)
2
= (

1)
k
2
k
2
+ 4
k
+ 2

k
2

k
2
= (

1)
k
k
2
+ 3
k
+ 2
2
= (

1)
k
(
k
+ 1)(
k
+ 2)
2
which is what we wanted to show.
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 Spring '08
 STRAIN
 Math, Integers, Mathematical Induction, Natural number, base case, induction step, positive integers