MC0063(a) Unit2 Final

DiscreteMathematics Unit2

SikkimManipalUniversity PageNo:41

Unit2 MathematicalInduction

Structure

2.1 Introduction

Objectives

2.2 Progressions

2.3 PrincipleofMathematicalInduction

2.4 Summationofseriesusing n, n2and n3.

2.5 ArithmeticoGeometricseries(A.G.P)

2.6 Summation of series by the Method of Differences and partial

fractions

SelfAssessmentQuestions

2.7 Summary

2.8 TerminalQuestions

2.9 Answers

2.1Introduction

Herewediscussedwaysinwhichstatementcanbelinkedtoformalogically

validargument.TheprincipleofMathematical inductionhasaveryspecial

place in mathematics because of its simplicity and vast amount of

applications. This unit acts as foundations on which all mathematical

knowledgeisbuilt.

Objectives

Attheendofthisunitthestudentshouldableto:

Identifydifferenttypesofprogressions

Stateandapplytheconceptofmathematicallyinduction

Findthesumoftheseries



2.2Progressions

2.2.1Definition

Afunction RN:f iscalledasequenceofrealnumbersdenotedby

( ) { } ( ) ( ) ( ) { } ....nf....2f,1fnf = Where ( ) nf is called the nth term of thesequence.Asequencemayalsobedenotedby ( ) na or ( ) nu where na or

nu isthe nthtermofthesequence.

2.2.2Definition

If ( ) nu isasequence, then un =u1 +u2 ++un+ iscalledaserieswhichmaybefiniteorinfiniteaccordingasthenumberoftermsinitisfinite

orinfinite.

2.2.3ArithmeticProgression(A.P.)

Asequenceoftheform ,...d1na......,d2a,da,a - + + + iscalled

an P.A . whose first term is a, common difference is d and nth term

denotedby d1natn - + = .

If nS denotesthesumofthefirstn termsoftheabove P.A ,then

( ) ( ) d1na......daaSn - + + + + + = [ ] ( ) la

2nord1na2

2n + - + =

where = l thelastterm ntd1na = - + =

Observations:

i) 1nn ttd - - = isindependentof n ,aconstant.

ii) If each term of an P.A is multiplied or added by a constant the

resultingsequenceisalsoin P.A



iii) If the corresponding terms of two P.A s are added, the resulting

sequenceisalsoin P.A .

iv) If b,A,a are in P.A , thenA is called thearithmeticmeanbetween

theextremes banda isgivenby2

baA

+ =

v) If b,x...,x,x,a n21 arein P.A , then n21 x...,x,x arethe n AMs

between banda ,andtheirsum ( ) ba2nSn + =

2.2.4GeometricProgression(G.P.)

A sequence of the form ...,ar.......ar,ar,a 1n2 - is called a P.G .

Whosefirsttermisa ,commonratiois randthe nth term 1nn art - = .

If nS denotesthesumofthefirstn termsoftheabove P.G ,then

1nn ar.........araS

- + + + =

1r,1r

1raor

r1

r1a nn

-

-

-

-

=

OrSn 1rifna = = .

Also ( ) 1r1ralr

r1lra

Sn - -

= - -

= where = l thelastterm 1nn art - = =

If < nand1r , then the sum S of infinite P.G isr1

aS -

=

0rSince n



Observations:

i) att

r1n

n = = -

termindependentof an = constantratio

ii) If each term of a P.G is multiplied by a constant, the resulting

sequenceisalsoin P.G

iii) If b,G,a arein P.G ,theGiscalledthegeometricmeanbetweenthe

extremesa andb andisgivenby abG = .

iv) If b,x......x,x,a n21 arein P.G ,then n21 x.......x,x arethe

n M.G sbetweena andb andthe

v) 1kk

thabaM.Gk +

=

vi) Alsoproductofthe n G.Ms ( )2nab =

2.2.5HarmonicProgression(H.P.)

Asequenceoftheform ,d1na

1,...,d2a

1,da

1,a1

- + + +

isa P.H whosenth term ( ) d1na1

tn - +

=

Observations:

i) Thereisnoformulatofindthesumofthefirst n termsofthe P.H

ii) If b,H,a are in P.H ., then H is the harmonicmeanbetween the

extremesa andb andisgivenbyba

ab2H +

=

iii) Problems on P.H . are solved by dealing with the corresponding

P.A .



2.2.6Example Findthe nthtermofthefollowingseries

(a) 1+3+5+7+..

(b) 7+315+..

(c) 1+23 +2+

25+...

(d) 1+3+9+27+..

