Pascals triangle and the binomial theorem mcTYpascal - PDF document

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Pascal'striangleandthebinomialtheorem mc-TY-pascal-2009-1.1 Abinomialexpressionisthesum,ordierence,oftwoterms.Forexample,x+1;3x+2y;abareallbinomialexpressions.Ifwewanttoraiseabinomialexpressiontoapowerhigherthan2(forexampleifwewanttond(x+1)7)itisverycumbersometodothisbyrepeatedlymultiplyingx+1byitself.Inthisunityouwilllearnhowatriangularpatternofnumbers,knownasPascal'striangle,canbeusedtoobtaintherequiredresultveryquickly.Inordertomasterthetechniquesexplainedhereitisvitalthatyouundertakeplentyofpracticeexercisessothattheybecomesecondnature.Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldbeableto:generatePascal'striangleexpandabinomialexpressionusingPascal'striangleusethebinomialtheoremtoexpandabinomialexpressionContents1.Introduction22.Pascal'striangle23.UsingPascal'striangletoexpandabinomialexpression34.Thebinomialtheorem6 www.mathcentre.ac.uk1c\rmathcentre2009 1.IntroductionAbinomialexpressionisthesum,ordierence,oftwoterms.Forexample,x+1;3x+2y;abareallbinomialexpressions.Youwillbefamiliaralreadywiththeneedtoexpandbracketswhensquaringsuchquantities.Youwillknow,forexample,that(x+1)2=(x+1)(x+1)=x2+x+x+1=x2+2x+1Ifwewanttoraiseabinomialexpressiontoapowerhigherthan2(forexampleifwewanttond(x+1)7)itisverycumbersometodothisbyrepeatedlymultiplyingx+1byitself.Inthisunityouwilllearnhowatriangularpatternofnumbers,knownasPascal'striangle,canbeusedtoobtaintherequiredresultveryquickly.2.Pascal'striangleWestarttogeneratePascal'strianglebywritingdownthenumber1.Thenwewriteanewrowwiththenumber1twice:111Wethengeneratenewrowstobuildatriangleofnumbers.Eachnewrowmustbeginandendwitha1:1111*11**1Theremainingnumbersineachrowarecalculatedbyaddingtogetherthetwonumbersintherowabovewhichlieabove-leftandabove-right.So,addingthetwo1'sinthesecondrowgives2,andthisnumbergoesinthevacantspaceinthethirdrow:111&.1211**1 www.mathcentre.ac.uk2c\rmathcentre2009 Thetwovacantspacesinthefourthrowareeachfoundbyaddingtogetherthetwonumbersinthethirdrowwhichlieabove-leftandabove-right:1+2=3,and2+1=3.Thisgives:111&.121&.&.1331Wecancontinuetobuildupthetriangleinthiswaytowritedownasmanyrowsaswewish.TheKeyPointbelowshowstherstsixrowsofPascal'striangle. KeyPointPascal'striangle11112113311464115101051...... Exercise11.Generatetheseventh,eighth,andninthrowsofPascal'striangle.3.UsingPascal'striangletoexpandabinomialexpressionWewillnowseehowusefulthetrianglecanbewhenwewanttoexpandabinomialexpression.Considerthebinomialexpressiona+b,andsupposewewishtond(a+b)2.Weknowthat(a+b)2=(a+b)(a+b)=a2+ab+ba+b2=a2+2ab+b2Thatis,(a+b)2=1a2+2ab+1b2Observethefollowinginthenalresult: www.mathcentre.ac.uk3c\rmathcentre2009 