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90
Chap. 14 Analysis Techniques
14.5
F
(
n
) = 2+4+
···
+2
n
2
F
(
n
) = 4+8+
···
+2
n
+1
When we subtract, we get
2
F
(
n
)
−
F
(
n
)=
F
(
n
)=2
n
+1
−
2
. Thus,
n
X
i
=1
2
i
=2
n
+1
−
2
.
14.6
Call our summation
G
(
n
)
, then
G
(
n
)=
n
X
i
=1
i
2
n
−
i
=2
n
−
1
+2
∗
2
n
−
2
+3
∗
2
n
−
3
+
···
+
n
∗
2
0
.
2
G
(
n
)=2
n
X
i
=1
i
2
n
−
i
=2
n
+2
∗
2
n
−
1
+3
∗
2
n
−
2
+
···
+
n
∗
2
1
.
Subtracting, we get
2
G
(
n
)
−
G
(
n
)=
G
(
n
)=2
n
+2
n
−
1
+2
n
−
2
+
···
+2
1
−
n
∗
2
0
.
This is simply
2
n
+1
−
2
−
n
.
14.7
TOH has recurrence relation
F
(
n
)=2
F
(
n
−
1)+1
,F
(
1
)=1
.S
i
n
c
e
the problem gives us the closed form solution, we can easily prove it by
induction.
14.8
The closed form solution is
F
(
n
)=
nc
, which can easily be proved using
induction.
14.9
Pick the constants
c
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This note was uploaded on 12/27/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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