M413 Test 1 Solutions
1
. Use induction to prove that 5 + 11 + 17 +
···
+(6
n

1)=3
n
2
+2
n
for all
n
∈
N
.
First, this is true for
n
= 1: 6(1)

1 = 3(1
2
) + 2(1).
Now assume the statement is true for
n
. So we assume that
5+11+17+
···
+(6
n

1) = 3
n
2
+2
n.
Our goal is to show the statement is true for
n
+ 1. That would be the statement
5+11+17+
···
+(6
n

1) + (6(
n
+1)

1) = 3(
n
+1)
2
+2(
n
+1)
.
We have
5+11+17+
···
+(6
n

1) + (6(
n
+1)

1) = 3
n
2
+2
n
+ (6(
n
+1)

1)
=3
n
2
+2
n
+6
n
+5
=3
n
2
+6
n
+3+2
n
+2
=3(
n
+1)
2
+2(
n
+1)
so we are done. (The ±rst equality uses the induction hypothesis.)
2
. Use the Rational Zeros Theorem to prove that
b
=
3
±
1+
√
2 is irrational.
We have
b
3
=1+
√
2so
b
3

1=
√
2 and (
b
3

1)
2
= 2, which is
b
6

2
b
3
+1=2o
r
b
6

2
b
3

1 = 0. By the rational zeros theorem, any rational solution of this equation must
be of the form
b
=
p
q
where
q
divides the coeﬃcient of
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 Spring '09
 WEISBART
 Rational Zeros Theorem, Supremum, Prime number, Rational number

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