{"id":98,"date":"2021-03-25T02:21:00","date_gmt":"2021-03-25T02:21:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/sequences-2\/"},"modified":"2022-01-03T18:29:21","modified_gmt":"2022-01-03T18:29:21","slug":"sequences-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/sequences-2\/","title":{"raw":"Problem Set: Sequences","rendered":"Problem Set: Sequences"},"content":{"raw":"

Find the first six terms of each of the following sequences, starting with [latex]n=1[\/latex].<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]{a}_{n}=1+{\\left(-1\\right)}^{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"756543\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"756543\"][latex]{a}_{n}=0[\/latex] if [latex]n[\/latex] is odd and [latex]{a}_{n}=2[\/latex] if [latex]n[\/latex] is even[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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2.\u00a0<\/strong>[latex]{a}_{n}={n}^{2}-1[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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3.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n}={a}_{n - 1}+n[\/latex] for [latex]n\\ge 2[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"957886\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"957886\"][latex]\\left\\{{a}_{n}\\right\\}=\\left\\{1,3,6,10,15,21\\text{,... }\\right\\}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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4.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=1[\/latex] and [latex]{a}_{n+2}={a}_{n}+{a}_{n+1}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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5.\u00a0<\/strong>Find an explicit formula for [latex]{a}_{n}[\/latex] where [latex]{a}_{1}=1[\/latex] and [latex]{a}_{n}={a}_{n - 1}+n[\/latex] for [latex]n\\ge 2[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"664489\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"664489\"][latex]{a}_{n}=\\frac{n\\left(n+1\\right)}{2}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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6.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the arithmetic sequence whose first term is [latex]{a}_{1}=1[\/latex] such that [latex]{a}_{n - 1}-{a}_{n}=17[\/latex] for [latex]n\\ge 1[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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7.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the arithmetic sequence whose first term is [latex]{a}_{1}=-3[\/latex] such that [latex]{a}_{n - 1}-{a}_{n}=4[\/latex] for [latex]n\\ge 1[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"376785\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"376785\"][latex]{a}_{n}=4n - 7[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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8.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the geometric sequence whose first term is [latex]{a}_{1}=1[\/latex] such that [latex]\\frac{{a}_{n+1}}{{a}_{n}}=10[\/latex] for [latex]n\\ge 1[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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9.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the geometric sequence whose first term is [latex]{a}_{1}=3[\/latex] such that [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{1}{10}[\/latex] for [latex]n\\ge 1[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"735660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"735660\"][latex]{a}_{n}={3.10}^{1-n}={30.10}^{\\text{-}n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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10.\u00a0<\/strong>Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence whose first several terms are [latex]\\left\\{0,3,8,15,24,35,48,63,80,99\\text{,... }\\right\\}[\/latex]. (Hint:<\/em> First add one to each term.)<\/div>\r\n<\/div>\r\n<\/div>\r\n
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11.\u00a0<\/strong>Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence satisfying [latex]{a}_{1}=0[\/latex] and [latex]{a}_{n}=2{a}_{n - 1}+1[\/latex] for [latex]n\\ge 2[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"231098\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"231098\"][latex]{a}_{n}={2}^{n}-1[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

Find a formula for the general term [latex]{a}_{n}[\/latex] of each of the following sequences.<\/p>\r\n\r\n

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12.\u00a0<\/strong>[latex]\\left\\{1,0,-1,0,1,0,-1,0\\text{,... }\\right\\}[\/latex] (Hint:<\/em> Find where [latex]\\sin{x}[\/latex] takes these values)<\/div>\r\n<\/div>\r\n<\/div>\r\n
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13.\u00a0<\/strong>[latex]\\left\\{1,\\text{-}\\frac{1}{3},\\frac{1}{5},\\text{-}\\frac{1}{7}\\text{,... }\\right\\}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"589164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589164\"][latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n - 1}}{2n - 1}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

Find a function [latex]f\\left(n\\right)[\/latex] that identifies the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] of the following recursively defined sequences, as [latex]{a}_{n}=f\\left(n\\right)[\/latex].<\/p>\r\n\r\n

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14.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\text{-}{a}_{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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15.\u00a0<\/strong>[latex]{a}_{1}=2[\/latex] and [latex]{a}_{n+1}=2{a}_{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"188550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"188550\"][latex]f\\left(n\\right)={2}^{n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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16.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\left(n+1\\right){a}_{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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17.\u00a0<\/strong>[latex]{a}_{1}=2[\/latex] and [latex]{a}_{n+1}=\\frac{\\left(n+1\\right){a}_{n}}{2}[\/latex] for [latex]n\\ge 1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"23498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"23498\"][latex]f\\left(n\\right)=\\frac{n\\text{!}}{{2}^{n - 2}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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18.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\frac{{a}_{n}}{{2}^{n}}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

Plot the first [latex]N[\/latex] terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.<\/p>\r\n\r\n

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19. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], and for [latex]n\\ge 2[\/latex], [latex]{a}_{n}=\\frac{1}{2}\\left({a}_{n - 1}+{a}_{n - 2}\\right)[\/latex]; [latex]N=30[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n[reveal-answer q=\"749048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"749048\"]Terms oscillate above and below [latex]\\frac{5}{3}[\/latex] and appear to converge to [latex]\\frac{5}{3}[\/latex]. \r\n<\/span>\r\n\r\n\r\n\"This<\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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20. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], [latex]{a}_{3}=3[\/latex] and for [latex]n\\ge 4[\/latex], [latex]{a}_{n}=\\frac{1}{3}\\left({a}_{n - 1}+{a}_{n - 2}+{a}_{n - 3}\\right)[\/latex], [latex]N=30[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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21. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], and for [latex]n\\ge 3[\/latex], [latex]{a}_{n}=\\sqrt{{a}_{n - 1}{a}_{n - 2}}[\/latex]; [latex]N=30[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n[reveal-answer q=\"999702\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"999702\"]Terms oscillate above and below [latex]y\\approx 1.57..[\/latex]. and appear to converge to a limit.\r\n<\/span>\r\n\r\n\r\n\"This<\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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22. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], [latex]{a}_{3}=3[\/latex], and for [latex]n\\ge 4[\/latex], [latex]{a}_{n}=\\sqrt{{a}_{n - 1}{a}_{n - 2}{a}_{n - 3}}[\/latex]; [latex]N=30[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

Suppose that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=1[\/latex], [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=-1[\/latex], and [latex]0<\\text{-}{b}_{n}<{a}_{n}[\/latex] for all [latex]n[\/latex]. Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.<\/p>\r\n\r\n

