MATH 187 – Quiz 7 – Key
1) Prove
1
3
+ 2
3
+ 3
3
+
···
+
n
3
=
b
n
(
n
+ 1)
2
B
2
(1)
for all positive integers
n
.
a) Proof by smallest counterexample.
So assume otherwise. That is, we as
sume there is a smallest positive integer
x
such that (1) is false for
n
=
x
.
Thus
1
3
+ 2
3
+ 3
3
+
+
x
3
n
=
b
x
(
x
+ 1)
2
B
2
.
(2)
Note that
x
n
= 1 because
n
= 1 renders (1) to be true. Thus
x
≥
2 and thus
x
−
1
< x
and
x
−
1 is a positive integer. Thus
n
=
x
−
1 makes (1) true. So
we get
1
3
+ 2
3
+ 3
3
+
+ (
x
−
1)
3
=
b
(
x
−
1)
x
2
B
2
.
(3)
Now add
x
3
to both sides of (3). Thus
1
3
+ 2
3
+ 3
3
+
+ (
x
−
1)
3
+
x
3
=
b
(
x
−
1)
x
2
B
2
+
x
3
=
x
2
±
p
x
−
1
2
P
2
+
x
²
=
x
2
±
(
x
−
1)
4
2
+
x
²
=
x
2
(
x
−
1)
2
+ 4
x
4
=
x
2
4
³(
x
2
−
2
x
+ 1
)
+ 4
x
´
=
x
2
4
(
x
2
+ 2
x
+ 1
)
=
x
2
4
(
x
+ 1)
2
=
b
x
(
x
+ 1)
2
B
2
,
which contradicts (2).
QED.
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View Full Documentb) Proof by induction on
n
.
1. Base case. Prove (1) is true for
n
= 1. Note LHS = 1
3
= 1 and RHS
=
(
1
·
2
2
)
2
= 1. So LHS = RHS. Done with base case.
2. Induction Hypothesis. Assume (1) is true for
n
=
k
. So we assume
1
3
+ 2
3
+ 3
3
+
···
+
k
3
=
b
k
(
k
+ 1)
2
B
2
.
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 Fall '08
 HOLMES
 Math, Integers, Mathematical Induction, Natural number, Prime number, base case, 1 1 2k

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