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# Test bank for Discrete Mathematics with Application 4th Edition

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• ISBN-10 ‏ : ‎ 0495391328
• ISBN-13 ‏ : ‎ 978-0495391326

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SKU:tb1002111

## Test bank for Discrete Mathematics with Application 4th Edition

Discrete Mathematics with Applications, 4th Edition
by Susanna S. Epp
Answers for Test Bank Questions: Chapters 5-8
Please use caution when using these answers. Small differences in wording, notation, or choice of
examples or counterexamples may be acceptable.
Chapter 5
1. ∑
3
k=0
1
2
k
=
1
2
0
+
1
2
1
+
1
2
2
+
1
2
3
(= 1 + 1
2
+
1
4
+
1
8
=
15
8
)
2. ∑
4
k=1
k
2 = 12 + 22 + 32 + 42
(= 1 + 4 + 9 + 16 = 30)
3. 1
3 − 2
3 + 33 − 4
3 + 53 =

5
k=1
(−1)k−1k
2
(This is just one of a number of correct answers to this
question.)
4. 1 −
1
2
+
1
3

1
4
+
1
5

1
6
=

5
i=1
(−1)i+1 1
i
(This is just one of a number of correct answers to this
question.)
5. When k = 1, j = 1 + 1 = 2. When k = n, j = n + 1. Since j = k + 1, then k = j − 1. So
k
2
n
=
(j − 1)2
n
.
Therefore,
∑n
k=1
k
2
n
=
n∑
+1
j=2
(j − 1)2
n
.
6. When k = 0, i = 0 + 1 = 1. When k = n, i = n + 1. Since i = k + 1, then k = i − 1. So
k
2
k + n
=
(i − 1)2
i − 1 + n
.
Therefore,
∑n
k=0
k
2
k + n
=
n∑
+1
i=1
(i − 1)2
i − 1 + n
.
7.
2
2
2
2
2
2
2
0
1
3
6
12
25
51
103
remainder = r[6] = 1
remainder = r[5] = 1
remainder = r[4] = 0
remainder = r[3] = 0
remainder = r[2] = 1
remainder = r[1] = 1
remainder = r[0] = 1
Hence 10310 = 11001112.
8. 2 + 22 + 23 + · · · + 2m = 2(1 + 2 + 22 + · · · + 2m−1
) = 2 (
2
(m−1)+1 − 1
2 − 1
)
= 2(2m − 1).
2 + 22 + 23 + · · · + 2m = (1 + 2 + 22 + · · · + 2m) − 1 = (
2
m+1 − 1
2 − 1
)
− 1
= 2m+1 − 1 − 1 = 2m+1 − 2 [= 2(2m − 1)]
9. a. P(3) : ∑
3
i=3
i =
(3 − 2)(3 + 3)
2
P(3) is true because the left-hand side equals ∑
3
i=3
i = 3 and the right-hand side equals
(3 − 2)(3 + 3)
2
=
1 · 6
2
= 3 also.
b. (i) P(k): 3 + 4 + 5 + · · · + k =
(k − 2)(k + 3)
2
;
(ii) P(k + 1): 3 + 4 + 5 + · · · + (k + 1) = ((k + 1) − 2)((k + 1) + 3)
2
Or, equivalently, P(k + 1) is 3 + 4 + 5 + · · · + (k + 1) = (k − 1)(k + 4)
2
c. Proof that for all integers k ≥ 3, if P(k) is true then P(k + 1) is true:
Let k be any integer that is greater than or equal to 3, and suppose that
3 + 4 + 5 + · · · + k =
(k − 2)(k + 3)
2

