Theory of Interest Solution Manual

Theory of Interest Solution Manual

ID:40476251

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頁數(shù):159頁

時間:2019-08-03

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1、TheTheoryofInterest-SolutionsManualChapter11.(a)Applyingformula(1.1)2Att()=++2t3andA(0)=3sothatAt()At()1()2at()===+t23.t+kA()03(b)Thethreepropertiesarelistedonp.2.1(1)a()03==()1.31(2)at′()=+()2t2>0fort≥0,3sothatat()isanincreasingfunction.(3)at()isapolynomialandthusiscon

2、tinuous.(c)Applyingformula(1.2)22IA=??()nA(n12)=+[nn+31]??+???()()n2n1+3??n22=++?+??+?nnnnn2321223=+21n.2.(a)Applingformula(1.2)II+++=…"IAA[()1021?()]+[A()?A()]++[AnAn()()??1]12n=?AnA()()0.(b)TheLHSistheincrementinthefundoverthenperiods,whichisentirelyattributabletothei

3、nterestearned.TheRHSisthesumoftheinterestearnedduringeachofthenperiods.3.Usingratioandproportion5000()12,153.9611,575.20?=$260.11,13024.Wehaveatatb()=+,sothatab()01==aa()3=+=91.72.b1TheTheoryofInterest-SolutionsManualChapter1Solvingtwoequationsintwounknownsab=.08and=1.T

4、hus,()2()a55.=0813+=()2()a10=10.08+=19.a(10)9andtheansweris100==100300.a()535.(a)Fromformula(1.4b)andA(tt)=1005+AA()5??(4)12512051i====.5A()412012024AA(10)??(9)15014551(b)i====.10A()914514529t6.(a)At()=1001.1and()54AA()5??()41001.1????()()1.1i===1.11.1.?=54A()41001.1()1

5、09AA()()10??91001.1????()()1.1(b)i===1.11.1.?=109A()91001.1()7.Fromformula(1.4b)AnAn()?(?1)i=nAn()?1sothatAnAn()?(?=11)iAn(?)nandAn()=(11+?iAn)().n8.Wehaveiii===.05,.06,.07,andusingtheresultderivedinExercise7567AAiii()74=+()(111)(+)(+)567==10001.051.061.07()()()$1190.91

6、.9.(a)Applyingformula(1.5)61550012.5=+=+(ii)5001250sothat2TheTheoryofInterest-SolutionsManualChapter11250ii===115and115/1250.092,or9.2%.(b)Similarly,6305001.078=+=+(tt)50039sothat39tt==130and130/3910/33==13years.10.Wehave111010001=+(it)=+10001000it1000it=110andit=.11sot

7、hat????35001??+=??()()it250011.5[]+it????4=+5001[]()1.5.11()=$582.50.11.Applyingformula(1.6)i.04i==and.025n11+?in()1+?.04()n1sothat.025.001+?()nnn1=.04,.001=.016,and=16.12.Wehaveiiiii=====.01.02.03.04.0512345andadaptingformula(1.5)10001??+++++=(iiiii)10001.15()=$1150.??

8、1234513.Applyingformula(1.8)26001()+=+=i600264864whichgives2()1+=ii864/6001.44,1=+=1.2,andi=.2sothat3320001()+

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