138
Solutions to Exercises
44. We use the strong form of the Principle of Mathematical Induction. When
n
=
0,
ao
=
0
:S
20(0)
=
0,
so the result holds. Now let
k
>
0 and assume the result holds for all 0
:S
n
<
k.
We wish to prove
the result for
n
=
k;
that is, we wish to prove that
ak
:S
20k.
Now,
ak
=
aLk/5J
+
aL3k/5J
+
k
<
20Lk/5j +20L3k/5j
+k
applying the induction hypothesis to Lk/5j and L3k/5j
<
20(~)
+
20C:)
+
k
=
4k
+
12k
+
k
=
17k
<
20k
as desired. By the Principle of Mathematical Induction, we conclude that the inequality is true for all
n
~
O.
45. Observe that <1>2
=
1 + <1>, hence, <1>3
=
<1> + <1>2, <1>4
=
<1>2 + <1>3 and, in general,
<1>n+l
=
<1>n
+ <1>nl
for all
n
~
1. This says that
an
=
<1>n
satisfies the Fibonacci recurrence relation.
46. [BB] Rewriting
Ik+l
=
Ik
+
Ikl
as
ikl
=
Ik+l

ik,
we see that
10
=
h

h
=
1  1
=
O.
Similarly,
11
=
1,12
=
1,13
=
2,14
=
3,
15
=
5,16
=
8. In general,
In
=
(_I)n+l In.
47. [BB] The Fibonacci sequence is the sequence 1,1,2,3,5,8
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 Summer '10
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 Graph Theory, Mathematical Induction, Natural number, 1 K, Fibonacci number, 1 2K, 1 7k

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