P(k)
inductive hypothesis
We must show that
3 + 4 + 5 + · · · + (k + 1) = ((k + 1) − 2)((k + 1) + 3)
2
,
or, equivalently,
3 + 4 + 5 + · · · + (k + 1) = (k − 1)(k + 4)
2
. ← P(k + 1)
Now the left-hand side of P(k + 1) is
3 + 4 + 5 + · · · + (k + 1) = 3 + 4 + 5 + · · · + k + (k + 1)
by making the next-to-last term explicit
=
(k − 2)(k + 3)
2
+ (k + 1)
by inductive hypothesis
=
(k − 2)(k + 3)
2
+
2(k + 1)
2
by creating a common denominator
=
k
2 + k − 6
2
+
2k + 2
2
by multiplying out
=
k
2 + k − 6 + 2k + 2
2
=
k
2 + 3k − 4
2
by combining like terms.
And the right-hand side of P(k + 1) is
2
(k − 1)(k + 4)
2
=
k
2 + 4k − k − 4
2
=
k
2 + 3k − 4
2
.
Thus the left-hand and right-hand sides of P(k + 1) are equal [as was to be shown].
10. a. P(3): ∑
3
i=3
(i − 1) · i =
(3 − 2)(32 + 2 · 3 + 3)
3
P(3) is true because the left-hand side equals ∑
3
i=3
(i − 1) · i = (3 − 1) · 3 = 2 · 3 = 6, and the
right-hand side equals (3 − 2)(32 + 2 · 3 + 3)
3
=
1 · (9 + 6 + 3)
3
=
1 · 18
3
= 6 also.
b. P(k): 2 · 3 + 3 · 4 + · · · + (k − 1) · k =
(k − 2)(k
2 + 2k + 3)
3
;
P(k + 1): 2 · 3 + 3 · 4 + · · · + ((k + 1) − 1) · (k + 1) = ((k + 1) − 2)((k + 1)2 + 2(k + 1) + 3)
3
Or, equivalently, P(k + 1) is 2 · 3 + 3 · 4 + · · · + k(k + 1) = (k − 1)(k
2 + 2k + 1 + 2k + 2 + 3)
3
=
(k − 1)(k
2 + 4k + 6)
3
c. Proof that for all integers k ≥ 3, if P(k) is true then P(k + 1) is true:
Let k be any integer that is greater than or equal to 3, and suppose that
2 · 3 + 3 · 4 + · · · + (k − 1) · k =
(k − 2)(k
2 + 2k + 3)
3
. ←
P(k)
inductive hypothesis
We must show that
2 · 3 + 3 · 4 + · · · + k(k + 1) = (k − 1)(k
2 + 4k + 6)
3
. ← P(k + 1)
Now the left-hand side of P(k + 1) is
2 · 3 + 3 · 4 + · · · + k(k + 1) = 2 · 3 + 3 · 4 + · · · + (k − 1)k + k(k + 1)
by making the next-to-last term explicit
=
(k − 2)(k
2 + 2k + 3)
3
+ k(k + 1)
by inductive hypothesis
=
k
3 + 2k
2 + 3k − 2k
2 − 4k − 6
3
+
3k(k + 1)
3
by creating a common denominator and multiplying out
=
k
3 − k − 6
3
+
3k
2 + 3k
3
by multiplying out and combining like terms
=
k
3 + 3k
2 + 2k − 6
3
by adding the fractions and combining like terms.
And the right-hand side of P(k + 1) is
(k − 1)(k
2 + 4k + 6)
3
=
k
3 + 4k
2 + 6k − k
2 − 4k − 6
3
=
k
3 + 3k
2 + 2k − 6
3
.
Thus the left-hand and right-hand sides of P(k + 1) are equal [as was to be shown].
11. For each integer n ≥ 0, let P(n) be the equation
1 + 3 + 32 + · · · + 3n =
3
n+1 − 1
2
. ← P(n)
3
(Recall that by definition 1 + 3 + 32 + · · · + 3n =
∑n
i=0
3
i
.)
(a) P(0): ∑
0
i=0
3
i =
3
0+1 − 1
2
P(0) is true because the left-hand side equals ∑
0
i=0
3
i = 30 = 1, and the right-hand side
equals 3
0+1 − 1
2
=
3 − 1
2
= 1 also.