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23.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\left(3{a}_{n}-4{b}_{n}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"861296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"861296\"][latex]7[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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24.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\left(\\frac{1}{2}{b}_{n}-\\frac{1}{2}{a}_{n}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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25.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}+{b}_{n}}{{a}_{n}-{b}_{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"747950\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"747950\"][latex]0[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

26.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}-{b}_{n}}{{a}_{n}+{b}_{n}}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n

Find the limit of each of the following sequences, using L\u2019H\u00f4pital\u2019s rule when appropriate.<\/p>\r\n\r\n

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27.\u00a0<\/strong>[latex]\\frac{{n}^{2}}{{2}^{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"581461\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"581461\"][latex]0[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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28.\u00a0<\/strong>[latex]\\frac{{\\left(n - 1\\right)}^{2}}{{\\left(n+1\\right)}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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29.\u00a0<\/strong>[latex]\\frac{\\sqrt{n}}{\\sqrt{n+1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"954\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"954\"][latex]1[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

30.\u00a0<\/strong>[latex]{n}^{\\frac{1}{n}}[\/latex] (<\/span>Hint:<\/em> [latex]{n}^{\\frac{1}{n}}={e}^{\\frac{1}{n}}\\text{ln}n[\/latex])<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n

For each of the following sequences, whose [latex]n\\text{th}[\/latex] terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.<\/p>\r\n\r\n

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31.\u00a0<\/strong>[latex]\\frac{n}{{2}^{n}}[\/latex], [latex]n\\ge 2[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"962850\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"962850\"]bounded, decreasing for [latex]n\\ge 1[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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32.\u00a0<\/strong>[latex]\\text{ln}\\left(1+\\frac{1}{n}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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33.\u00a0<\/strong>[latex]\\sin{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"389753\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"389753\"]bounded, not monotone[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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34.\u00a0<\/strong>[latex]\\cos\\left({n}^{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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35.\u00a0<\/strong>[latex]{n}^{\\frac{1}{n}}[\/latex], [latex]n\\ge 3[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"927481\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"927481\"]bounded, decreasing[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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36.\u00a0<\/strong>[latex]{n}^{\\frac{-1}{n}}[\/latex], [latex]n\\ge 3[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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37.\u00a0<\/strong>[latex]\\tan{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"350782\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350782\"]not monotone, not bounded[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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38.\u00a0<\/strong>Determine whether the sequence defined as follows has a limit. If it does, find the limit.<\/p>\r\n

[latex]{a}_{1}=\\sqrt{2}[\/latex], [latex]{a}_{2}=\\sqrt{2\\sqrt{2}}[\/latex], [latex]{a}_{3}=\\sqrt{2\\sqrt{2\\sqrt{2}}}[\/latex] etc.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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39.\u00a0<\/strong>Determine whether the sequence defined as follows has a limit. If it does, find the limit.<\/p>\r\n

[latex]{a}_{1}=3[\/latex], [latex]{a}_{n}=\\sqrt{2{a}_{n - 1}}[\/latex], [latex]n=2,3\\text{,... }[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"168909\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"168909\"][latex]{a}_{n}[\/latex] is decreasing and bounded below by [latex]2[\/latex]. The limit [latex]a[\/latex] must satisfy [latex]a=\\sqrt{2a}[\/latex] so [latex]a=2[\/latex], independent of the initial value.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

Use the Squeeze Theorem to find the limit of each of the following sequences.<\/p>\r\n\r\n

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40.\u00a0<\/strong>[latex]n\\sin\\left(\\frac{1}{n}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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41.\u00a0<\/strong>[latex]\\frac{\\cos\\left(\\frac{1}{n}\\right)-1}{\\frac{1}{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"25509\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"25509\"][latex]0[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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42.\u00a0<\/strong>[latex]{a}_{n}=\\frac{n\\text{!}}{{n}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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43.\u00a0<\/strong>[latex]{a}_{n}=\\sin{n}\\sin\\left(\\frac{1}{n}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"151303\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"151303\"][latex]0:|\\sin{x}|\\le |x|[\/latex] and [latex]|\\sin{x}|\\le 1[\/latex] so [latex]-\\frac{1}{n}\\le {a}_{n}\\le \\frac{1}{n}[\/latex]).[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

For the following sequences, plot the first [latex]25[\/latex] terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.<\/p>\r\n\r\n

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44. [T]<\/strong> [latex]{a}_{n}=\\sin{n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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45. [T]<\/strong> [latex]{a}_{n}=\\cos{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n[reveal-answer q=\"347599\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"347599\"]Graph oscillates and suggests no limit.\r\n<\/span>\r\n\r\n\r\n\"This<\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.<\/p>\r\n\r\n

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46.\u00a0<\/strong>[latex]{a}_{n}={\\tan}^{-1}\\left({n}^{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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47.\u00a0<\/strong>[latex]{a}_{n}={\\left(2n\\right)}^{\\frac{1}{n}}-{n}^{\\frac{1}{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"53633\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"53633\"][latex]{n}^{\\frac{1}{n}}\\to 1[\/latex] and [latex]{2}^{\\frac{1}{n}}\\to 1[\/latex], so [latex]{a}_{n}\\to 0[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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48.\u00a0<\/strong>[latex]{a}_{n}=\\frac{\\text{ln}\\left({n}^{2}\\right)}{\\text{ln}\\left(2n\\right)}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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49.\u00a0<\/strong>[latex]{a}_{n}={\\left(1-\\frac{2}{n}\\right)}^{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"782186\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"782186\"]Since [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}\\to e[\/latex], one has [latex]{\\left(1 - \\frac{2}{n}\\right)}^{n}\\approx {\\left(1+k\\right)}^{-2k}\\to {e}^{-2}[\/latex] as [latex]k\\to \\infty [\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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50.\u00a0<\/strong>[latex]{a}_{n}=\\text{ln}\\left(\\frac{n+2}{{n}^{2}-3}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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51.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{2}^{n}+{3}^{n}}{{4}^{n}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"121462\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"121462\"][latex]{2}^{n}+{3}^{n}\\le 2\\cdot {3}^{n}[\/latex] and [latex]\\frac{{3}^{n}}{{4}^{n}}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex], so [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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52.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{\\left(1000\\right)}^{n}}{n\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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53.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{\\left(n\\text{!}\\right)}^{2}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\r\n[reveal-answer q=\"254122\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"254122\"][latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{n\\text{!}}{\\left(n+1\\right)\\left(n+2\\right)}\\text{... }\\left(2n\\right)[\/latex] [latex]=\\frac{1\\cdot 2\\cdot 3\\text{... }n}{\\left(n+1\\right)\\left(n+2\\right)\\text{... }\\left(2n\\right)}<\\frac{1}{{2}^{n}}[\/latex]. In particular, [latex]\\frac{{a}_{n+1}}{{a}_{n}}\\le \\frac{1}{2}[\/latex], so [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n

