Solutions to the Sample Final
1.
Find general solutions to the following two recurrences:
(a)
g(n+2)=3g(n+1)+g(n),
0
n
≥
, where g(0) = 1, g(1)=1.
Using the Theorem 9 in the book on page 135 to solve this problem:
The two solutions to
2
310
xx
−−
=
are
31
3
2
±
.
So, the general form is
11
2 2
()
nn
g
nK
rK
r
=+
, and we solve the values to
12
,
K
K
using the
values g(0) = 1, g(1)=1.
(0) 1
g
KK
== +
,
(1) 1 3/2*(
)
13/2*(
)
g
==
+
+
−
.
13 1
13 1
,
21
3
3
−+
.
13 1 3
13
13 1 3
13
(
)
(
)
22
3
3
gn
+−
(b)
h(n+2)=8h(n+1)16h(n),
0
n
≥
, where h(0) = 1, h(1)=1.
The only solutions to
2
81
60
−+=
is 4.
So, the general form is
2
1
hn
Kr
Knr
, and we solve the values of
,
K
K
using the
values h(0) = 1, h(1)=1.
1
(0) 1
hK
,
2
(1)
1
4
4
==+
.
1,
3 / 4
−
.
3
() 4
(
4
)
4
n
=−
2. We could have k = 0, 1, 2, 3 possible defective bulbs to be chosen. For each k (k
defective bulbs), the possibilities that k bulbs are chosen can calculated as follows:
stage 1: choose k defective bulbs.
Number of choices:
5
k
⎛⎞
⎜⎟
⎝⎠
stage 2: choose 3
−
k more bulbs from good bulbs. Number of choices:
7
3
k
⎛
⎞
⎜
−
⎝
⎠
Total number of possible choices:
12
3
The expected number of defective bulbs are
1..3
57
3
5
12
4
3
k
kk
k
=
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 Spring '08
 Graham
 Lefthandedness, Spanning tree

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