Practice Workbook Answers
(continued)
d. A(2,3) + A(4,15) 2(4x 15) 3
8x 27
A(8,27);
A(4,15) + A(2,3) 4(2x 3) 15
8x 27
A(8,27); the result of the
compositions are the same even
though the affine transformations
are not in the same order.
6. a.
3. a. A(1,8)(4) 1(4) 8 4 8
4;
A(3,8)(4) 3(4) 8 12 8
4
b. AA(1,8) + A(3,8)B(4) A(1,8)(4) 4 because each
step in the composition is in S4
and therefore returns an output of
4. Similarly, AA(3,8) + A(1,8)B(4)
A(3,8)(4) 4.
c. From part (a), A(1,8)(4) 4.
Apply AA(1,8)B1 to each side of
the equation to get
4 AA(1,8)B1(4). Similarly,
apply AA(3,8)B1 to each side of
A(3,8)(4) 4 to get
4 AA(3,8)B1. So AA(1,8)B1
and AA(3,8)B1 are both in S4.
b.
4
y
2
x
4
O
2
2
4
2
4
2
4
4
y
2
x
4
O
2
2
c.
y
4
2
x
4
2 O
2
4
2
4. a. AT2 + D5 + AT2B1B(2) AT2 + D5B(0)
T2(0) 2
b. AT2 + D3 + AT2B1B(2) AT2 + D3B(0)
T2(0) 2
c. AT2 + D1 + AT2B1B(2) AT2 + D1B(0)
T2(0) 2
d. AT2 + Da + AT2B1B(2) AT2 + DaB(0)
T2(0) 2
Chapter 7
Lesson 7.2 Additional Practice
1. a. 21
c. 125,250
b. 78
2. 2n 2 n
5. a. A(81,80)
b. To compute f (4), start with f (0)
and apply the recursion step,
A(3,2), to it four times.
c. 1
3. 74
n(n 1 1)
2
3n(3n 1 1)
c.
2
4. a.
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b. n(2n 1)
d.
kn(kn 1 2)
8
Practice Workbook Answers
5. n
5. <0.9999999404
g(n)
0
1
1
4
3
2
13
9
3
40
27
4
121
81
5
364
243
6. a. 20
d. 51
2. a. 619
4. a. 420
b. 255
n(n 1 1)
2
2
n(n 1 1)
b. 2 1
2
d. 6630
7. a b. 1,594,296
7
7
c. 10
3 a 10 2 Q 10 R b < 2.267
Lessons 7.3 and 7.4
Additional Practice
r (n)
0
1
1
1
4
5
2
7
12
3
10
22
c. 5n
n(n 1 1)
n(n 1 1)
b
2
2
1. a. 153
c. 2n 2 3n 2
e. 2186
b. 125
d. 1093
f. 3n1 1
2. a. 280
c. 40
b. 224
d. 89,153
3. r(n) 6x 5
(n 1 1)(3n 1 2)
2
4. No; since you want the polynomial
to agree with s for all integers n 0,
there is only one way to get the set
of running totals with a polynomial
function.
n
2. Answers may vary. Sample: 1 Q 16 R
1 ; 21
b. 64
64
c. 1n ; 13 a1 1nb
4
4
5. 83,166
n
d. 5n
Lessons 7.7 and 7.8
Additional Practice
b. r(n) 3n 1
c. Answers may vary. Sample:
1; 5
3. a. 16
16
b. 40
8. a. 11,718
b. 22,888,183,593
c. 22,888,171,875
d. 34 (5n1 1)
11
n
b. 72
5. a. 3 6. a. 40
c. 1824
7. a. 797,148
4. a. a 1j
j=1 4
b. 103
3. n(n 1)
1 2 Q1
3R
g(n) 1 2 Q1
3R
12
c. 50
f. 32
2
1. 3n 2 24n 2 1
n
1. a.
b. 80
e. 165
Lesson 7.6 Additional Practice
Answers may vary. Sample:
6. a. 35
(continued)
b. a 1j
j=1 4
6. F(n) 16 `
c. a 1j
j=1 4
n(3n 2 1)
2
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Practice Workbook Answers
7. a.
n
t (n)
0
6
1
1
7
3
2
10
5
3
15
t(n) e
Lessons 7.12 and 7.13
Additional Practice
32 64
1. a. sequence: 12, 8, 16
3 , 9 , 27 ;
7 ; yes; 36.