Solution:theseriesgivenin

(a) Formsan P.A inwhich

( ) 1n221n1tand2d,1a n - = - + = = = .(b) Forms an P.A in which

( ) ( ) n41141n7t,4d,7a n - = - - + = - = = (c) Formsan P.A inwhich

( ) ( ) .1n21

211n1t,

21d,1a n + = - + = = =

(d) Formsa P.G inwhich 1n1nn 33.1t,3r,1a - - = = = =

2.2.7Example:Findthe thn termofthefollowingseries

(a) + + + 222 531 . (b) 1.4 + 4.7 + 7.10 +

(c) + + + 11.8

18.5

15.2

1 ..

Solution:Thegivenseriesareneitherin P.A ., P.G .norin P.H .These

arespecialtypesofseries.

In(a) + + + 222 531 .,thebasesineachtermnamely1,3,5,form

an P.A . whose nth term 1n2 - = by Ex. 1(a) above. Hence, nth term

ofthegivenseriesis ( ) 2n 1n2t - = i.e., ( ) 2222 1n2.......531 - + + + + .In(b)1.4+4.7+7.10+.., thefirstfigure ineachtermviz.,1,4,7, ...



forms an P.A whose nth term ( ) 2n331n1 - = - + = and the secondfigure in each term viz., 4, 7, 10, forms an P.A . whose nth term

( ) 1n331n4 + = - + = . Hence nth term of the given series is ( ) ( ) 1n32n3tn + - =

In (c) + + + 11.81

8.51

5.21 , in each term numerator is 1 and in

denominator thecorrespondingfiguresare2.5,5.8,8.11,...As before the

first figure in each term viz 2, 5, 8, forms an P.A whose nth term

( ) 1n331n2 - = - + = andthesecondfigure ineachtermviz.,5,8,11,formsan P.A .whose nth term ( ) 2n331n5 + = - + = .Hencethe nthtermofthegivenseriesis

( )( ) 2n31n31

tn + -

=

2.2.8Example: Ifthe thp termofan P.A is q and thq termis p ,show

that ( ) thqp + termiszero.Solution: Let ...,d2a,da,a + + be the given P.A whose nth term

( ) .d1natn - + = Byhypothesis,wehave

( ) ( ) pd1qatandqd1pat qp = - + = = - + = Solvingtheseequations,weget

1d,1qpa - = - + =

Hence ( ) d1qpat qp - + + = + ( ) ( ) ( ) 11qp1qp - - + + - + = 0 =



2.2.9 Example: If ththth r,qp terms of an P.A are respectively a, b, c

provethat ( ) ( ) ( ) 0qpcprbrqa = - + - + - Solution:LetA andD bethefirsttermandcommondifferencerespectively

( ) )i(aD1pAtHence p = - + = ( ) )ii(bD1qAtq = - + = ( ) )iii(cD1rAtr = - + =

Multiplying (i), (ii), and (iii) respectively by pr,rq - - and qp - and

addingweget

( ) ( ) ( ) { } qpprrqAqpcprbrqa - + - + - = - + - + - ( )( ) ( )( ) ( )( ) { } qp1rpr1qrq1pD - - + - - + - - +

( ) ( ) ( ) 0qpcprbrqaTherefore = - + - + - 2.2.10Example:The4th,7thand10th termsof P.G .area,b,crespectively

showthat acb2 =

Solution:LetAbethefirsttermandRthecommonratioofthegiven P.G .,

where 1nn ARt - =

Thenbyhypothesis,

ct,bt,at 1074 = = =

i.e. cAR,bAR,aAR 963 = = =

Now, acARARRAb 931222 =

= =

2.3PrincipleofMathematicalInduction

Mathematical induction is the process of proving a general theorem or

formulainvolvingthepositiveinteger n fromparticularcases.

Aproofbymathematicalinductionconsistsofthefollowingtwosteps.

(i) Showbyactualsubstitutionthatthetheoremistruefor 1n =



(ii)Assumingthetheoremtobetruefor mn = ,provethatitisalsotruefor

1mn + =

Notethatherem isaparticularvalueof n .From(i)thetheoremistruefor

1n = and from (ii) it is true for 211n = + = since it is true for 2n = it

follows from (iii) that it is also true for 312n = + = and so on. Hence

theoremistrueforallpositiveintegralvaluesof n .

2.3.1Example:Provebymathematicalinductionthatthesumofthefirst n

naturalnumbersis ( ) 2

1nn +

Solution

Thatistoprovethat1+2+3+.+ = n ( ) 2

1nn +

(i) For 1n = ,leftside=1,rightside ( )

12

111 =

+ =

Hencetheresultistruefor 1n =

(ii) Nowassumethattheresulttobetruefor mn = ,then1+2+3+.

( ) 2

1mmm

+ = (Inductionhypothesis)

We now show that the above result is true for 1mn + = . Adding the

( ) th1m + termviz., 1m + tobothsidesweobtain.