1.Aswemovethrougheachtermfromlefttoright,thepowerofadecreasesfrom2downtozero.2.Thepowerofbincreasesfromzeroupto2.3.Thecoecientsofeachterm,(1,2,1),arethenumberswhichappearintherowofPascal'strianglebeginning1,2.4.Theterm2abarisesfromcontributionsof1aband1ba,i.e.1ab+1ba=2ab.Thisisthelinkwiththewaythe2inPascal'striangleisgenerated;i.e.byadding1and1inthepreviousrow.Ifwewanttoexpand(a+b)3weselectthecoecientsfromtherowofthetrianglebeginning1,3:theseare1,3,3,1.Wecanimmediatelywritedowntheexpansionbyrememberingthatforeachnewtermwedecreasethepowerofa,thistimestartingwith3,andincreasethepowerofb.So(a+b)3=1a3+3a2b+3ab2+1b3whichwewouldnormallywriteasjust(a+b)3=a3+3a2b+3ab2+b3Thinkingof(a+b)3as(a+b)(a2+2ab+b2)=a3+2a2b+ab2+ba2+2ab2+b3=a3+3a2b+3ab2+b3wenotethattheterm3ab2,forexample,arisesfromthetwotermsab2and2ab2;againthisisthelinkwiththeway3isgeneratedinPascal'striangle-byaddingthe1and2inthepreviousrow.ExampleSupposewewishtond(a+b)4.Tondthisweusetherowbeginning1,4,andcanimmediatelywritedowntheexpansion.(a+b)4=a4+4a3b+6a2b2+4ab3+b4 Wecanapplythesameproceduretoexpandanybinomialexpression,evenwhenthequantitiesaandbaremorecomplicated.Considerthefollowingexamples.ExampleSupposewewanttoexpand(2x+y)3.WepickthecoecientsintheexpansionfromtherelevantrowofPascal'striangle:(1,3,3,1).Aswemovethroughthetermsintheexpansionfromlefttorightweremembertodecreasethepowerof2xandincreasethepowerofy.So,(2x+y)3=1(2x)3+3(2x)2y+3(2x)1y2+1y3=8x3+12x2y+6xy2+y3ExampleSupposewewanttoexpand(1+p)4. www.mathcentre.ac.uk4c\rmathcentre2009 Wepickthecoecientsintheexpansionfromtherowofthetrianglebeginning1,4;thatis(1,4,6,4,1).Aswemovethroughthetermsintheexpansionfromlefttorightweremembertoincreasethepowerofp.Thisexampleissimplerthanthepreviousonebecausetherstterminbracketsis1,and1toanypowerisstill1.So,(1+p)4=1(1)4+4(1)3p+6(1)2p2+4(1)p3+1p4=1+4p+6p2+4p3+p4: Eitherorbothofthetermsinthebinomialexpressioncanbenegative.Whenraisinganegativenumbertoanevenpowertheresultispositive.Whenraisinganegativenumbertoanoddpowertheresultisnegative.Considerthefollowingexample.ExampleExpand(3a2b)5.WepickthecoecientsintheexpansionfromtherowofPascal'strianglebeginning1,5;thatis1,5,10,10,5,1.Powersof3adecreasefrom5aswemovelefttoright.Powersof2bincrease.(3a2b)5=1(3a)5+5(3a)4(2b)+10(3a)3(2b)2+10(3a)2(2b)3+5(3a)(2b)4+1(2b)5=243a5810a4b+1080a3b2720a2b3+240ab432b5 Eitherorbothofthetermscouldbefractions.ExampleExpand1+2 x3.WepickthecoecientsintheexpansionfromtherowofPascal'striangle(1,3,3,1).Powersof2 xincreaseaswemovelefttoright.Anypowerof1isstill1.1+2 x3=1(1)3+3(1)22 x+3(1)12 x2+12 x3=1+6 x+12 x2+8 x3Exercises2UsePascal'striangletoexpandthefollowingbinomialexpressions:1.(1+3x)22.(2+x)33.(1x)34.(15x)55.(x+6)36.(ab)77.1+3 a48.x1 x6. www.mathcentre.ac.uk5c\rmathcentre2009 4.ThebinomialtheoremIfwewantedtoexpandabinomialexpressionwithalargepower,e.g.