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6.
Experiment:
Step 1: The inequality holds for
n
5
6, because
720
.
216.
Step 2: Suppose
k
!
.
k
3
. Then (
k
1
1)!
.
(
k
1
1)
k
3
.
For
k
$
4,
k
.
1
1 }
2
k
}
1 }
k
1
2
}
(since
}
2
k
}
,
2 and
}
k
1
2
} ,
1), and so
k
3
.
k
2
1
2
k
1
1
5
(
k
1
1)
2
. So
(
k
1
1)
k
3
.
(
k
1
1)
3
, and thus by the transitivity
of
.
,(
k
1
1)!
.
(
k
1
1)
3
.
The mathematical induction principle now guarantees the
original inequality for all
n
$
6.
7.
Experiment:
Step 1: The inequality holds for
n
5
5, because 32
.
25.
Step 2: Suppose 2
k
.
k
2
. For
k
$
3,
k
.
2
1 }
1
k
}
1
since
}
1
k
}
,
1
2
, and so
k
2
.
2
k
1
1. Then by the
transitivity of
.
,2
k
.
2
k
+ 1. And thus
2
k
1
1
5
2
k
1
2
k
.
k
2
1
2
k
1
1
5
(
k
1
1)
2
.
The mathematical induction principle now guarantees the
original inequality for all
n
$
5.
8.
Step 1: The inequality holds for
n
52
3, because 2
2
3
5 }
1
8
}
.
Step 2: Suppose 2
k
$ }
1
8
}
. Then 2
k
1
1
5
2
?
2
k
$ }
2
8
}
5 }
1
4
}
, and
by the transitivity of
$
and the fact that
}
1
4
}
$ }
1
8
}
k
1
1
$ }
1
8
}
.
The mathematical induction principle now guarantees the
original inequality for
n
$2
3.
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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