Download Math 3005 – Chapter 6 Bonus Homework

Math 3005 – Chapter 6 Bonus Homework
6.38 We proved in class that for every positive integer n,
12 + 22 + 32 + · · · + n2 =
n(n + 1)(2n + 1)
.
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(a) Use this result to determine a formula for 22 + 42 + 62 + · · · + (2n)2 for every positive
integer n.
Solution. Note that
22 + 42 + 62 + · · · + (2n)2 = (2 · 1)2 + (2 · 2)2 + (2 · 3)2 + · · · + (2 · n)2
= 22 · 12 + 22 · 22 + 22 · 32 + · · · + 22 · n2
= 4(12 + 22 + 32 + · · · + n2 )
n(n + 1)(2n + 1)
=4
6
=
2n(n + 1)(2n + 1)
.
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(b) Use the result we proved in class and the formula from (a) to determine a formula for
12 + 32 + 52 + · · · + (2n − 1)2 for every positive integer n.
Solution. Note that
12 + 22 + 32 + · · · + (2n)2 = (11 + 32 + 52 + · · · + (2n − 1)2 ) + (22 + 42 + 62 + · · · + (2n)2 ),
which implies that
11 + 32 + 52 + · · · + (2n − 1)2 = (12 + 22 + 32 + · · · + (2n)2 ) − (22 + 42 + 62 + · · · + (2n)2 ).
Now,
12 + 22 + 32 + · · · + (2n)2 =
n(2n + 1)(4n + 1)
2n(2n + 1)(4n + 1)
=
6
3
due to the result we proved in class, and
22 + 42 + 62 + · · · + (2n)2 =
1
2n(n + 1)(2n + 1)
3
due to the result in part (a). Thus,
n(2n + 1)(4n + 1) 2n(n + 1)(2n + 1)
−
3
3
n
=
(2n + 1)(4n + 1) − 2(n + 1)(2n + 1)
3
n
= (8n2 + 6n + 1 − 4n2 − 6n − 2)
3
n
= (4n2 − 1)
3
11 + 32 + 52 + · · · + (2n − 1)2 =
=
n(2n − 1)(2n + 1)
.
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6.48 A proof is given in the textbook. What result is being proved and which proof technique
is being used?
Solution. This is a proof using the Strong Principle of Mathematical Induction. The result
being proven is
Result. A sequence {an } is defined recursively by
a1 = 8,
a2 = 11,
and an = 5an−1 − 4an−2 − 9 for n ≥ 3.
Then an = 3n − 5 for all positive integers n.
6.49 A proof is given in the textbook. What result is being proved and which proof technique
is being used?
Solution. This is a proof by minimum counterexample. The result being proven is
Result. For all positive integers n, 8 | (32n − 1).
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6.51 By an n-gon, we mean an n-sided polygon† . It is well-known that the sum of the
interior angles of a triangle is 180◦ . Use induction to prove that for every integer n ≥ 3, the
sum of the interior angles of an n-gon is (n − 2) · 180◦ .
Proof. (by induction) Let S = {n ∈ Z : n ≥ 3} and for all n ∈ S, let P (n) be the statement
The sum of the interior angles of an n-gon is (n − 2) · 180◦ .
Since the sum of the interior angles of a triangle (a 3-gon) is (3 − 2) · 180◦ = 180◦ , the
base case is true. Assume that the sum of the interior angles of a k-gon is (k − 2) · 180◦ ,
where k ≥ 3. We must show that the sum of the interior angles of an (k + 1)-gon is
((k + 1) − 2) · 180◦ = (k − 1) · 180◦ . Let K be a (k + 1)-gon and let X, Y , and Z be vertices
of K such that XY and Y Z are edges of K, and the line segment drawn from X to Z lies
in the interior† of K. Then K consists of a k-gon and a triangle that share a common side,
XZ. By the induction hypothesis, the sum of the interior angles of the k-gon is (k − 2) · 180◦ .
The sum of the interior angles of triangle XY Z is 180◦ . The sum of the interior angles of K
is the sum of the interior angles of the k-gon and the triangle, which is
(k − 2) · 180◦ + 180◦ = (k − 1) · 180◦ .
By the Principle of Mathematical Induction, P (n) is true for all n ∈ S.
†
In this problem, we are not told that the n-gon must be convex, which would mean that
every interior angle is less than 180◦ . For a convex polygon, any vertices X, Y , and Z such
that XY and Y Z are sides automatically implies that a segment drawn from X to Z would
lie in the interior of the polygon. But what if the n-gon is not convex? How do we know
that we will be able to find three vertices X, Y , and Z such that XY and Y Z are sides and
the segment drawn from X to Z lies in the interior of the polygon? That is, if we do not
know that the n-gon is convex, we must also establish that there exist such vertices X, Y ,
and Z, or else we will have to find another way to prove the result than the method given
above! By the way, we call a segment drawn from vertices X to Z that lies in the interior of
a polygon, convex or not, a diagonal. Did you even consider the possibility that our n-gon
might not be convex?
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