1+2+3+.+ ( ) ( ) ( ) 1m2

1mm1mm + +

+ = + +

( ) ( ) ( ) 2

2m1m1

2

m1m

+ + =

+ + =

( ) ( ) 2

11m1m + + + =

Whichisthesameasthegivenresultfor 1mn + =



Hence by mathematical induction, the result is true for all + ve integral

valuesof n .

2.3.2Example:Provebymathematicalinductionthat

( )( ) 6

1n21nnn......321 2222

+ + = + + + +

Solution:

(i) If 1n = ,leftside 112 = = .

rightside ( )( )

16

3.2.1

6

11.2111 = =

+ + =

Hencetheresultistruefor 1n = .

(ii) Nowassumethattheresulttobetruefor mn =

Then ( ) ( )

6

1m21mmm.........321 2222

+ + = + + +

(inductionhypothesis)

Adding the ( ) th1m + term i.e. ( ) 21m + to both sides of the aboveequation,weget,

( ) ( ) ( ) ( ) 22222 1m6

1m21mm1mm.......21 + +

+ + = + + + + +

( ) ( ) ( ) { } 1m61m2m6

1m + + +

+ =

( )

+ +

+ = 6m7m2

6

1m 2

( )( ) ( ) 6

3m22m1m + + + =

( ) ( ) ( ) ( ) 6

11m211m1m + + + + + =

Thereforetheresultistruefor 1mn + = .Hencebymathematicalinduction

thegivenresultistrueforallpositiveintegers n .



2.3.3Example:Provebymathematicalinductionthat

( ) 4

1nnn........321

223333 + = + + + +

Solution:

(i) For 1n = ,leftside 113 = =

rightside ( )

144.1

4111 22

= = +

=

Henceitistruefor 1n =

(ii) Assumetheresulttobetruefor mn =

Then ( )

4

1mmm........321

223333 + = + + + +

(inductionhypothesis)

Addingthe ( ) th1m + termviz., ( ) 31m + tobothsides,

( ) ( ) ( ) 322

3333 1m4

1mm1mm........21 + +

+ = + + + + +

( )

+ +

+ = 4m4m

4

1m 22

( ) ( ) 4

2m1m 22 + + =

( ) ( ) 4

11m1m 22 + + + =

Therefore The result is true for 1mn + = . Hence by mathematical

inductionthegivenresultisestablishedforall+veintegers.

2.3.4Example:Provebymathematicalinduction

( ) ( ) 4n6n

2n31n3

1......

11.8

1

8.5

1

5.2

1

+ =

+ - + + + +



Solution:

(i) If 1n = ,leftside10

1

5.2

1 = =

rightside10

1

41.6

1 =

+ =

Thereforetheresultistruefor 1n =

(ii) Assumetheresulttobetruefor mn =

Then

( ) ( ) 4m6m

2m31m3

1.....

11.8

1

8.5

1

5.2

1

+ =

+ -

+ + + +

Addingthe ( ) th1m + term,viz., ( ) ( ) 5m32m3

1

+ + tobothsides.

Wehave,

( )( ) 5m32m31

......8.5

1

5.2

1

+ + + + +

( )( ) 5m32m31

4m6

m

+ + +

+ =

( ) ( ) ( ) 5m32m32

25m3m

+ +

+ + =

( ) ( ) 5m32m322m5m3 2

+ +

+ + =

( )( ) ( ) ( ) 5m32m32

2m31m

+ +

+ + =

10m61m

+ +

=



Thisisthevalueof4n6

n

+ when 1m + issubstitutedfor n

Thereforethepropositionistrueforall+veintegralvaluesof n

2.3.5Example:Provebymathematicalinductionthat n2n > forallpositive

integern

Solution:

Let ( ) nP be the given proposition. Now ( ) 1P implies 2 > 1which is true.Hence ( ) 1P istrueLetusassumethat ( ) mP istrue.Thatis m2m > (inductionhypothesis)Consider m22.22 m1m > = + byinductionhypothesis.

Weknowthat 1mmmm2 + + = forall Nm

1m2Therefore 1m + > +

Hence ( ) 1mP + istrue.Thereforebyinduction ( ) nP istrueforall n .2.3.6Example:Showbyinductionthat ( ) ( ) 1n21nn + + isdivisibleby6.Solution:

Let ( ) ( ) ( ) 1n21nnnP + + = Now ( ) ( ) ( ) 61211.11P = + + = whichisdivisibleby6.Assumethat ( ) mP isdivisibleby6.

i.e., ( ) ( ) 1m21mm + + isdivisibleby6.i.e. ( ) ( ) k61m21mm = + + forsomeintegerk.

