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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
File size: 2791 byte(s)
`Rogawski problems contributed by publisher WHFreeman. These are a subset of the problems available to instructors who use the Rogawski textbook. The remainder can be obtained from the publisher.`

```    1 ## DBsubject('Calculus')
2 ## DBchapter('Infinite Sequences and Series')
3 ## DBsection('Infinite Sequences and Series')
4 ## KEYWORDS('calculus', 'series')
5 ## TitleText1('Calculus: Early Transcendentals')
6 ## EditionText1('2')
7 ## AuthorText1('Rogawski')
8 ## Section1('10.1')
9 ## Problem1('13')
10 ## Author('Danny Glin')
11 ## Institution('W.H.Freeman')
12
13 DOCUMENT();
17
18 Context()->texStrings;
19 Context()->flags->set(reduceConstants=>0);
20 Context()->variables->are(n=>'Real');
21
22 #\$k=3
23 \$k=random(2,4,1);
24
25 \$andenom=Formula("n^(\$k)");
26
27 foreach \$i (1..3) {
28 \$denom[\$i]=\$andenom->eval(n=>\$i);
29 \$numer[\$i]=(-1)**(\$i+1);
30 \$a[\$i]=Formula("\$numer[\$i]/\$denom[\$i]");
31 }
32
33 #\$a1=Formula("1/1");
34 #
35 #\$d2=2**\$k;
36 #\$a2=Formula("-1/\$d2");
37 #
38 #\$d3=3**\$k;
39 #\$a3=Formula("1/\$d3");
40
41 \$an=Formula("(-1)^(n+1)/n^(\$k)");
42 \$an->{test_points} = [[1],[2],[3],[4],[5],[6]];
43
44 BEGIN_TEXT
45 \{ beginproblem() \}
46 \{ textbook_ref_exact("Rogawski ET 2e", "10.1","13") \}
47 \$PAR
48 Find a formula for the \(n\)th term of the following sequences:
49
50 \$PAR
51 (a) \(\$a[1],\$a[2],\$a[3],\ldots\)
52 \$PAR
53 \( a_n = \) \{ans_rule()\}\$BR
54 END_TEXT
55 Context()->normalStrings;
56
57 ANS(\$an->cmp);
58
59 Context()->texStrings;
60 @ord=('second','third','fourth');
61 \$m=\$k-2;
62 \$neg=(-1)**\$k;
63 \$antrans=Formula("\$neg/n^(\$k)");
64
65 SOLUTION(EV3(<<'END_SOLUTION'));
66 \$PAR
67 \$SOL
68 The denominators are the \$ord[\$m] powers of the positive integers starting with \(n=1\).  Also, the sign of the terms is alternating with the sign of the first term being positive.  Thus,
69 \[
70 a_1=\{\$antrans->substitute(n=>1)\}=\{\$an->substitute(n=>1)\},\
71 a_2=\{\$antrans->substitute(n=>2)\}=\{\$an->substitute(n=>2)\},\
72 a_3=\{\$antrans->substitute(n=>3)\}=\{\$an->substitute(n=>3)\}
73 \]
74 This rule leads to the following formula for the \(n\)th term:
75 \[
76 a_n=\$an
77 \]
78 END_SOLUTION
79
80 #\$a=1
81 #\$b=5
82
83 \$p = random(1,3,1);
84 \$q = \$p + random(1,3,1);
85
86 \$bnumer=Formula("n+\$p");
87 \$bdenom=Formula("n+\$q");
88
89 \$bn=Formula("(\$bnumer)/(\$bdenom)");
90 \$bn->{test_points} = [[1],[2],[3],[4],[5],[6]];
91
92 foreach \$j (1..3) {
93 \$bnum[\$j]=\$bnumer->eval(n=>\$j);
94 \$bden[\$j]=\$bdenom->eval(n=>\$j);
95 \$b[\$j]=Formula("\$bnum[\$j]/\$bden[\$j]");
96 }
97
98 BEGIN_TEXT
99 \$PAR
100 (b) \(\$b[1],\$b[2],\$b[3],\ldots\)
101 \$PAR
102 \( b_n = \) \{ans_rule()\}\$BR
103 \$BITALIC Assume a starting index of \(n=1\). \$EITALIC
104
105 END_TEXT
106 Context()->normalStrings;
107
108 ANS(\$bn->cmp);
109
110 Context()->texStrings;
111 \$diff = \$q-\$p;
112
113 SOLUTION(EV3(<<'END_SOLUTION'));
114 \$PAR
115 \$SOL
116 Assuming a starting index of \(n=1\), we see that each numerator is \$p more than the index and the denominator is \$diff more than the numerator.  Thus, the general term \(b_n\) is:
117 \[
118 b_n=\$bn
119 \]
120 END_SOLUTION
121
122
123 ENDDOCUMENT();
```