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Section 5.2
Exercises 5.2
1. (a) [BB]
a1
=
1;
ak+1
=
5ak
for
k
2':
1.
(b)
=
5;
=
ak
 2 for
k
2':
(c)
=
4;
a2
=
1;
ak+2
=

for
k
2':
(d) There are two possibilities:
= 1;
=
+
(_1)k+1k
for
k
Also,
=
1,
=
2,
=
ak2
+
(_1)k
for
k
3.
2. (a) [BB] 16,8,4,2,1,1,1;
(c)
18,9,4,2,1,1,1;
(b) 17,8,4,2,1,1,1;
(d) 100,50,25,12,6,3,1.
129
3. When
n
=
3
n

1
=
30
=
1, which agrees with
a1.
Now assume the result is true for
n
=
k;
that is, assume
=
3
k

1
.
We must prove the result for
n
=
k
+
that is, we must prove that
=
3(k+1)1
=
3
k
.
Now
=
3ak
=
3(3
k

1
)
(by the induction hypothesis)
=
3
k
,
as
required. By the Principle of Mathematical Induction, the result is true for all
n
4. [BB] When
n
=
1, the formula becomes (11)(1+
2
1(1
2
+1+
2
)
=
0, which is
Now assume that the
formula is correct when
n
=
k.
When
n
=
k
+
1,
(k
1)
3
(k
1)3
(k

1)(k
+
2)(k
2
+
k
+
2)
=
+
+
=
+
+
4
4k
3
+
12k2
+
12k
+
4
+
k4
+
2k
3
+
k
2

4
4
+
6k
3
+
13k
2
+
12k
4
k(k
+
3)(k
2
+
3k
+
4)
=
''''
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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