<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

Newton\u2019s method seeks to approximate a solution [latex]f\\left(x\\right)=0[\/latex] that starts with an initial approximation [latex]{x}_{0}[\/latex] and successively defines a sequence [latex]{x}_{n+1}={x}_{n}-\\frac{f\\left({x}_{n}\\right)}{{f}^{\\prime }\\left({x}_{n}\\right)}[\/latex]. For the given choice of [latex]f[\/latex] and [latex]{x}_{0}[\/latex], write out the formula for [latex]{x}_{n+1}[\/latex]. If the sequence appears to converge, give an exact formula for the solution [latex]x[\/latex], then identify the limit [latex]x[\/latex] accurate to four decimal places and the smallest [latex]n[\/latex] such that [latex]{x}_{n}[\/latex] agrees with [latex]x[\/latex] up to four decimal places.<\/p>\r\n\r\n

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54. [T]<\/strong> [latex]f\\left(x\\right)={x}^{2}-2[\/latex], [latex]{x}_{0}=1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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55. [T]<\/strong> [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}-2[\/latex], [latex]{x}_{0}=2[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"207231\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"207231\"][latex]{x}_{n+1}={x}_{n}-\\frac{\\left({\\left({x}_{n}-1\\right)}^{2}-2\\right)}{2}\\left({x}_{n}-1\\right)[\/latex]; [latex]x=1+\\sqrt{2}[\/latex], [latex]x\\approx 2.4142[\/latex], [latex]n=5[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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56. [T]<\/strong> [latex]f\\left(x\\right)={e}^{x}-2[\/latex], [latex]{x}_{0}=1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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57. [T]<\/strong> [latex]f\\left(x\\right)=\\text{ln}x - 1[\/latex], [latex]{x}_{0}=2[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"587075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"587075\"][latex]{x}_{n+1}={x}_{n}-{x}_{n}\\left(\\text{ln}\\left({x}_{n}\\right)-1\\right)[\/latex]; [latex]x=e[\/latex], [latex]x\\approx 2.7183[\/latex], [latex]n=5[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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58. [T]<\/strong> Suppose you start with one liter of vinegar and repeatedly remove [latex]0.1\\text{L,}[\/latex] replace with water, mix, and repeat.<\/p>\r\n\r\n

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  1. Find a formula for the concentration after [latex]n[\/latex] steps.<\/li>\r\n \t
  2. After how many steps does the mixture contain less than [latex]10\\text{%}[\/latex] vinegar?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    59. [T]<\/strong> A lake initially contains [latex]2000[\/latex] fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by [latex]6\\text{%}[\/latex] each month. However, factoring in all causes, [latex]150[\/latex] fish are lost each month.<\/p>\r\n\r\n

      \r\n \t
    1. Explain why the fish population after [latex]n[\/latex] months is modeled by [latex]{P}_{n}=1.06{P}_{n - 1}-150[\/latex] with [latex]{P}_{0}=2000[\/latex].<\/li>\r\n \t
    2. How many fish will be in the pond after one year?<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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      [reveal-answer q=\"156653\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"156653\"]a. Without losses, the population would obey [latex]{P}_{n}=1.06{P}_{n - 1}[\/latex]. The subtraction of [latex]150[\/latex] accounts for fish losses. b. After [latex]12[\/latex] months, we have [latex]{P}_{12}\\approx 1494[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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      60. [T]<\/strong> A bank account earns [latex]5\\text{%}[\/latex] interest compounded monthly. Suppose that [latex]\\$\\text{1000}[\/latex] is initially deposited into the account, but that [latex]\\$\\text{10}[\/latex] is withdrawn each month.<\/p>\r\n\r\n

        \r\n \t
      1. Show that the amount in the account after [latex]n[\/latex] months is [latex]{A}_{n}=\\left(1+.\\frac{05}{12}\\right){A}_{n - 1}-10[\/latex]; [latex]{A}_{0}=1000[\/latex].<\/li>\r\n \t
      2. How much money will be in the account after [latex]1[\/latex] year?<\/li>\r\n \t
      3. Is the amount increasing or decreasing?<\/li>\r\n \t
      4. Suppose that instead of [latex]\\$\\text{10}[\/latex], a fixed amount [latex]d[\/latex] dollars is withdrawn each month. Find a value of [latex]d[\/latex] such that the amount in the account after each month remains [latex]\\$\\text{1000}[\/latex].<\/li>\r\n \t
      5. What happens if [latex]d[\/latex] is greater than this amount?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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        61. [T]<\/strong> A student takes out a college loan of [latex]\\$\\text{10,000}[\/latex] at an annual percentage rate of [latex]6\\text{%}[\/latex], compounded monthly.<\/p>\r\n\r\n