(1+x)32,useofPascal'strianglewouldnotberecommendedbecauseoftheneedtogeneratealargenumberofrowsofthetriangle.Analternativemethodistousethebinomialtheorem.Thetheoremenablesustoexpand(a+b)ninincreasingpowersofbanddecreasingpowersofa.Wewilllookatexpandingexpressionsoftheform(a+b)2,(a+b)3,...,(a+b)32,...,thatiswhenthepowerisapositivewholenumber.Undercertainconditionsthetheoremcanbeusedwhennisnegativeorfractionalandthisisusefulinmoreadvancedapplications,buttheseconditionswillnotbestudiedhere. KeyPointThebinomialtheorem:Whennisapositivewholenumber(a+b)n=an+nan1b+n(n1) 2!an2b2+n(n1)(n2) 3!an3b3+n(n1)(n2)(n3) 4!an4b4+:::+bn Notethatthisisaniteseries(thatis,itstopsafteranitenumberofterms)andthelasttermisbn.Asimplerformofthetheoremisoftenquotedbytakingthespecialcaseinwhicha=1andb=x.Itisstraightforwardtoverifythatthetheorembecomes: KeyPointThebinomialtheorem:Whennisapositivewholenumber(1+x)n=1+nx+n(n1) 2!x2+n(n1)(n2) 3!x3+n(n1)(n2)(n3) 4!x4+:::+xn www.mathcentre.ac.uk6c\rmathcentre2009 ExampleWeshallapplythebinomialtheoremtoexpand(1+x)2.Weusethetheoremwithn=2andstopwhenwehavewrittendowntheterminx2.(1+x)2=1+2x+(2)(21) 2!x2=1+2x+x2whichisthefamiliarandwell-knownresult.ExampleWenowapplythebinomialtheoremtoexpand(1+x)3.Weusethetheoremwithn=3.(1+x)3=1+3x+(3)(31) 2!x2+(3)(31)(32) 3!x3=1+3x+3x2+x3ExampleSupposewewishtoapplythebinomialtheoremtondtherstthreetermsinascendingpowersofxof(1+x)32.Weusethetheoremwithn=32andjustwritedowntherstthreeterms.(1+x)32=1+32x+(32)(321) 2!x2=1+32x+496x2+::: Withsomeingenuitywecanusethetheoremtoexpandotherbinomialexpressions.ExampleSupposewewishtondtherstfourtermsintheexpansionof(1+1 3y)10.Weusethetheorem,replacingxwithy 3andlettingn=10.Thisgives(1+1 3y)10=1+10y 3+(10)(101) 2!y 32+(10)(101)(102) 3!y 33+:::=1+10 3y+5y2+40 9y3+:::ExampleSupposewewishtondtherstthreetermsintheexpansionof(35z)14.Weshallapplythebinomialtheoremintheoriginalformgivenonpage6witha=3,b=5zandn=14.(35z)14=314+14(313)(5z)+(14)(13) 2!(312)(5z)2=314(313)70z+(312)2275z2=314170z 3+2275 9z2::: www.mathcentre.ac.uk7c\rmathcentre2009 Exercises31.Usethebinomialtheoremtoexpand(a)(1+x)4and(b)(1+x)5.2.Usethebinomialtheoremtoexpand(1+2x)3.3.Usethebinomialtheoremtoexpand(13x)4.4.Usethebinomialtheoremtondtherstthreetermsinascendingpowersofxof(1x 2)8.5.Findthecoecientofx5intheexpansionof(1+4x)9.6.Intheexpansionof(1x)8ndthecoecientofx7.7.Findtherstfourtermsintheexpansionof2+x 3
12.AnswersExercise11.Seventhrow:1,6,15,20,15,6,1.Eighthrow:1,7,21,35,35,21,7,1.Ninthrow:1,8,28,56,70,56,28,8,1.Exercise21.1+6x+9x22.8+12x+6x2+x33.13x+3x2x34.125x+250x21250x3+3125x43125x55.x3+18x2+108x+2166.a77a6b+21a5b235a4b3+35a3b421a2b5+7ab6b7.7.1+12 a+54 a2+108 a3+81 a48.x66x4+15x220+15 x26 x4+1 x6Exercise31.(a)1+4x+6x2+4x3+x4(b)1+5x+10x2+10x3+5x4+x5.2.1+6x+12x2+8x3.3.112x+54x2108x3+81x4.4.14x+7x2:::.5.129024.6.8.7.4096+8192x+22528 3x2+112640 27x3::: www.mathcentre.ac.uk8c\rmathcentre2009