1
CMPS 101
Summer 2009
Homework Assignment 3
Solutions
1.
(1 Point)
The last exercise in the handout entitled
Some Common Functions
.
Use Stirling's formula to prove that
Θ
=
n
n
n
n
4
2
.
Proof:
By Stirling’s formula
( )
( )
2
2
2
)
/
1
(
1
2
)
2
/
1
(
1
2
2
2
)
!
(
)!
2
(
)!
2
(
!
)!
2
(
2
Θ
+
⋅
⋅
Θ
+
⋅
⋅
⋅
=
=

=
n
e
n
n
n
e
n
n
n
n
n
n
n
n
n
n
n
n
π
( )
( )
2
2
2
)
/
1
(
1
)
2
/
1
(
1
4
1
)
/
1
(
1
)
2
/
1
(
1
2
n
n
n
n
n
n
n
n
Θ
+
Θ
+
⋅
⋅
=
Θ
+
Θ
+
⋅
=
so that
( )
1
)
/
1
(
1
)
2
/
1
(
1
1
4
2
2
→
Θ
+
Θ
+
⋅
=
n
n
n
n
n
n
as
∞
→
n
The result now follows since
∞
<
<
1
0
.
///
2.
(2 Points) (Exercise 1 from the induction handout)
Prove that for all
1
≥
n
:
2
1
3
2
)
1
(
+
=
∑
=
n
n
i
n
i
.
Do this twice:
a.
(1 Point) using form
I
Ia of the induction step.
b.
(1 Point) using form IIb of the induction step.
Proof:
Let
)
(
n
P
be the equation
2
1
3
2
)
1
(
+
=
∑
=
n
n
i
n
i
.
I.
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 Spring '09
 AgoreBack
 Mathematical Induction, Natural number, induction hypothesis, Mathematical proof

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