series: 12, 20, 2513, 2889, 3127
6
t(n 2 1) 1 2n 2 1
b. sequence: 6, 12, 24, 48, 96;
series: 6, 6, 18, 30, 66; no
c. sequence: 16, 6.4, 2.56, 1.024,
0.4096;
series: 16, 22.4, 24.96, 25.984,
26.3936; yes; 26.67
d. sequence: 15, 22.5, 33.75,
50.625, 75.9375;
series: 15, 37.5, 71.25,
121.875, 197.8125; no
if n 0
if n 0
b. T(n) 6 n 2
n
8. a. r(n) a 3k
k=0
(continued)
b. 12 A3n1 1B
Lessons 7.10 and 7.11
Additional Practice
2. a.
1. a. 58
b. 7n 5
c. Yes; it is the 269th term.
2. 1, 4, 7, 10, 13, 16
3. 16, 12, 8, 4, 0
Stage 3
4. a. 6,144
b. <153.77
c. 128
125
b. Stage 1: 53 ; Stage 2: 25
9 ; Stage 3: 27
5. a. 12(2)n1
d. No; as n increases, Q 53 R gets larger
b. 4 Q 32 R
n
c. Q 53 R
n
and larger without bound.
n21
3. a. 16
7 < 2.286
b. 10
c. 4"2 A"2 B n1
6. a. 21 sit-ups
b. 107 sit-ups
c. 10(1.2)w1 sit-ups
45
4. a. a 1000
1
b. r 1000
5
c. 111
7. $25,848
8. a. <0.78 ft
b. <87.66 ft
c. <58.43 ft
d. 150 ft
26
5. a. 33
d. 17
99
b. 13
33
7
c. 333
7
e. 11
115
f. 333
Lesson 7.15 Additional Practice
1. a. 16
c. 512
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183
b. 256
d. 2048
Practice Workbook Answers
5. n 1 entries
2. There are 45 even numbers in row 60.
There are 75 even numbers in row 90.
Answers may vary. Sample: To find the
number of even numbers in a given
row, find how many numbers are in
that row and how many are odd.
Row k has a total of k 1 numbers.
In row k there are 2n odd numbers,
where n is the number of 1’s if k is
written in binary form. Thus, row k
has k 1 2n even numbers.
Row 60 has 61 numbers in it. To find
the number of odd numbers in row
60, write 60 inbinary as 111100
(32 16 8 4). There are four 1’s
in the binary representation, so there
are 24 16 odd numbers. Row 60
has 61 16 45 even numbers
(60 1 24 45).
6. a. row 0: 1, row 1: 2, row 2: 4, row 3: 8,
row 4: 16, row 5: 32; the sum of
each row n is 2n.
b. 2048
7. 5005
Lesson 7.16 Additional Practice
1. a. 256x 8 1024x7y 1792x 6y 2
1792x 5y 3 1120x 4y 4 448x 3y 5
112x 2y 6 16xy7 y 8
b. x 6 18x 5y 135x 4y 2 540x 3y 3
1215x 2y 4 1458xy 5 729y 6
c. 256x 8 3072x7y 16,128x 6y 2
48,384x 5y 3 90,720x 4y 4
108,864x 3y 5 81,648x 2y 6
34,992xy7 6561y 8
d. 729x 6 5832x 5y 19,440x 4y 2
34,560x 3y 3 34,560x 2y 4
18,432xy 5 4096y 6
3. a. m 4, 5
b. m 4, 1
c. m 5, 9
d. m 49, n 9, 40
4.
1
1
1
1
1
1
1
1
1
1
1
7
8
9
4
5
6
3
1
3
6
1
b. a 5b 4
c. none
3. a. 66
b. 495
c. 792
5. a. 0
4
10 10
2. a. 126
4. a. 1,031,250,000
b. 1,237,500,000
c. 792,000,000
1
2
(continued)
1
5
6. Answers may vary. Sample:
x 6 12x 5 60x 4 160x 3 240x 2
192x 64 factors to (x 2)6.
1
15 20 15
6
21 35 35 21
1
7
28 56 70 56 28
36 84 126 126 84 36
b. 0; 0; 0
1
8
7. a. 350
b. (3)100 3100
1
9
10 45 120 210 252 210 120 45 10
1
1
Answers may vary. Sample: The first
and last entry of each row is 1. To find
each remaining entry in the row, add
the two consecutive entries directly
above it.
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