From (1), we have
1
2
1
3
1
2 2
1
2 1
1
2 2
1
1
k
k
k
x
x p
k
k
k
x
+
+ +
+
+
−
= + +
+
+
+
− +
−
=
L
(
)
p
k
k
p
k
k
+
+
−
+
= +
+
+
1
2 1
1
2 2
1
2 1 2 2
(
)(
)
Hence the value of x is increasing for increase in n.
⇒
<
Maximum value of A
7
12
.
Ì Exam ple 0.1.41: For the sequence of integers 〈 〉 ≥
F n n 1 defined by
F 1 = 1, F 2 = 1 and F
F
F
n
n
n
n
=
+
≥
−
−
1
2
3
,
prove by Mathematical
Induction that
F
n
n
1
1
5
1
5
2
1
5
2
=
+
−
−
Proof: Basis: n = 1,
F 1
1
5
1
5
2
1
5
2
1
5
1
2
5
2
1
2
5
2
=
+
−
−
=
+
− +
= 1
Inductive Hypothesis: n = k,
F k
n
k
=
+
−
−
1
5
1
5
2
1
5
2
Inductive Step: n = k + 1,
F
F
F
k
k
k
k
k
+
−
=
+
=
+
−
−
+
1
1
1
5
1
5
2
1
5
2
1
5
1 +
−
−
−
−
5
2
1
5
2
1
1
k
k
¨
Introduction
31
1
2
1
3
1
2 2
1
2 1
1
2 2
1
1
k
k
k
x
x p
k
k
k
x
+
+ +
+
+
−
= + +
+
+
+
− +
−
=
L
(
)
p
k
k
p
k
k
+
+
−
+
= +
+
+
1
2 1
1
2 2
1
2 1 2 2
(
)(
)
Hence the value of x is increasing for increase in n.
⇒
<
Maximum value of A
7
12
.
Ì Exam ple 0.1.41: For the sequence of integers 〈 〉 ≥
F n n 1 defined by
F 1 = 1, F 2 = 1 and F
F
F
n
n
n
n
=
+
≥
−
−
1
2
3
,
prove by Mathematical
Induction that
F
n
n
1
1
5
1
5
2
1
5
2
=
+
−
−
Proof: Basis: n = 1,
F 1
1
5
1
5
2
1
5
2
1
5
1
2
5
2
1
2
5
2
=
+
−
−
=
+
− +
= 1
Inductive Hypothesis: n = k,
F k
n
k
=
+
−
−
1
5
1
5
2
1
5
2
Inductive Step: n = k + 1,
F
F
F
k
k
k
k
k
+
−
=
+
=
+
−
−
+
1
1
1
5
1
5
2
1
5
2
1
5
1 +
−
−
−
−
5
2
1
5
2
1
1
k
k
¨
Introduction
31