126
Solutions to Exercises
35. [BB]
(using the result of Exercise 34)
n
IT
(1
+
(±1)2r)
=
2n.
r=1
IT
n
2r
 1
3
5
2n
 1
36.

=  . 
...

=
2n

1.
r=2 2r
 3
1
3
2n
 3
Proof.
If
n
=
2,
Il~=2 ~;=~
=
f
=
3
=
2(2)  1 and so the result is true. Now suppose that
k
::::: 1
and the result is true for
n
=
k;
that is, suppose
Il~=2 ~;=~
=
2k

1. We must prove it is true for
n
=
k
+
1; that is, we must prove
Il~~~ ~;=~
=
2(k
+
1)  1
=
2k
+
1. We have
k+1
k
IT
2r1
=
(IT
2r1)(2(k+1)1)
=(2k_1)(2k+1)
=2k+1
r=2 2r

3
r=2 2r
 3
2(k
+
1)  3
2k
 1
as claimed. By the Principle of Mathematical Induction, the result is true for all
n
::::: 2.
•
nI
37. (a) For
n
=
1,
IT
Fr
=
Fo
=
3
=
5  2
=
FI

2, so the formula is correct.
r=O
kI
Now let
k
::::: 1 and assume that
IT
Fr
=
Fk

2. Then
r=O
k
kI
IT
Fr
=
Fk
IT
Fr
=
Fk(Fk

2)
(by the induction hypothesis)
r=O
r=O
=
Ff

2Fk
=
(22k
+
1)2 
2(22k
+
1)
=
22
k
+
1
+
2(22k)
+
1 _
2(22k)
_ 2
=
22
k
+
1
+
1  2
=
F
k
+
1

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 Summer '10
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 Graph Theory, Mathematical Induction, Natural number

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