Chapter 5
161
which is the desired formula for
n
=
k.
By the Principle of Mathematical Induction, the formula holds
for all
n
~
1.
14. (a) We have
al
=
f(1)
=
1,
a2
=
f(al)
=
f(
1)
=
7,
a3
=
f(7)
= 25,
a4
=
f(a3)
=
103 and
a5
=
409.
(b)
a2
=
al
implies
f(al)
=
ab
so
f(J(t))
=
f(t).
This yields 3 
4f(t)
=
3 
4t,
which gives
3  4(3 
4t)
=
3 
4t
or 3 
12
+
16t
=
3 
4t, 20t
=
12,
t
=
~.
Since
f(~)
=
3 
4(~)
=
3 
1;
=
~,
the sequence is constant.
(c) The answer is no.
al
<
a2
implies
f(t)
<
f(J(t)),
which is 3 
4t
<
9
+
16t,
so that
t
>
~.
But
a2
<
a3
implies
9
+
16t
<
f(
9
+
16t)
=
3  4(
9
+
16t)
=
39 
64t,
so
80t
< 48, that
is,
t
<
~.
No such
t
exists.
15. We have
a
=
5,
d
=
4.
(a)
a32
=
5
+
31(4)
=
129;
alOO
=
5
+
99(4)
=
401.
(b) Attempting to solve 125
=
5
+
(n
 1)4, we get
n
 1
=
30. Thus, 125 is the 31st term of this
sequence.
(c) Attempting to solve 429
=
5
+
(n
 1)4, we get
n
 1
=
106. Thus, 429 is the 107th term of the
sequence. Attempting to solve 1000
=
5
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 Summer '10
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 Graph Theory, Mathematical Induction, Zagreb, Zagreb bypass, Geometric progression, Angie Stone

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