          \r\n \t
        1. If the student makes payments of [latex]\\$\\text{100}[\/latex] per month, how much does the student owe after [latex]12[\/latex] months?<\/li>\r\n \t
        2. After how many months will the loan be paid off?<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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          [reveal-answer q=\"930694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930694\"]a. The student owes [latex]\\$\\text{9383}[\/latex] after [latex]12[\/latex] months. b. The loan will be paid in full after [latex]139[\/latex] months or eleven and a half years.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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          62. [T]<\/strong> Consider a series combining geometric growth and arithmetic decrease. Let [latex]{a}_{1}=1[\/latex]. Fix [latex]a>1[\/latex] and [latex]0<b<a[\/latex]. Set [latex]{a}_{n+1}=a.{a}_{n}-b[\/latex]. Find a formula for [latex]{a}_{n+1}[\/latex] in terms of [latex]{a}^{n}[\/latex], [latex]a[\/latex], and [latex]b[\/latex] and a relationship between [latex]a[\/latex] and [latex]b[\/latex] such that [latex]{a}_{n}[\/latex] converges.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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          63. [T]<\/strong> The binary representation [latex]x=0.{b}_{1}{b}_{2}{b}_{3}..[\/latex]. of a number [latex]x[\/latex] between [latex]0[\/latex] and [latex]1[\/latex] can be defined as follows. Let [latex]{b}_{1}=0[\/latex] if [latex]x<\\frac{1}{2}[\/latex] and [latex]{b}_{1}=1[\/latex] if [latex]\\frac{1}{2}\\le x<1[\/latex]. Let [latex]{x}_{1}=2x-{b}_{1}[\/latex]. Let [latex]{b}_{2}=0[\/latex] if [latex]{x}_{1}<\\frac{1}{2}[\/latex] and [latex]{b}_{2}=1[\/latex] if [latex]\\frac{1}{2}\\le x<1[\/latex]. Let [latex]{x}_{2}=2{x}_{1}-{b}_{2}[\/latex] and in general, [latex]{x}_{n}=2{x}_{n - 1}-{b}_{n}[\/latex] and [latex]{b}_{n - 1}=0[\/latex] if [latex]{x}_{n}<\\frac{1}{2}[\/latex] and [latex]{b}_{n - 1}=1[\/latex] if [latex]\\frac{1}{2}\\le {x}_{n}<1[\/latex]. Find the binary expansion of [latex]\\frac{1}{3}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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          [reveal-answer q=\"174628\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"174628\"][latex]{b}_{1}=0[\/latex], [latex]{x}_{1}=\\frac{2}{3}[\/latex], [latex]{b}_{2}=1[\/latex], [latex]{x}_{2}=\\frac{4}{3} - 1=\\frac{1}{3}[\/latex], so the pattern repeats, and [latex]\\frac{1}{3}=0.010101\\text{... }[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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          64. [T]<\/strong> To find an approximation for [latex]\\pi [\/latex], set [latex]{a}_{0}=\\sqrt{2+1}[\/latex], [latex]{a}_{1}=\\sqrt{2+{a}_{0}}[\/latex], and, in general, [latex]{a}_{n+1}=\\sqrt{2+{a}_{n}}[\/latex]. Finally, set [latex]{p}_{n}={3.2}^{n}\\sqrt{2-{a}_{n}}[\/latex]. Find the first ten terms of [latex]{p}_{n}[\/latex] and compare the values to [latex]\\pi [\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n

          For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of [latex]N[\/latex] integers [latex]{a}_{1},{a}_{2}\\text{,... },{a}_{N}[\/latex] by fixing two special integers [latex]K[\/latex] and [latex]M[\/latex] and letting [latex]{a}_{n+1}[\/latex] be the remainder after dividing [latex]K.{a}_{n}[\/latex] into [latex]M[\/latex], then creates a bit sequence of zeros and ones whose [latex]n\\text{th}[\/latex] term [latex]{b}_{n}[\/latex] is equal to one if [latex]{a}_{n}[\/latex] is odd and equal to zero if [latex]{a}_{n}[\/latex] is even. If the bits [latex]{b}_{n}[\/latex] are pseudorandom, then the behavior of their average [latex]\\frac{\\left({b}_{1}+{b}_{2}+\\cdots +{b}_{N}\\right)}{N}[\/latex] should be similar to behavior of averages of truly randomly generated bits.<\/p>\r\n\r\n

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          65. [T]<\/strong> Starting with [latex]K=16,807[\/latex] and [latex]M=2,147,483,647[\/latex], using ten different starting values of [latex]{a}_{1}[\/latex], compute sequences of bits [latex]{b}_{n}[\/latex] up to [latex]n=1000[\/latex], and compare their averages to ten such sequences generated by a random bit generator.<\/p>\r\n\r\n<\/div>\r\n

          \r\n

          [reveal-answer q=\"321766\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"321766\"]For the starting values [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2\\text{,... }[\/latex], [latex]{a}_{1}=10[\/latex], the corresponding bit averages calculated by the method indicated are [latex]0.5220[\/latex], [latex]0.5000[\/latex], [latex]0.4960[\/latex], [latex]0.4870[\/latex], [latex]0.4860[\/latex], [latex]0.4680[\/latex], [latex]0.5130[\/latex], [latex]0.5210[\/latex], [latex]0.5040[\/latex], and [latex]0.4840[\/latex]. Here is an example of ten corresponding averages of strings of [latex]1000[\/latex] bits generated by a random number generator: [latex]0.4880[\/latex], [latex]0.4870[\/latex], [latex]0.5150[\/latex], [latex]0.5490[\/latex], [latex]0.5130[\/latex], [latex]0.5180[\/latex], [latex]0.4860[\/latex], [latex]0.5030[\/latex], [latex]0.5050[\/latex], [latex]0.4980[\/latex]. There is no real pattern in either type of average. The random-number-generated averages range between [latex]0.4860[\/latex] and [latex]0.5490[\/latex], a range of [latex]0.0630[\/latex], whereas the calculated PRNG bit averages range between [latex]0.4680[\/latex] and [latex]0.5220[\/latex], a range of [latex]0.0540[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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          66. [T]<\/strong> Find the first [latex]1000[\/latex] digits of [latex]\\pi [\/latex] using either a computer program or Internet resource. Create a bit sequence [latex]{b}_{n}[\/latex] by letting [latex]{b}_{n}=1[\/latex] if the [latex]n\\text{th}[\/latex] digit of [latex]\\pi [\/latex] is odd and [latex]{b}_{n}=0[\/latex] if the [latex]n\\text{th}[\/latex] digit of [latex]\\pi [\/latex] is even. Compute the average value of [latex]{b}_{n}[\/latex] and the average value of [latex]{d}_{n}=|{b}_{n+1}-{b}_{n}|[\/latex], [latex]n=1\\text{,...},999[\/latex]. Does the sequence [latex]{b}_{n}[\/latex] appear random? Do the differences between successive elements of [latex]{b}_{n}[\/latex] appear random?<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

          Find the first six terms of each of the following sequences, starting with [latex]n=1[\/latex].<\/p>\n

