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130
Solutions to Exercises
6. [BB] The first six terms are 1,3,7,15,31,63. Our guess is that
an
=
2
n
1. When
n
= 1,2
1
1
=
1,
agreeing with
al.
Now assume that
k
~
1 and the result is true for
n
=
k;
that is, assume that
ak
=
2k
1. We must prove the result is true for
n
=
k+
1;
that is, we must prove that
ak+
1
=
2
k
+
1_1.
NOwak+l
=
2ak
+
1
=
2(2k

1)
+
1 (by the induction hypothesis)
=
2k+l

2
+
1
=
2k+l

1, as
required. By the Principle of Mathematical Induction, we conclude that the result is true for all
n
~
1.
7. The first six terms
are~,
1
2
3, 6
2
3,
3~3,
15
2
63
and
78
2
13. We guess that
an
=
5
n
tl.
Note that when
n
=
1,
the formula gives
2¥
=
~,agreeing
with
al.
Now assume that
ak
=
5
k
tl
for some
k
~
1. We must
th
5k+l+1
prove
atak+l
=
4.
Now
ak+l
=
5ak1
__
5(5
k
4
+
1)
1
by the induction hypothesis
5k+l
+
5
5k+l
+
1
4
1= 4
'
as desired.
8. (a) [BB]
a2
=
l(al
+
ao)
=
1(1
+
1)
=
2;
a3
=
2(a2
+
al)
=
2(2
+
1)
=
2(3)
=
6;
a4
=
3(a3
+
a2)
=
3(6
+
2)
=
3(8)
=
24;
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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