CSE 260
Homework 9 Induction ANSWER
October 22, 2007
1. Section 4.3: 20
max
(
a
1
, a
2
, ...a
n
+1
) =
max
(
max
(
a
1
, a
2
, ...a
n
)
, a
n
+1
)
min
(
a
1
, a
2
, ...a
n
+1
) =
min
(
min
(
a
1
, a
2
, ...a
n
)
, a
n
+1
)
2. Prove that 5
n

4
n

1 is divisible by 16 for
n²Z
+
p
(
n
) : 5
n

4
n

1 is divisible by 16 for
n²Z
+
p
(1) : 5
1

4
.
1

1 is divisible by 16 ?
0 is divisible by 16? true.
p
(
n
+ 1) : 5
n
+1

4(
n
+ 1)

1 is divisible by 16?
5
n
+1

4(
n
+ 1)

1 = 5(5
n

4
n

1) + 16
n
is divisible by 16? yes.
3. Recursively deFne
b
0
=
b
1
=
b
2
= 1 and
b
n
=
b
n

1
+
b
n

3
for
n
≥
3.
(a) Calculate
b
n
for
n
= 3
,
4
,
5
b
3
=
b
2
+
b
0
= 1 + 1 = 2
b
4
=
b
3
+
b
1
= 2 + 1 = 3
b
5
=
b
+ 4 +
b
2
= 3 + 1 = 4
(b) Show by using the second principle (strong) of mathematical induc
tion that
b
n
≥
2
b
n

2
for
n
≥
3
proof:
b
n
+1
≥
2
b
n

1
for
n
+ 1
≥
3
b
n
+1
=
b
n
+
b
n

2
≥
2
b
n

2
+ 2
b
n

4
= 2(
b
n

2
+
b
n

4
) = 2
b
n

1
4. Prove by mathematical induction
1
3
+ 2
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 Fall '08
 SaktiPramanik
 Recursion, East Lansing, rooted trees

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