Discrete Mathematics with Graph Theory (3rd Edition) 159 -...

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
i(i!)
=
(k
+
2)! - 1. We have
i=l
k+1 k
L
i(i!)
=
[L
i(i!)]
+
(k
+
l)(k
+
I)!
i=l i=l
=
[(k
+
I)! - 1J
+
(k
+
l)(k
+
I)!
=
(k
+
1)!(1
+
k
+
1) - 1
by the induction hypothesis
=
(k
+
l)!(k
+
2) - 1
=
(k
+
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
2:
1
i=l
2
k k
and the result is true for
n
=
k;
that is, assume that
~
3
i
-
1
=
3 ; 1.