114
Solutions to Exercises
k+1
k
Now
L(i
+
1)2i
=
+
1)2i
+
(k
+
2)2k+1
i=1
= k2k+1
+
+
2)2k+1(using the induction hypothesis)
=
(k
+
k
+
=
(2k
+
=
2(k
+
1)2k+1
=
+
1)2k+2
as required. By the Principle of Mathematical Induction, the given assertion is true for all
n
2: 1.
(b) When
n
=
1,
n
i2
12
1
1
8
(2i 1)(2i
+
1)
=
[2(1) 1][2(1)
+
1]
=
1(3)
=
3'
while
~~';:,~V)
=
~
=
i
and so the result holds. Now suppose that
k
2: 1 and the result is true
for
n
=
k;
thus, we suppose that
k
k(k+1)
8
(2i  1)(2i
+
1)
=
2(2k
+
1) .
We have to show that the result holds for
n
=
k
+
1, that is, that
i
2
+
l)[(k
+
+
8
(2i  1)(2i
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 Summer '10
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 Graph Theory, Mathematical Induction, Natural number, Mathematical logic, 1 k, Mathematical proof, 1 L

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