Consider

P(m+1)=(m+1)[(m+1)+1][2(m+1)+1]

( ) ( ) ( ) 3m22m1m + + + =



( )( ) ( ) 21m22m1m + + + + = ( ) ( ) ( ) ( ) ( ) 2m1m21m22m1m + + + + + + =

( )( ) ( ) ( ) ( ) ( ) 2m1m21m21m21m21mm + + + + + + + + = ( ) ( ) 3m31m2k6 + + + = byinductionhypothesis ( ) 21m6k6 + + =

Sinceeachterm ontheR.H.Sisdivisibleby6theirsumisalsodivisibleby6.

Hence ( ) 1mP + isdivisibleby6.Thereforebyinduction ( ) nP isdivisibleby6forall Nn

Example 2.3.7: Using the principle of mathematical induction show that

110 1n2 + - isdivisibleby11forall Nn

Solution:

Let ( ) 110nP 1n2 + = -

Now ( ) 111101P 12 = + = - ,whichisdivisibleby11.

Assume that ( ) 110mP 1m2 + = - which is divisible by 11. Therefore ( ) k11110mP 1m2 = + = - forsomeintegerk.

Consider ( ) ( ) 1101101mP 1m211m2 + = + = + + - +

( ) ( ) 1m221m2 101010mP1mPTherefore - + - = - + 1m221m2 101010 - - - =

- = - 11010 21m2

,9910 1m2 = - whichisclearlydivisibleby11.

Therefore ( ) ( ) 9910mP1mP 1m2 + = + -

9910k11 1m2 + = -



Therefore ( ) 1mP + isdivisibleby11.Hencebytheprincipleofmathematicalinduction ( ) nP isdivisibleby11forall n .

2.4Summationofseriesusing n, n2and n3.

Nowweknowthefollowingresults:

( )

+ = = + + + +

2

1nnnn..........321

( ) ( )

+ + = = + + + +

6

1n21nnnn......321 22222

( ) ( ) 4

1nn

2

1nnnn....321

22233333 + =

+ = = + + +

2.4.1Note: ( ) = 23 nnThefollowingexampleswillillustratesomesimpleapplicationsoftheabove

resultstofindthesumoftheseries.

2.4.2Example: Findthesumof

(i) 3333 20.....321 + + + + , (ii) 3333 30....232221 + + + +

Solution:

Weknowthat ( )

+

= + + + + = 4

1nnn....321n

2233333 .

Therefore,onputting ,20n = weget

( ) ( ) 223333 21202120......321 = + + + +

44100 = (i)

andonputting 30n = ,weget

( ) ( ) 22333333 31304130.......2120......321 = + + + + + + +

216225 = (ii)



From(i)and(ii),weget

Therefore 4410021622530.....2221 333 - = + + +

172125 =

2.4.3Example:Findthesumofthesquaresofthefirst n oddintegers.

Solution:

Thefirstn oddintegersare1,3,5,7 ( ) 1n2 - .Therefore ( ) 2222n 1n2.....531S - + + + + =

( ) 21n2 - =

+ - =

+ - = 1n4n41n4n4 22

( ) ( ) ( ) n

2

1nn4

6

1n21nn4 +

+ -

+ + =

( ) ( ) ( ) { } 31n61n21n23n + + - + + =

- = 1n4

3n 2

2.4.4Example: Sumtontermstheseries.

......7.55.33.1 222 + + +

Solution:

Herethenthtermoftheseries ( ) ( ) 21n21n2 + - = .Therefore, ( ) ( ) 2222n 1n21n2.....7.55.33.1S + - + + + + =

( ) ( ) + - = 21n21n2

- - + = 1n2n4n8 23

- - + = 1n2n4n8 23



( ) ( ) ( ) ( ) n

21nn2

61n21nn4

41nn8 22

- +

- + +

+ +

=

( ) ( ) ( ) ( )

- + - + + + + = 31n31n21n21nn6

3n 2

- - - = + + + + = 33n32n6n4n6n12n6

3n 223

- - + = 4n9n16n6

3n 23

2.4.5Example: Sumthefollowingserieston terms

.....4.33.22.1 222 + + +

Solution:

Herethenthtermoftheseriesis ( ) 1nnt 2n + = .Therefore, ( ) 1nn......4.33.22.1S 2222n + + + + + =

( ) + = 1nn2

+ = 23 nn

+ = 23 nn

( ) ( )( ) 6

1n21nn4

1nn 22 + + +

+ =

( ) ( ) ( ) [ ] 1n221nn312

1nn + + +

+ =

( ) 12

2n7n31nn 2

+ + +

=

( ) ( ) ( ) 12

1n32n1nn + + + =



2.4.6Example: Sumtheseries1.2+3.5+5.8+.ton terms.

Solution:

Here,the nth termoftheseriesis ( )( ) 31n212n1tn - + - + = ( ) ( ) 1n31n2 - - =

Let nS denotethesumto n termsofthegivenseries.

