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.
This is the end of the preview.
Sign up
to
access the rest of the document.
- Summer '10
- any
- Graph Theory, Mathematical Induction, Inductive Reasoning, Natural number, Mathematical logic, Mathematical proof
-
Click to edit the document details