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          1.\u00a0<\/strong>[latex]{a}_{n}=1+{\\left(-1\\right)}^{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{a}_{n}=0[\/latex] if [latex]n[\/latex] is odd and [latex]{a}_{n}=2[\/latex] if [latex]n[\/latex] is even<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          2.\u00a0<\/strong>[latex]{a}_{n}={n}^{2}-1[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          3.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n}={a}_{n - 1}+n[\/latex] for [latex]n\\ge 2[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]\\left\\{{a}_{n}\\right\\}=\\left\\{1,3,6,10,15,21\\text{,... }\\right\\}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          4.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=1[\/latex] and [latex]{a}_{n+2}={a}_{n}+{a}_{n+1}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          5.\u00a0<\/strong>Find an explicit formula for [latex]{a}_{n}[\/latex] where [latex]{a}_{1}=1[\/latex] and [latex]{a}_{n}={a}_{n - 1}+n[\/latex] for [latex]n\\ge 2[\/latex].<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{a}_{n}=\\frac{n\\left(n+1\\right)}{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          6.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the arithmetic sequence whose first term is [latex]{a}_{1}=1[\/latex] such that [latex]{a}_{n - 1}-{a}_{n}=17[\/latex] for [latex]n\\ge 1[\/latex].<\/div>\n<\/div>\n<\/div>\n
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          7.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the arithmetic sequence whose first term is [latex]{a}_{1}=-3[\/latex] such that [latex]{a}_{n - 1}-{a}_{n}=4[\/latex] for [latex]n\\ge 1[\/latex].<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{a}_{n}=4n - 7[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          8.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the geometric sequence whose first term is [latex]{a}_{1}=1[\/latex] such that [latex]\\frac{{a}_{n+1}}{{a}_{n}}=10[\/latex] for [latex]n\\ge 1[\/latex].<\/div>\n<\/div>\n<\/div>\n
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          9.\u00a0<\/strong>Find a formula [latex]{a}_{n}[\/latex] for the [latex]n\\text{th}[\/latex] term of the geometric sequence whose first term is [latex]{a}_{1}=3[\/latex] such that [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{1}{10}[\/latex] for [latex]n\\ge 1[\/latex].<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{a}_{n}={3.10}^{1-n}={30.10}^{\\text{-}n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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          10.\u00a0<\/strong>Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence whose first several terms are [latex]\\left\\{0,3,8,15,24,35,48,63,80,99\\text{,... }\\right\\}[\/latex]. (Hint:<\/em> First add one to each term.)<\/div>\n<\/div>\n<\/div>\n
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          11.\u00a0<\/strong>Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence satisfying [latex]{a}_{1}=0[\/latex] and [latex]{a}_{n}=2{a}_{n - 1}+1[\/latex] for [latex]n\\ge 2[\/latex].<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{a}_{n}={2}^{n}-1[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          Find a formula for the general term [latex]{a}_{n}[\/latex] of each of the following sequences.<\/p>\n

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          12.\u00a0<\/strong>[latex]\\left\\{1,0,-1,0,1,0,-1,0\\text{,... }\\right\\}[\/latex] (Hint:<\/em> Find where [latex]\\sin{x}[\/latex] takes these values)<\/div>\n<\/div>\n<\/div>\n
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          13.\u00a0<\/strong>[latex]\\left\\{1,\\text{-}\\frac{1}{3},\\frac{1}{5},\\text{-}\\frac{1}{7}\\text{,... }\\right\\}[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n - 1}}{2n - 1}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          Find a function [latex]f\\left(n\\right)[\/latex] that identifies the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] of the following recursively defined sequences, as [latex]{a}_{n}=f\\left(n\\right)[\/latex].<\/p>\n

          \n
          \n
          14.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\text{-}{a}_{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          \n
          \n

          15.\u00a0<\/strong>[latex]{a}_{1}=2[\/latex] and [latex]{a}_{n+1}=2{a}_{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/p>\n<\/div>\n

          \n

          \n

          Show Solution<\/span><\/p>\n
          [latex]f\\left(n\\right)={2}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          16.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\left(n+1\\right){a}_{n}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          \n
          \n

          17.\u00a0<\/strong>[latex]{a}_{1}=2[\/latex] and [latex]{a}_{n+1}=\\frac{\\left(n+1\\right){a}_{n}}{2}[\/latex] for [latex]n\\ge 1[\/latex]<\/p>\n<\/div>\n

          \n

          \n

          Show Solution<\/span><\/p>\n
          [latex]f\\left(n\\right)=\\frac{n\\text{!}}{{2}^{n - 2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          18.\u00a0<\/strong>[latex]{a}_{1}=1[\/latex] and [latex]{a}_{n+1}=\\frac{{a}_{n}}{{2}^{n}}[\/latex] for [latex]n\\ge 1[\/latex]<\/div>\n<\/div>\n<\/div>\n

          Plot the first [latex]N[\/latex] terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.<\/p>\n

          \n
          \n
          \n
          \n

          19. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], and for [latex]n\\ge 2[\/latex], [latex]{a}_{n}=\\frac{1}{2}\\left({a}_{n - 1}+{a}_{n - 2}\\right)[\/latex]; [latex]N=30[\/latex]<\/p>\n<\/div>\n

          \n
          Show Solution<\/span><\/p>\n
          Terms oscillate above and below [latex]\\frac{5}{3}[\/latex] and appear to converge to [latex]\\frac{5}{3}[\/latex].
          \n<\/span><\/p>\n


          \n\"This<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n
          20. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], [latex]{a}_{3}=3[\/latex] and for [latex]n\\ge 4[\/latex], [latex]{a}_{n}=\\frac{1}{3}\\left({a}_{n - 1}+{a}_{n - 2}+{a}_{n - 3}\\right)[\/latex], [latex]N=30[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          \n
          \n

          21. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], and for [latex]n\\ge 3[\/latex], [latex]{a}_{n}=\\sqrt{{a}_{n - 1}{a}_{n - 2}}[\/latex]; [latex]N=30[\/latex]<\/p>\n<\/div>\n

          \n
          Show Solution<\/span><\/p>\n
          Terms oscillate above and below [latex]y\\approx 1.57..[\/latex]. and appear to converge to a limit.
          \n<\/span><\/p>\n


          \n\"This<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n
          \n
          22. [T]<\/strong> [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2[\/latex], [latex]{a}_{3}=3[\/latex], and for [latex]n\\ge 4[\/latex], [latex]{a}_{n}=\\sqrt{{a}_{n - 1}{a}_{n - 2}{a}_{n - 3}}[\/latex]; [latex]N=30[\/latex]<\/div>\n<\/div>\n<\/div>\n

          Suppose that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=1[\/latex], [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=-1[\/latex], and [latex]0<\\text{-}{b}_{n}<{a}_{n}[\/latex] for all [latex]n[\/latex]. Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.<\/p>\n

          \n
          \n
          \n
          \n

          23.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\left(3{a}_{n}-4{b}_{n}\\right)[\/latex]<\/p>\n<\/div>\n

          \n

          \n

          Show Solution<\/span><\/p>\n
          [latex]7[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          24.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\left(\\frac{1}{2}{b}_{n}-\\frac{1}{2}{a}_{n}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          \n
          \n
          \n

          25.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}+{b}_{n}}{{a}_{n}-{b}_{n}}[\/latex]<\/p>\n<\/div>\n