Therefore, - - = )1n3()1n2(Sn

+ - = 1n5n6 2

+ - = 1n5n6 2

( )( ) ( ) n

2

1nn.5

6

1n21nn.6 +

+ -

+ + =

( )( ) ( ) n1nn251n21nn + + - + + =

( )( ) ( ) [ ] 21n51n21n22n + + - + + =

- + = 1nn4

2n 2

2.4.7Example: Findthesumton termsoftheseries

1.2.3+2.3.4+3.4.5+.

Solution:

Herethenthtermoftheseriesis ( )( ) 2n1nntn + + = .Therefore, = nn tS

( ) ( ) + + = 2n1nn

+ + = n2n3n 23

+ + = n2n3n 23



( ) ( )( ) ( ) 2

1nn.2

61n21nn

.34

1nn 22 + +

+ + +

+ =

( ) ( ) ( ) ( ) 1

1nn2

1n21nn4

1nn 22 + +

+ + +

+ =

( ) ( ) ( ) [ ] 41n221nn.4

1nn + + + +

+ =

( ) 4

6n5n1nn 2

+ + +

=

( ) ( ) ( ) 4

3n2n1nn + + + =

2.4.8Example: Findthesumton bracketsoftheseries

(1)+(1+2)+(1+2+3)+

Solution:

Forthegivenseries, nthbracket n....321 + + + + =

( ) n

21n

21

2

1nn 2 + = +

=

Thereforesumton terms = + n21n

21 2

( ) ( ) ( ) 4

1nn

12

1n21nn + +

+ + =

( ) ( ) [ ] 31n212

1nn + +

+ =

( ) ( ) 6

2n1nn + + =

2.4.9Example: Findthesumofthefollowingserieston terms.

....531

321

31

21

11

333333 +

+ +

+ + +

+

+ +



Solution:

The nth termofthegivenseriesis

( )

( )

( ) [ ] ( )

4

1n

1n212n

4

1nn

1n2....531

n......321t

2

22

3333

n +

= - +

+

= - + + + +

+ + + + =

41n

21n

411n2n

41 22 + + =

+ + = .

Therefore ( ) ( ) ( )

n41

2

1nn.

21

6

1n21nn.

41Sn +

+ +

+ + =

( )( ) ( ) [ ] 61n61n21n24n + + + + + =

+ + = 13n9n2

24n 2

2.5ArithmeticoGeometricseries(A.G.P)

Aseriesoftheform

( ) ( ) ( ) ...rd1na...rd2arda1.a n2 + - + + + + + + + Inwhicheach term is theproduct of corresponding terms inanarithmetic

andgeometricseriesiscalledanArithmeticoGeometricseries.

Forexample

i) Theseries + + + + 32 x7x5x31 ...formsanA.G.P.sinceeachtermis

the corresponding terms of the A. P 1, 3, 5, 7 and the G.P

32 x,x,x,1 ,

ii) Since1, 4, 7, 10 ,are inA.Pand2,4,8,16are inG.P the

series1x2+4x4+7x8+10x16+..i.e.,2+16+56+160+

...formsanA.G.P.



iii) The series ......647

165

431 + + + + formsanA.G.P since 1, 3, 5, 7

areinA.Pand32 4

1,4

1,41,1 ,.areinG.P.

2.5.1ThesumoftheFirstntermsofanArithmeticoGeometricseries

ThefirstntermsofthegeneralA.G.Pare

( ) ( ) ( ) 1n2 rd1na,...,rd2a,rda,a - - + + + LetSnbethesumofntermsofthisseries.Then,wehave

( ) ( ) ( ) ( ) 1n2n2n rd1nard2na....rd2ardaaS - - - + + - + + + + + + + = Multiplyingbothsidesbyr,

( ) ( ) ( ) n1n2n rd1nard2na.....rd2aarS.r - + + - + + + + + = -

Bysubtraction

( ) ( ) n1n2n rd1nadr...........drdraSr1 - + -

+ + + + = - -

( ) n2n rd1nar.....r1dra - + -

+ + + + = -

( ) n1n

rd1nar1

r1dra - + -

-

-

+ =

-

( ) ( ) ( )

r1

rd1na

r1

dr

r1

drr1

aSThereforen

2

n

2n -

- + -

- -

- +

- = (1)

Ifrisnumericallylessthan1,then nas0rn

ThereforeS,thesumtoinfinityis ( ) 2r1

drr1

a

- +

- (2)



2.5.2Example:

sum nto...............x7x5x31 32 + + + + terms.