(b) P(k): 1 + 3 + 32 + · · · + 3k =
3
k+1 − 1
2
P(k + 1): 1 + 3 + 32 + · · · + 3k+1 =
3
(k+1)+1 − 1
2
,
Or, equivalently, P(k + 1) is 1 + 3 + 32 + · · · + 3k+1 =
3
k+2 − 1
2
.
(c) Proof that for all integers k ≥ 0, if P(k) is true then P(k + 1) is true:
Let k be any integer that is greater than or equal to 0, and suppose that
1 + 3 + 32 + · · · + 3k =
3
k+1 − 1
2
. ←
P(k)
inductive hypothesis
We must show that
1 + 3 + 32 + · · · + 3k+1 =
3
k+2 − 1
2
. ← P(k + 1)
Now the left-hand side of P(k + 1) is
1 + 3 + 32 + · · · + 3k+1 = 1 + 3 + 32 + · · · + 3k + 3k+1
by making the next-to-last term explicit
=
3
k+1 − 1
2
+ 3k+1
by inductive hypothesis
=
3
k+1 − 1
2
+
2 · 3
k+1
2
by creating a common denominator
=
3 · 3
k+1 − 1
2
by adding the fractions and combining like terms
=
3
k+2 − 1
2
by a law of exponents,
which equals the right-hand side of P(k + 1).
Thus the left-hand and right-hand sides of P(k + 1) are equal [as was to be shown].
12. Proof (by mathematical induction): Let the property P(n) be the equation
4 + 8 + 12 + · · · + 4n = 2n
2 + 2n. ← P(n)
Note that 4 + 8 + 12 + · · · + 4n =
∑n
i=1
4i.
4
Show that P(1) is true: P(1) is true because the left-hand side is ∑
1
i=1
4i = 4 · 1 = 4 and the
right-hand side is 2 · 1
2 + 2 · 1 = 4 also.
Show that for all integers k ≥ 1, if P(k) is true then P (k + 1) is true: Let k be
any integer with k ≥ 1, and suppose that
4 + 8 + 12 + · · · + 4k = 2k
2 + 2k. ←
P(k)
inductive hypothesis
We must show that
4 + 8 + 12 + · · · + 4(k + 1) = 2(k + 1)2 + 2(k + 1). ← P(k + 1)
Now the left-hand side of P(k + 1) is
4 + 8 + 12 + · · · + 4(k + 1) = 4 + 8 + 12 + · · · + 4k + 4(k + 1)
by making the next-to-last term explicit
= (2k
2 + 2k) + 4(k + 1)
by inductive hypothesis
= 2k
2 + 6k + 4
by multiplying out and combining like terms.
And the right-hand side of P(k + 1) is
2(k + 1)2 + 2(k + 1) = 2(k
2 + 2k + 1) + 2k + 2 = 2k
2 + 4k + 2 + 2k + 2 = 2k
2 + 6k + 4.
Thus the left-hand and right-hand sides of P(k + 1) are equal [as was to be shown].
13. Proof (by mathematical induction): Let the property P(n) be the equation
3 + 4 + 5 + · · · + n =
(n − 2)(n + 3)
2
. ← P(n)
Note that 3 + 4 + 5 + · · · + n =
∑n
i=3
i.
Show that P(3) is true: P(3) is true because the left-hand side is ∑
3
i=3
i = 3 and the righthand side is (3 − 2)(3 + 3)
2
= 3 also.
Show that for all integers k ≥ 3, if P(k) is true then P (k + 1) is true: Let k be
any integer with k ≥ 1, and suppose that
3 + 4 + 5 + · · · + k =
(k − 2)(k + 3)
2
. ←
P(k)
inductive hypothesis
We must show that
3 + 4 + 5 + · · · + (k + 1) = [(k + 1) − 2][(k + 1) + 3]
2
. ← P(k + 1)
Now the left-hand side of P(k + 1) is
5