          \n

          \n

          Show Solution<\/span><\/p>\n
          [latex]0[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          26.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}-{b}_{n}}{{a}_{n}+{b}_{n}}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n

          Find the limit of each of the following sequences, using L\u2019H\u00f4pital\u2019s rule when appropriate.<\/p>\n

          \n
          \n
          \n
          \n

          27.\u00a0<\/strong>[latex]\\frac{{n}^{2}}{{2}^{n}}[\/latex]<\/p>\n<\/div>\n

          \n

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          Show Solution<\/span><\/p>\n
          [latex]0[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          28.\u00a0<\/strong>[latex]\\frac{{\\left(n - 1\\right)}^{2}}{{\\left(n+1\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          \n
          \n
          \n

          29.\u00a0<\/strong>[latex]\\frac{\\sqrt{n}}{\\sqrt{n+1}}[\/latex]<\/p>\n<\/div>\n

          \n

          \n

          Show Solution<\/span><\/p>\n
          [latex]1[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          30.\u00a0<\/strong>[latex]{n}^{\\frac{1}{n}}[\/latex] (<\/span>Hint:<\/em> [latex]{n}^{\\frac{1}{n}}={e}^{\\frac{1}{n}}\\text{ln}n[\/latex])<\/span><\/div>\n<\/div>\n<\/div>\n

          For each of the following sequences, whose [latex]n\\text{th}[\/latex] terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.<\/p>\n

          \n
          \n
          \n
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          31.\u00a0<\/strong>[latex]\\frac{n}{{2}^{n}}[\/latex], [latex]n\\ge 2[\/latex]<\/p>\n<\/div>\n

          \n

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          Show Solution<\/span><\/p>\n
          bounded, decreasing for [latex]n\\ge 1[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          32.\u00a0<\/strong>[latex]\\text{ln}\\left(1+\\frac{1}{n}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          \n
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          33.\u00a0<\/strong>[latex]\\sin{n}[\/latex]<\/p>\n<\/div>\n

          \n

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          Show Solution<\/span><\/p>\n
          bounded, not monotone<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          34.\u00a0<\/strong>[latex]\\cos\\left({n}^{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          35.\u00a0<\/strong>[latex]{n}^{\\frac{1}{n}}[\/latex], [latex]n\\ge 3[\/latex]<\/p>\n<\/div>\n

          \n

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          Show Solution<\/span><\/p>\n
          bounded, decreasing<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          36.\u00a0<\/strong>[latex]{n}^{\\frac{-1}{n}}[\/latex], [latex]n\\ge 3[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          \n
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          37.\u00a0<\/strong>[latex]\\tan{n}[\/latex]<\/p>\n<\/div>\n

          \n

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          Show Solution<\/span><\/p>\n
          not monotone, not bounded<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
          \n
          \n

          38.\u00a0<\/strong>Determine whether the sequence defined as follows has a limit. If it does, find the limit.<\/p>\n

          [latex]{a}_{1}=\\sqrt{2}[\/latex], [latex]{a}_{2}=\\sqrt{2\\sqrt{2}}[\/latex], [latex]{a}_{3}=\\sqrt{2\\sqrt{2\\sqrt{2}}}[\/latex] etc.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          \n
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          39.\u00a0<\/strong>Determine whether the sequence defined as follows has a limit. If it does, find the limit.<\/p>\n

          [latex]{a}_{1}=3[\/latex], [latex]{a}_{n}=\\sqrt{2{a}_{n - 1}}[\/latex], [latex]n=2,3\\text{,... }[\/latex].<\/p>\n<\/div>\n

          \n

          \n

          Show Solution<\/span><\/p>\n
          [latex]{a}_{n}[\/latex] is decreasing and bounded below by [latex]2[\/latex]. The limit [latex]a[\/latex] must satisfy [latex]a=\\sqrt{2a}[\/latex] so [latex]a=2[\/latex], independent of the initial value.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          Use the Squeeze Theorem to find the limit of each of the following sequences.<\/p>\n

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          40.\u00a0<\/strong>[latex]n\\sin\\left(\\frac{1}{n}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          41.\u00a0<\/strong>[latex]\\frac{\\cos\\left(\\frac{1}{n}\\right)-1}{\\frac{1}{n}}[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]0[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
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          42.\u00a0<\/strong>[latex]{a}_{n}=\\frac{n\\text{!}}{{n}^{n}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          \n
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          43.\u00a0<\/strong>[latex]{a}_{n}=\\sin{n}\\sin\\left(\\frac{1}{n}\\right)[\/latex]<\/p>\n<\/div>\n

          \n

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          Show Solution<\/span><\/p>\n
          [latex]0:|\\sin{x}|\\le |x|[\/latex] and [latex]|\\sin{x}|\\le 1[\/latex] so [latex]-\\frac{1}{n}\\le {a}_{n}\\le \\frac{1}{n}[\/latex]).<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          For the following sequences, plot the first [latex]25[\/latex] terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.<\/p>\n

          \n
          \n
          44. [T]<\/strong> [latex]{a}_{n}=\\sin{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          45. [T]<\/strong> [latex]{a}_{n}=\\cos{n}[\/latex]<\/p>\n<\/div>\n

          \n
          Show Solution<\/span><\/p>\n
          Graph oscillates and suggests no limit.
          \n<\/span><\/p>\n


          \n\"This<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.<\/p>\n

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          46.\u00a0<\/strong>[latex]{a}_{n}={\\tan}^{-1}\\left({n}^{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          47.\u00a0<\/strong>[latex]{a}_{n}={\\left(2n\\right)}^{\\frac{1}{n}}-{n}^{\\frac{1}{n}}[\/latex]<\/p>\n<\/div>\n

          \n

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          Show Solution<\/span><\/p>\n
          [latex]{n}^{\\frac{1}{n}}\\to 1[\/latex] and [latex]{2}^{\\frac{1}{n}}\\to 1[\/latex], so [latex]{a}_{n}\\to 0[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
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          48.\u00a0<\/strong>[latex]{a}_{n}=\\frac{\\text{ln}\\left({n}^{2}\\right)}{\\text{ln}\\left(2n\\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          49.\u00a0<\/strong>[latex]{a}_{n}={\\left(1-\\frac{2}{n}\\right)}^{n}[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          Since [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}\\to e[\/latex], one has [latex]{\\left(1 - \\frac{2}{n}\\right)}^{n}\\approx {\\left(1+k\\right)}^{-2k}\\to {e}^{-2}[\/latex] as [latex]k\\to \\infty[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
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          50.\u00a0<\/strong>[latex]{a}_{n}=\\text{ln}\\left(\\frac{n+2}{{n}^{2}-3}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          51.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{2}^{n}+{3}^{n}}{{4}^{n}}[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{2}^{n}+{3}^{n}\\le 2\\cdot {3}^{n}[\/latex] and [latex]\\frac{{3}^{n}}{{4}^{n}}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex], so [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
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          52.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{\\left(1000\\right)}^{n}}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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          53.\u00a0<\/strong>[latex]{a}_{n}=\\frac{{\\left(n\\text{!}\\right)}^{2}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\n

