132
Solutions to Exercises
as desired. By the Principle of Mathematical Induction, we conclude that the formula is correct for all
n?1.
14. We have
a2
=
1
2
a1
= 1 . 1 =
1,
a3
=
22a2
= 4 . 1 = 4 = 2
2
,
a4
=
32a3
= 3
2
. 22 = 36,
a5
=
42a4
= 4
2
.3
2
.22 = 16·9·4 = 576, and
a6
=
52a5
= 5
2
.4
2
.3
2
.22 = 25 ·16·9·4 = 14400.
It
seems that
an
=
(n

1)2(n

2)2
... 3
2
.2
2
=
[(n

1)!]2. So we make the conjecture that
an
=
[(n

1)!]2 for
n
?
1.
For
n
=
1, [(1 
1)!]2
=
0!2
=
12
=
1, in agreement with the conjectured value of
a1.
Now we
assume the conjecture is true for
n
=
k;
that is, we assume that
ak
=
[(k

1)!]2. We must prove the
conjecture is true for
n
=
k
+ 1; that is, we must prove that
ak+1
=
[((k
+ 1) 
1)!]2, that is, that
ak+1
=
(k!)2.
Now
ak+1
k2ak
=
k
2
[(k

1)!]2
(using the induction hypothesis)
[k(k
 1)!]2
=
(k!)2
as desired. We conclude that our conjecture is correct for all
n
? 1 by the Principle of Mathematical
Induction.
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