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Chapter 5
157
Chapter 5 Review
1.
When
n
=
1, the sum is 1(1!)
=
1, while the right hand side is 2! 
1
=
1, so the result holds. Now
k
assume that
k
2:
1 and that the result is true for
n
=
k;
that is, assume
L
i(i!)
=
(k
+
I)! 
1. We
i=l
k+1
must prove that
L
=
+
2)! 
1. We have
k+1
k
L
=
[L
i(i!)]
+
+
l)(k
+
I)!
i=l i=l
=
[(k
+
1J
+
+
+
=
(k
+
1)!(1
+
k
+
1) 
1
by the induction hypothesis
= (k
+
l)!(k
+
2) 
1
=
+
2)! 
1
as desired. So the result is true, by the Principle of Mathematical Induction.
n
.
3n
1 3
1
1
2. For
n
=
1,
L
3,1
=
3°
=
1 while

=
2
=
1, so the result holds. Now suppose
k
1
2
k
k
and the result is true for
n
=
k;
that is, assume that
~
3
i

1
=
3 ; 1. We wish to prove that the
k+1. 3k+1
_
1
result is true for
n
=
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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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