          Show Solution<\/span><\/p>\n
          [latex]\\frac{{a}_{n+1}}{{a}_{n}}=\\frac{n\\text{!}}{\\left(n+1\\right)\\left(n+2\\right)}\\text{... }\\left(2n\\right)[\/latex] [latex]=\\frac{1\\cdot 2\\cdot 3\\text{... }n}{\\left(n+1\\right)\\left(n+2\\right)\\text{... }\\left(2n\\right)}<\\frac{1}{{2}^{n}}[\/latex]. In particular, [latex]\\frac{{a}_{n+1}}{{a}_{n}}\\le \\frac{1}{2}[\/latex], so [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex].<\/div>\n<\/div>\n<\/div>\n
          <\/div>\n<\/div>\n<\/div>\n<\/div>\n

          Newton\u2019s method seeks to approximate a solution [latex]f\\left(x\\right)=0[\/latex] that starts with an initial approximation [latex]{x}_{0}[\/latex] and successively defines a sequence [latex]{x}_{n+1}={x}_{n}-\\frac{f\\left({x}_{n}\\right)}{{f}^{\\prime }\\left({x}_{n}\\right)}[\/latex]. For the given choice of [latex]f[\/latex] and [latex]{x}_{0}[\/latex], write out the formula for [latex]{x}_{n+1}[\/latex]. If the sequence appears to converge, give an exact formula for the solution [latex]x[\/latex], then identify the limit [latex]x[\/latex] accurate to four decimal places and the smallest [latex]n[\/latex] such that [latex]{x}_{n}[\/latex] agrees with [latex]x[\/latex] up to four decimal places.<\/p>\n

          \n
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          54. [T]<\/strong> [latex]f\\left(x\\right)={x}^{2}-2[\/latex], [latex]{x}_{0}=1[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          55. [T]<\/strong> [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}-2[\/latex], [latex]{x}_{0}=2[\/latex]<\/p>\n<\/div>\n

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          Show Solution<\/span><\/p>\n
          [latex]{x}_{n+1}={x}_{n}-\\frac{\\left({\\left({x}_{n}-1\\right)}^{2}-2\\right)}{2}\\left({x}_{n}-1\\right)[\/latex]; [latex]x=1+\\sqrt{2}[\/latex], [latex]x\\approx 2.4142[\/latex], [latex]n=5[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
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          56. [T]<\/strong> [latex]f\\left(x\\right)={e}^{x}-2[\/latex], [latex]{x}_{0}=1[\/latex]<\/div>\n<\/div>\n<\/div>\n
          \n
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          57. [T]<\/strong> [latex]f\\left(x\\right)=\\text{ln}x - 1[\/latex], [latex]{x}_{0}=2[\/latex]<\/p>\n<\/div>\n

          \n

          \n

          Show Solution<\/span><\/p>\n
          [latex]{x}_{n+1}={x}_{n}-{x}_{n}\\left(\\text{ln}\\left({x}_{n}\\right)-1\\right)[\/latex]; [latex]x=e[\/latex], [latex]x\\approx 2.7183[\/latex], [latex]n=5[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n
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          58. [T]<\/strong> Suppose you start with one liter of vinegar and repeatedly remove [latex]0.1\\text{L,}[\/latex] replace with water, mix, and repeat.<\/p>\n

            \n
          1. Find a formula for the concentration after [latex]n[\/latex] steps.<\/li>\n
          2. After how many steps does the mixture contain less than [latex]10\\text{%}[\/latex] vinegar?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
            \n
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            59. [T]<\/strong> A lake initially contains [latex]2000[\/latex] fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by [latex]6\\text{%}[\/latex] each month. However, factoring in all causes, [latex]150[\/latex] fish are lost each month.<\/p>\n

              \n
            1. Explain why the fish population after [latex]n[\/latex] months is modeled by [latex]{P}_{n}=1.06{P}_{n - 1}-150[\/latex] with [latex]{P}_{0}=2000[\/latex].<\/li>\n
            2. How many fish will be in the pond after one year?<\/li>\n<\/ol>\n<\/div>\n
              \n

              \n

              Show Solution<\/span><\/p>\n
              a. Without losses, the population would obey [latex]{P}_{n}=1.06{P}_{n - 1}[\/latex]. The subtraction of [latex]150[\/latex] accounts for fish losses. b. After [latex]12[\/latex] months, we have [latex]{P}_{12}\\approx 1494[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
              \n
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              \n

              60. [T]<\/strong> A bank account earns [latex]5\\text{%}[\/latex] interest compounded monthly. Suppose that [latex]\\$\\text{1000}[\/latex] is initially deposited into the account, but that [latex]\\$\\text{10}[\/latex] is withdrawn each month.<\/p>\n

                \n
              1. Show that the amount in the account after [latex]n[\/latex] months is [latex]{A}_{n}=\\left(1+.\\frac{05}{12}\\right){A}_{n - 1}-10[\/latex]; [latex]{A}_{0}=1000[\/latex].<\/li>\n
              2. How much money will be in the account after [latex]1[\/latex] year?<\/li>\n
              3. Is the amount increasing or decreasing?<\/li>\n
              4. Suppose that instead of [latex]\\$\\text{10}[\/latex], a fixed amount [latex]d[\/latex] dollars is withdrawn each month. Find a value of [latex]d[\/latex] such that the amount in the account after each month remains [latex]\\$\\text{1000}[\/latex].<\/li>\n
              5. What happens if [latex]d[\/latex] is greater than this amount?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
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                61. [T]<\/strong> A student takes out a college loan of [latex]\\$\\text{10,000}[\/latex] at an annual percentage rate of [latex]6\\text{%}[\/latex], compounded monthly.<\/p>\n

                  \n
                1. If the student makes payments of [latex]\\$\\text{100}[\/latex] per month, how much does the student owe after [latex]12[\/latex] months?<\/li>\n
                2. After how many months will the loan be paid off?<\/li>\n<\/ol>\n<\/div>\n
                  \n