Solution

Here1,3,5,7..areinA.Pwhosenth term ( ) 21n1 - + = 1n2 - =

and1, 2x,x ,areinG.P.whosenth term 1nx.1 - = .Hence,nth termof

thegivenseriesis ( ) .x1n2t 1nn - - = Let nS bethesumof n termsofthegivenseries.Then,wehave

( ) ( ) 1n2n32n x1n2x3n2.....x7x5x31S - - - + - + + + + + = Multiplybothsidesbythecommonratiox

( ) ( ) n1n32n x1n2x3n2.....x5x3xS.x - + - + + + + = - Bysubtraction,

( ) ( ) n1n2n x1n2x2.....x2x21x1S - - + + + = - -

( ) n2n2 x1n2x.....xx1x21 - -

+ + + + + = -

( ) n1n

x1n2x1

x1x21 - -

-

-

+ =

-

Therefore ( ) ( )

x1x1n2

x1

x1x2

x11S

n

2

1n

n - -

- -

-

+ -

=

-

Aliter:

Here xr,2d,1a = = = .HenceusingtheformulaforSngivenin(1),we

have



( ) ( ) ( )

x1

x1n2

x1

x2

x1

x2x1

1Sn

2

n

2n -

- -

- -

- +

- =

If x isnumericallylessthan1,then nas0xn

ThereforeS,thesumtoinfinity ( ) 2x1

x2x1

1

- +

- =

2.5.3Example:sum ......a3a21 2 + + + tonterms.Alsofindthesumto

infinity,if 1a <

Solution

Let nS bethesumtontermsofthegivenseries.Then

1n2n a.n.....a3a21S

- + + + + =

Multiplyingbythecommonratioa bothsides

n2n na....a2aS.a + + + =

Bysubtraction,

( ) n1n2n naa.....aa1Sa1 - + + + + = - -

nn

naa1a1

- -

- =

Therefore ( ) a1na

a1

a1S

n

2

n

n - -

-

- =

If ,1a < then 0an as n

Therefore,thesumtoinfinityis ( ) 2a11

-

2.5.4Example: Sum .....647

165

431 + + + + toinfinity



Solution:

LetSbethesumtoinfinityofthegivenseries.Then

+ + + + = 32 4

7

4

5431S ..

Multiplyingbothsidesbyc.r.41

32 4

5

4

341S

41 + + = +.

Subtracting,weget

+ + + + =

- infinityot.....

4

2

4

2

4

21S

4

11

32

211

4

21

4

2

1 = + =

-

+ =

i.e. 2S43 = Therefore

38S =

2.5.5Example: Sumtontermstheseries

.....9

10

9

7941

32 + + + + andshowthat

thesumtoinfinityoftheseriesis6499 .

Solution:

Let nS bethesumofthegivenseries.Then

Sn= .....9

10

9

7941

32 + + + +

1nn 9

2n3

9

5n3 -

- +

- +

Multiplyingbythec.r.91 bothsides,



n1n32n 9

2n3

9

5n3........

9

7

9

491S

91 - +

- + + + + =

-

Bysubtraction,

n1n32n 9

2n3

9

3......9

3

9

3931S

911

- -

+ + + + + =

-

-

n2n2 9

2n3

9

1......9

1911

931

- -

+ + + + + =

-

n

1n

9

2n3

911

9

11.

931

- -

-

- + =

-

n1nn 9

2n3

9

11

8

31S

9

8Therefore

- -

- + =

-

8

9

9

2n3

9

11

64

27

8

9STherefore

n1nn

- -

- + =

-

.9.8

2n3

9.64

36499

1n2n

- + - =

- -

Since < nas,191 ,thesecondtermontherighttendstozero.

Therefore S ,thesuminfinityis6499 .

2.6SummationofseriesbytheMethodofDifferences

Let .....uuuuu 4321n + + + + = be given series in which the law offormationoftheseriesisnotgiven.Insuchcaseweobtainanewseriesby

subtractingeachtermfromthesucceedingtermofthegivenseriesnamely

( ) ( ) ( ) ......uuuuuu 342312 + - + - + -



Which is called the series of the first order of differences which may be

eitheran P.A ora P.G .Thesedifferencesmaybeconvenientlydenotedby

n1nn321 uuuwhere......uuu - = + + + + D D D D

ThisiscalledtheMethodofDifferences.

2.6.1Example: Findthenth termandsumton termsoftheseries

......114805230144 + + + + + +

Solution

Thegivenseriesis4+14+30+52+80+114+..inwhichthelawofa

seriesisnotgiven.Theseriesofthefirstorderofdifferencesis

(144)+(3014)+(5230)+(8052)+(11480)+..

i.e.,10+16+22+28+34+.

