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162
provethatfk
< G)kl.
Now
fk
=
fkl
+
ik2
< (D(kl)l
+
G)(k2)l
=
(D
k

3
G
+ 1)
=
G)k3
(11)
<
G)k3
(f~)
= G)kl
Solutions to Review Exercises
(applying the induction hypothesis to
k

1 and to
k

2, both of which are in the range 2
~
e
<
k
where the result holds)
as required. By the Principle of Mathematical Induction (strong form), we conclude that the result
holds for all
n
::::
l.
21. For
n
=
1,12
=
1
=
2  1
=
h

1, so the formula holds. Now assume
k
:::: 1 and that the formula
holds for
n
=
k.
We have
12
+!4 +
...
+
f2(k+l)
=
12
+
f4
+
...
+
hk+2
=
(12
+
f4
+ .
.. +
12k)
+
hk+2
=
(f2k+l

1)
+
hk+2
=
hk+3 1
by the induction hypothesis
using the definition of
hk+3
and this is the desired formula with
n
=
k
+
1 since
2(k
+
1)
+
1
=
2k
+
3. By the Principle of
Mathematical Induction, the formula is correct for all
n
:::: 1.
22. Let
ai
be the number of ways in which you can reach stair i. Then
al
=
1,
a2
=
2 (one single step
or one double step) and, in general, you can reach stair
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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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