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Base case: When
n
= 1
,
a
1
= 2
and
a
2
= 3
. So
1
≤
a
1
≤
a
2
<
4
.
Induction hypothesis: Assume that the statement is true for some
k
≥
1
, i.e.
1
≤
a
k
≤
a
k
+1
<
4
.
Induction step: We want to prove that
1
≤
a
k
+1
≤
a
k
+2
<
4
by breaking it into three
cases.
1. We already know that
1
≤
a
k
+1
by the induction hypothesis.
2. For the second inequality, we have
a
k
+2
=
5
a
k
+1

4
a
k
+1
= 5

4
a
k
+1
≥
5

4
a
k
since
a
k
≤
a
k
+1
by hypothesis
=
5
a
k
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Unformatted text preview: 4 a k = a k +1 3. For the third inequality, we have a k +2 = 5 a k +14 a k +1 = 54 a k +1 < 54 4 since a k +1 < 4 by hypothesis = 4 . By combining these three inequalities, we see that 1 ≤ a k +1 ≤ a k +2 < 4 . By mathematical induction, the statement is true for all n ≥ 1 . 1...
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This note was uploaded on 07/24/2011 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.
 Winter '08
 ANDREWCHILDS
 Math

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