                  \n

                  Show Solution<\/span><\/p>\n
                  a. The student owes [latex]\\$\\text{9383}[\/latex] after [latex]12[\/latex] months. b. The loan will be paid in full after [latex]139[\/latex] months or eleven and a half years.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
                  \n
                  \n
                  62. [T]<\/strong> Consider a series combining geometric growth and arithmetic decrease. Let [latex]{a}_{1}=1[\/latex]. Fix [latex]a>1[\/latex] and [latex]0\n<\/div>\n<\/div>\n
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                  63. [T]<\/strong> The binary representation [latex]x=0.{b}_{1}{b}_{2}{b}_{3}..[\/latex]. of a number [latex]x[\/latex] between [latex]0[\/latex] and [latex]1[\/latex] can be defined as follows. Let [latex]{b}_{1}=0[\/latex] if [latex]x<\\frac{1}{2}[\/latex] and [latex]{b}_{1}=1[\/latex] if [latex]\\frac{1}{2}\\le x<1[\/latex]. Let [latex]{x}_{1}=2x-{b}_{1}[\/latex]. Let [latex]{b}_{2}=0[\/latex] if [latex]{x}_{1}<\\frac{1}{2}[\/latex] and [latex]{b}_{2}=1[\/latex] if [latex]\\frac{1}{2}\\le x<1[\/latex]. Let [latex]{x}_{2}=2{x}_{1}-{b}_{2}[\/latex] and in general, [latex]{x}_{n}=2{x}_{n - 1}-{b}_{n}[\/latex] and [latex]{b}_{n - 1}=0[\/latex] if [latex]{x}_{n}<\\frac{1}{2}[\/latex] and [latex]{b}_{n - 1}=1[\/latex] if [latex]\\frac{1}{2}\\le {x}_{n}<1[\/latex]. Find the binary expansion of [latex]\\frac{1}{3}[\/latex].<\/p>\n<\/div>\n

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                  [latex]{b}_{1}=0[\/latex], [latex]{x}_{1}=\\frac{2}{3}[\/latex], [latex]{b}_{2}=1[\/latex], [latex]{x}_{2}=\\frac{4}{3} - 1=\\frac{1}{3}[\/latex], so the pattern repeats, and [latex]\\frac{1}{3}=0.010101\\text{... }[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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                  64. [T]<\/strong> To find an approximation for [latex]\\pi[\/latex], set [latex]{a}_{0}=\\sqrt{2+1}[\/latex], [latex]{a}_{1}=\\sqrt{2+{a}_{0}}[\/latex], and, in general, [latex]{a}_{n+1}=\\sqrt{2+{a}_{n}}[\/latex]. Finally, set [latex]{p}_{n}={3.2}^{n}\\sqrt{2-{a}_{n}}[\/latex]. Find the first ten terms of [latex]{p}_{n}[\/latex] and compare the values to [latex]\\pi[\/latex].<\/div>\n<\/div>\n<\/div>\n

                  For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of [latex]N[\/latex] integers [latex]{a}_{1},{a}_{2}\\text{,... },{a}_{N}[\/latex] by fixing two special integers [latex]K[\/latex] and [latex]M[\/latex] and letting [latex]{a}_{n+1}[\/latex] be the remainder after dividing [latex]K.{a}_{n}[\/latex] into [latex]M[\/latex], then creates a bit sequence of zeros and ones whose [latex]n\\text{th}[\/latex] term [latex]{b}_{n}[\/latex] is equal to one if [latex]{a}_{n}[\/latex] is odd and equal to zero if [latex]{a}_{n}[\/latex] is even. If the bits [latex]{b}_{n}[\/latex] are pseudorandom, then the behavior of their average [latex]\\frac{\\left({b}_{1}+{b}_{2}+\\cdots +{b}_{N}\\right)}{N}[\/latex] should be similar to behavior of averages of truly randomly generated bits.<\/p>\n

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                  65. [T]<\/strong> Starting with [latex]K=16,807[\/latex] and [latex]M=2,147,483,647[\/latex], using ten different starting values of [latex]{a}_{1}[\/latex], compute sequences of bits [latex]{b}_{n}[\/latex] up to [latex]n=1000[\/latex], and compare their averages to ten such sequences generated by a random bit generator.<\/p>\n<\/div>\n

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                  For the starting values [latex]{a}_{1}=1[\/latex], [latex]{a}_{2}=2\\text{,... }[\/latex], [latex]{a}_{1}=10[\/latex], the corresponding bit averages calculated by the method indicated are [latex]0.5220[\/latex], [latex]0.5000[\/latex], [latex]0.4960[\/latex], [latex]0.4870[\/latex], [latex]0.4860[\/latex], [latex]0.4680[\/latex], [latex]0.5130[\/latex], [latex]0.5210[\/latex], [latex]0.5040[\/latex], and [latex]0.4840[\/latex]. Here is an example of ten corresponding averages of strings of [latex]1000[\/latex] bits generated by a random number generator: [latex]0.4880[\/latex], [latex]0.4870[\/latex], [latex]0.5150[\/latex], [latex]0.5490[\/latex], [latex]0.5130[\/latex], [latex]0.5180[\/latex], [latex]0.4860[\/latex], [latex]0.5030[\/latex], [latex]0.5050[\/latex], [latex]0.4980[\/latex]. There is no real pattern in either type of average. The random-number-generated averages range between [latex]0.4860[\/latex] and [latex]0.5490[\/latex], a range of [latex]0.0630[\/latex], whereas the calculated PRNG bit averages range between [latex]0.4680[\/latex] and [latex]0.5220[\/latex], a range of [latex]0.0540[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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                  66. [T]<\/strong> Find the first [latex]1000[\/latex] digits of [latex]\\pi[\/latex] using either a computer program or Internet resource. Create a bit sequence [latex]{b}_{n}[\/latex] by letting [latex]{b}_{n}=1[\/latex] if the [latex]n\\text{th}[\/latex] digit of [latex]\\pi[\/latex] is odd and [latex]{b}_{n}=0[\/latex] if the [latex]n\\text{th}[\/latex] digit of [latex]\\pi[\/latex] is even. Compute the average value of [latex]{b}_{n}[\/latex] and the average value of [latex]{d}_{n}=|{b}_{n+1}-{b}_{n}|[\/latex], [latex]n=1\\text{,...},999[\/latex]. Does the sequence [latex]{b}_{n}[\/latex] appear random? Do the differences between successive elements of [latex]{b}_{n}[\/latex] appear random?<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
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