Whichformsan P.A whosefirstterm=10c.d=6

LetSnbethesumtofirst n termsoftheseries

= \ nS 4+14+30+52+80+114+..+ nu

Also nS =4+14+30+52+80++ n1n uu + -

Bysubtraction,0=4+[10+16+22+28+..to 1n - terms] nu -

( ) [ ] 62n202

1n4un - +

- + =

( ) ( ) 4n31n4 + - + = nn3 2 + =

Therefore

+ = = nn3uS 2nn

+ = nn3 2

( ) ( ) ( ) 2

1nn6

1n21nn.3

+ +

+ + =

( ) ( ) ( ) 21nn2

2n21nn + =

+ + =



2.6.2Example: Findthenth termandthesumtofirst n termsoftheseries

6+13+24+39+58+..

Solution:

Thegivenseriesis6+13+24+39+58+...Theseriesofthefirstorder

ofdifferencesis

(136)+(2413)+(3924)+(5839)+

i.e.,7+11+15+19+

WhichformsanA.P.whosefirstterm=7c.d.=4.

LetSn bethesumtofirst n termsoftheseries.

Therefore = nS 6+13+24+39+58+..+ nu

AlsoSn=6+13+24++ n1n uu + -

Bysubtraction,

0=6+[7+11+15+to 1n - terms] nu -

Therefore ( ) [ ] 42n142

1n6un - +

- + =

( ) ( ) 3n21n6 + - + = 3nn2 2 + + =

Therefore

+ + = = 3nn2uS 2nn

+ + = 13nn2 2

( ) ( ) ( ) n3

21nn

61n21nn

.2 + +

+ + +

=

+ + = 23n9n4

6n 2

2.6.3Example: Find thesum to n termsof theseries1+4+13+40+

..



Solution:

Thegivenseriesis

1+4+13+40+.

Theseriesofthefirstorderofdifferencesis

(41)+(134)+(4013)+..

i.e.,3+9+27+..

whichformsaG.P.whosefirsttermis3andc.r.=3

LetSn bethesumtofirst n termsoftheseries.Then,

Sn=1+4+13+40+..+ nu

Also, Sn=1+4+13+.. n1n uu + -

Subtracting,0=1+(3+9+27+.to 1n - terms) nu -

Therefore23

231

13

1331u

n1n

n - + = -

-

+ =

-

- = 13

21 n

Therefore n21

13

133.

211

213

21S

n

nn - -

-

= - =

n2113

43 n -

- =

2.6.4SummationofseriesbyPartialFractions

Lettheseriesbedenotedby

n321 u......uuu + + + +

andits nth termby nu anditssumby nS .

Thesumoftheseriescaneasilybefoundbyexpressing nu asthedifference

oftwoquantitiesintheform n1nn vvu - = + ,where nv isafunctionofn



Byputting n.......3,2,1n = successivelyintherelation

Wehave

n1nn vvu - = +

121 vvu - =

232 vvu - =

343 vvu - =

..

and n1nn vvu - = +

Addinguptheserelations,wehave

11nn vvS - = +

Wedetermine nv eitherbyinspectionorusingpartialfractions.

2.6.5Example: sumtontermstheseries

.....11.8

18.5

15.2

1 + + +

Solution:

Here the nth term of 2, 5, 8,. is ( ) 1n331n2 - = - + and thenth termof5,8 ,11,is 5+ ( ) 2n331n + = - .Let nu bethe nthtermofthegivenseries.Then

( ) ( ) 2n31n31un

+ - =

Byinspection,wehave

+

- -

= 2n3

11n3

131un



Byputting n.....3,2,1n = successivelyintheaboveresult,wehave

- =

51

21

31u1

- =

71

51

31u2

- =

111

81

31u3

.

+

- -

= 2n3

11n3

131un

Addinguptheseresults,wehave

+

- = 2n3

1.2

131Sn

( ) 4n6n

2n32

n3.31

+ =

+ =

2.6.6Example: Sumtontermstheseries

......31.22

122.13

113.4

1 + +

Solution

Here the nth termof 4,13,22,.. is ( ) 5n991n4 - = - + and thenth termof13,22,31,..is 13+(n1)9=9n+4

Thereforenth termofthegivenseriesis

( ) ( ) 4n95n91un

+ - =

ThereforeByinspection, ( )

+ -

- =

4n9

15n9

191un



Byputting n =1,2,3, n successivelyintheaboveresult,wehave

- =

131

41

91u1

- =

221

131

91u2

- =

311

221

91u3

.

+

- -

= 4n9

15n9

191un

Addinguptheseresults,wehave

+

- = 4n9

141

91Sn

( ) 4n9444n9

.91

+ - +

=

( ) 4n94n

+ =

2.6.7Example:Sumtontermstheseries

( ) ( ) ( ) ( ) ( ) ( ) 1nana1.....

