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Section 5.2
131
10. The first few tenns are 1, 2, 3·2
=
6,4(3·2)
=
24, .
... Our guess is that
an
=
nL When
n
=
1, 1!
=
1,
in agreement with
al.
Now assume the fonnula is correct for
n
=
k;
that is, assume that
ak
=
kL We
must prove the fonnula is correct for
n
=
k
+
1; that is, we must prove that
ak+l
=
(k
+
l)L But
ak+l
=
(k
+
l)ak
=
(k
+
l)k!
=
(k
+
1)!, as desired. By the Principle of Mathematical Induction,
the formula is true for all
n
~
1.
11. [BB] The first few tenns are 0, 1,0,4,0,16,
.... Our guess is that
{
0
ifn is odd
an
=
4 ¥1
if
n
is even.
We will prove this using the strong fonn of mathematical induction. Note first that
al
=
0 and
d
1
=
4
0
=
1
=
a2
so that our guess is correct for
al
and
a2.
Now let
k
>
2 and assume the
result is true for all
n,
1
~
n
<
k.
We must show that the result is true if
n
=
k.
If
k
is odd, then
ak
=
4ak2
=
4(0)
=
0 since
ak2
=
0 by the induction hypothesis,
k

2 being odd. If
k
is even,
then
k2
1
k2
k
1
ak
=
4ak2
= 4(4
2
  ) (using the induction hypothesis)
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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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