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)
=
42
=
4'2
as desired. By the Principle
of
Mathematical Induction, the result is true for all
n
~
1.
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 Summer '10
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 Graph Theory, Mathematical Induction, Natural number, Mathematical logic, Mathematical proof, 1  k

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