3a2a

12a1a

1

+ + + + +

+ + +

+ +

Solution:

Here,the nth termoftheseriesis

( ) ( ) 1nana1un

+ + + =

Byinspection,1na

1na

1un + + -

+ =



Byputtingn =1,2,3, n successively,intheaboveresult,wehave

2a1

1a1u1 +

- +

=

3a

1

2a

1u2 +

- +

=

4a1

3a1u3 +

- +

=

1na

1

na

1un + +

- +

=

Addingweget ( ) ( ) 1na1an

1na1

1a1Sn

+ + + =

+ + -

+ =


1) Findthen thtermoftheseries1+ + + + 81

41

21 .

2) If ththth r,qp terms of a P.G are x, y, z respectively, prove that

1z.y.x qpprrq = - - -

3) Showbymethod of mathematicalinductionthat 1n22n4 53 + + + is

divisibleby14,forallpositiveintegralvaluesofn .

4) Find the sum of n terms of the series

......3n32n21n 323232 +

- +

- +

-

5) Findthesumton termsoftheseries

.....2

1.

11.814

2

1.8.5

1121.

5.28

32 + + +



2.7Summary

InthisunitwestudiedthedifferenttypesprogressionslikeA.P.,G.P.and

H.P.ThePrincipleofMathematicalinductionwasstudiednextwithavariety

ofproblems.Summing theseriesofhigherpowersofn,with theconcepts

beingused to find sumofAGseries.Themethodofdifferencesand the

methodpartialfractionsisalsobeingusedheretosumtheseries.

2.8TerminalQuestions

1. Defineprogression

2. Brieflyexplainsummationofseriesbythemethodofdifferences

2.9Answers


1. Formsa P.G inwhich1n

1n

n2

121.1t,

21r,1a

-

-

=

= = =

2. Let A bethefirsttermand R thecommonratioof the P.G .Thenby

hypothesis,wehave

zARtyARtxARt 1rr1q

q1p

p = = = = = = - - -

Raisingtheseequationsrespectivelybythepowers qp,pr,rq - - -

andmultiplying,weget

( ) ( ) ( )( ) ( ) ( ) qp1rqppr1q

.prrq1prqqpprrq

R.A.R

A.R.Az.y.x - - - - -

- - - - - - - =

( )( ) ( )( ) ( ) ( ) qp1xpr1qrq1pqpprrq R.A - - + - - + - - - + - + - =

1R.A 00 = =



3. Let ( ) 1n22n4 53nP + + + = Now ( ) 611485412572953531P 361224 = = + = + = + = + + Therefore ( ) 1p isdivisibleby14.Letusassumethat ( ) 1m22m4 53mP + + + = isdivisibleby14.

Consider ( ) ( ) ( ) 11m221m4 531mP + + + + + = + ( ) 3m26m4 531mP + + + = +

Therefore

( ) ( ) 1m23m22m46m4 5533mP1mP + + + + - + - = - +

- +

- = + + 155133 21m242m4

1m22m4 5.243.80 + + + =

( ) ( ) 1m22m4 51014310145 + + + + + = 1m22m4 5143145 + + + =

+ + + + 1m22m4 5310

SinceeachtermontheR.H.Sisdivisibleby14theirsumisalsodivisibleby14.

Therefore ( ) ( ) mP1mP - + isdivisibleby14.Therefore ( ) ( ) k14mP1mP = - + forsomeintegerkTherefore ( ) ( ) k14mp1mP + = +

Since ( ) mP isdivisibleby14, ( ) 1mP + isdivisibleby14.Hencebyinduction ( ) nP isdivisibleby14forallpositiveintegers n .

4.

+ + + + - + + + + = 33332222n n....321n.n....n3n2nS

( )

+ + + + - + + + + = 33332 n.....321n.....321n

( ) ( ) 4

1nn

2

1nn.n

222 + -

+ =

- = 1nn

41 22



5. Herethenthtermof8,11,14,is ( ) 5n331n8 + = - + n th termof2,5,8,is2+ ( ) 1n331n - = - n th termof5,8,11,..is5+ ( ) 2n331n + = - n th termof 32 2,2,2 .is n2 =

Thereforen th termoftheseriesis

( ) ( ) nn 21.

2n31n3

5n3u

+ -

+ =

ByPartialFractions,

nn 2

1.2n3

11n3

2u

+

- -

=

n1n 2

1.2n3

1

2

1.1n3

1 +

- -

= -

Byputtingn =1,2,3successivelyintheaboveresult,wehave

21.

51

2

1.21u

01 - =

22 2

1.81

21.

51u - =

323 2

1.111

2

1.81u - =

..

n1nn 2

1.2n3

1

2

1.1n3

1u +

- -

= -

Byadding,

nn 2

1.2n3

121S

+ - =

TerminalQuestions

1. RefertoSection2.2

2. RefertoSection2.6

Documents

MC0063(a) Unit2 Final