Sequence, Series and Integral-solutions - lee(jsl2382...

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lee (jsl2382) – Sequence, Series and Integral – tran – (56320)1Thisprint-outshouldhave38questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This homework covers section 11.1, 11.2and 11.3. Concepts tested are sequence, lim-its, series, geometric series, thinking of limitas differentition (the ”log trick”), divergencetest, and comparing series to integrals.00110.0pointsFind a formula for the general termanofthe sequenceanofthe sequence{an}n=1=braceleftBig1,54,2516,12564, . . .bracerightBig,assuming that the pattern of the first fewterms continues.1.parenleftBigparenrightBign-1{an}n=1=,5,9,13, . . .an= 3n22.an=n+ 33.an= 4n3correct4.an=n+ 45.an= 5n4Explanation:By inspection, consecutive termsan-1andanin the sequence{an}n=1=braceleftBig1,5,9,13, . . .bracerightBighave the property thatanan-1=d= 4.Thusan=an-1+d=an-2+ 2d=. . .=a1+ (n1)d= 1 + 4(n1).Consequently,an= 4n3.an=parenleftBig65parenrightBign4.an=parenleftBig45parenrightBign-15.an=parenleftBig45parenrightBign6.an=parenleftBig54parenrightBignExplanation:By inspection, consecutive termsan-1andanin the sequence{an}n=1=braceleftBig1,54,2516,12564, . . .bracerightBighave the property thatan=ran-1=parenleftBig54parenrightBigan-1.Thusan=ran-1=r2an-2=. . .=rn-1a1=parenleftBig54parenrightBign-1a1.Consequently,an=parenleftBig54parenrightBign-1sincea1= 1.1,5,9,13, . . .an= 3n22.an=n+ 33.an= 4n3correct4.an=n+ 45.an= 5n4Explanation:By inspection, consecutive termsan-1andanin the sequence{an}n=1=braceleftBig1,5,9,13, . . .bracerightBighave the property thatanan-1=d= 4.Thusan=an-1+d=an-2+ 2d=. . .=a1+ (n1)d= 1 + 4(n1).Consequently,an= 4n3.an=parenleftBig65parenrightBign4.an=parenleftBig45parenrightBign-15.an=parenleftBig45parenrightBign6.an=parenleftBig54parenrightBignExplanation:By inspection, consecutive termsan-1andanin the sequence{an}n=1=braceleftBig1,54,2516,12564, . . .bracerightBighave the property thatan=ran-1=parenleftBig54parenrightBigan-1.Thusan=ran-1=r2an-2=. . .=rn-1a1=parenleftBig54parenrightBign-1a1.Consequently,an=parenleftBig54parenrightBign-1sincea1= 1.
,assuming that the pattern of the first fewterms continues.an=651.an= 3n22.an=n+ 33.an= 4n3correct4.an=n+ 45.an= 5n4Explanation:By inspection, consecutive termsan-1andanin the sequence{an}n=1=braceleftBig1,5,9,13, . . .bracerightBighave the property thatanan-1=d= 4.Thusan=an-1+d=an-2+ 2d=. . .=a1+ (n1)d= 1 + 4(n1).Consequently,an= 4n3.an=parenleftBig65parenrightBign4.an=parenleftBig45parenrightBign-15.an=parenleftBig45parenrightBign6.an=parenleftBig54parenrightBignExplanation:By inspection, consecutive termsan-1andanin the sequence{an}n=1=braceleftBig1,54,2516,12564, . . .bracerightBighave the property thatan=ran-1=parenleftBig54parenrightBigan-1.Thusan=ran-1=r2an-2=. . .=rn-1a1=parenleftBig54parenrightBign-1a1.Consequently,an=parenleftBig54parenrightBign-1sincea1= 1.
keywords:00210.0pointsFind a formula for the general termanofthe sequence{an}n=1=braceleftBig1,54,2516,12564, . . .bracerightBig,assuming that the pattern of the first fewterms continues.1.parenleftBigparenrightBign-1
2.an=65
3.an=parenleftBig65parenrightBign4.an=parenleftBig45parenrightBign-15.an=parenleftBig45parenrightBign6.an=parenleftBig54parenrightBignExplanation:By inspection, consecutive termsan-1andanin the sequence{an}n=1=braceleftBig1,54,2516,12564, . . .bracerightBighave the property thatan=ran-1=parenleftBig54parenrightBigan-1.Thusan=ran-1=r2an-2=. . .=rn-1a1=parenleftBig54parenrightBign-1a1.Consequently,an=parenleftBig54parenrightBign-1sincea1= 1.
lee (jsl2382) – Sequence, Series and Integral – tran – (56320)2keywords: sequence, common ratio00310.0pointsIf thenthpartial sum of an infinite series isSn=5n223n2+ 4,what is the sum of the series?1.series diverges2.sum =53correct3.sum =234.sum =545.sum =12Explanation:By definitionsum =limn→∞Sn=limn→ ∞parenleftBig5n223n2+ 4parenrightBig.Thussum =53.00410.0pointsIflimn→ ∞an= 4,determine the value, if any, oflimn→ ∞an+5.1.limit =452.limit = 93.limit =14.limit = 4correct5.limit doesn’t existExplanation:To say thatlimn→ ∞an= 4means thatangets as close as we please to 4for all sufficiently largen. But thenan+5getsas close as we please to 4 for all sufficientlylargen. Consequently,limn→ ∞an+5= 4.00510.0pointsDetermine if the sequence{an}nconvergeswhenan=4n5n2and if it does, find its limit when1.converges with limit =432.converges with limit = 03.converges with limit =544.converges with limit =45correct5.the sequence divergesExplanation:After division bynwe see thatan=452n−→45asn→ ∞. Thus{an}nconverges and haslimit =45.006
10.0points
lee (jsl2382) – Sequence, Series and Integral – tran – (56320)3Determine if{an}converges whenan= 6n34n2,and if it does, find its limit.1.limit = 22.limit = +03.sequence divergescorrect4.limit = 4

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