3 + 4 + 5 + · · · + (k + 1) = 3 + 4 + 5 + · · · + k + (k + 1)
by making the next-to-last term explicit
=
(k − 2)(k + 3)
2
+ (k + 1)
by inductive hypothesis
=
(k − 2)(k + 3)
2
+
2 · (k + 1)
2
by creating a common denominator
=
k
2 + k − 6
2
+
2k + 2
2
by multiplying out
=
k
2 + 3k − 4
2
by adding fractions and combining like terms.
And the right-hand side of P(k + 1) is
[(k + 1) − 2][(k + 1) + 3]
2
=
(k − 1)(k + 4)
2
=
k
2 + 3k − 4
2
.
Thus the left-hand and right-hand sides of P(k + 1) are equal [as was to be shown].
14. Proof (by mathematical induction): Let the property P(n) be the equation
2 · 3 + 3 · 4 + · · · + (n − 1) · n =
(n − 2)(n
2 + 2n + 3)
3
. ← P(n)
Note that 2 · 3 + 3 · 4 + · · · + (n − 1) · n =
∑n
i=3
(i − 1) · i.
Show that P(3) is true: P(3) is true because the left-hand side is ∑
3
i=3
(i−1)·i = (3−1)·3 = 6
and the right-hand side is (3 − 2)(32 + 2 · 3 + 3)
3
=
1 · (9 + 6 + 3)
3
= 6 also.
Show that for all integers k ≥ 3, if P(k) is true then P (k + 1) is true: Let k be
any integer with k ≥ 1, and suppose that
2·3+3·4+· · ·+(k−1)·k =
(k − 2)(k
2 + 2k + 3)
3
. ←
P(k)
inductive hypothesis
We must show that
2·3+3·4+· · ·+((k+1)−1)·(k+1) = ((k + 1) − 2)((k + 1)2 + 2(k + 1) + 3)
3
. ← P(k + 1)
6
Now the left-hand side of P(k + 1) is
2 · 3 + 3 · 4 + · · · + ((k + 1) − 1) · (k + 1)
= 2 · 3 + 3 · 4 + · · · + (k − 1) · k + k · (k + 1)
by simplifying the last term and making the next-to-last term explicit
=
(k − 2)(k
2 + 2k + 3)
3
+ k · (k + 1)
by inductive hypothesis
=
k
3 + 2k
2 + 3k − 2k
2 − 4k − 6
3
+ k
2 + k
by combining like terms
=
k
3 − k − 6
3
+
3 · (k
2 + k)
3
by creating a common denominator
=
k
3 − k − 6
3
+
3k
2 + 3k
3
by multiplying out
=
k
3 + 3k
2 + 2k − 6
3
by adding fractions and combining like terms.
And the right-hand side of P(k + 1) is
((k + 1) − 2)((k + 1)2 + 2(k + 1) + 3)
3
=
(k − 1)(k
2 + 2k + 1 + 2k + 2 + 3)
3
=
(k − 1)(k
2 + 4k + 6)
3
=
(k − 1)(k
2 + 4k + 6)
3
=
k
3 + 4k
2 + 6k − k
2 − 4k − 6
3
=
k
3 + 3k
2 + 2k − 6
3
.
Thus the left-hand and right-hand sides of P(k + 1) are equal [as was to be shown].
15. Proof (by mathematical induction): Let the property P(n) be the equation
1 + 3 + 32 + · · · + 3n =
3
n+1 − 1
2
. . ← P(n)
Note that 1 + 3 + 32 + · · · + 3n =
∑n
i=0
3
i
.
Show that P(0) is true: P(0) is true because the left-hand side is ∑
0
i=0
3
i = 30 = 1, and the
right-hand side is 3
0+1 − 1
2
=
3 − 1
2
= 1 also.
Show that for all integers k ≥ 0, if P(k) is true then P (k + 1) is true: Let k be
any integer with k ≥ 0, and suppose that
1 + 3 + 32 + · · · + 3k =
3
k+1 − 1
2
. ←
P(k)
inductive hypothesis

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