Nizovi-zadaci II deo.pdf

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1

Odrediti graničnu vrednost niza čiji je opšti član

a)2 2 2 2

3

1 2 3 n

na

n

+ + + +=

b)1 3 5 (2 1)

2 4 6 (2 )n

na

n

+ + + + minus=

+ + + +

c)2 2 2 2

1 1 2 3 3 5 (2 1)

1 2 3 n

n na

n

sdot + sdot + sdot + + sdot minus=

+ + + +

Rešenje

Profesori najčešće dozvoljavaju da se za ove i slične zbirove koriste gotove formule Njihovo

dokazivanje se radi matematičkom indukcijom ( pogledajte fajl kod nas )

Ali ima i profesora koji zahtevaju da se te formulice izvedu Evo nekoliko ideja

1 2 3 n+ + + + =

Postoji priča da je Gaus ovaj zbir našao još u I razredu osnovne škole pa ćemo u čast njemu

rešenje ovog zadatka zvati Gausova dosetka

Stvar je ustvari vrlo prosta ( ali se treba setiti )

Označimo početni zbir slovom S

1 2 3 n S + + + + =

On je okrenuo redosled brojeva napisao ispod početnog zbira i sabrao te dve jednakosti

Dakle

( 1) 2

( 1)

2

( 1)1 2 3

2

n n S

n nS

n nn

sdot + =

sdot +=

sdot ++ + + + =



983090

Sličan trik možemo primeniti i za zbir prvih n parnih ili neparnih brojeva

2 4 6 2 n+ + + + =

2 4 6 2n S + + + + =

2 (2 2) (2 4) 2n n n S + minus + minus + + =

Odavde je

(2 2) 2 2 ( 1) 22 ( 1)

( 1)2

2 4 6 2 ( 1)

n n S n n S

n nS n n

n n n

sdot + = rarr sdot + =sdot += = sdot +

+ + + + = sdot +

Za zbir prvih n neparnih brojeva bi bilo

( )

( )

( )

2

2

1 3 5 2 1

1 3 5 2 1

(2 1) (2 3) 1

2 2 2 2

2 2

1 3 5 2 1

n

n S

n n S

n n n S

n n S

S n

n n

+ + + + minus =

+ + + + minus =

minus + minus + + =

+ + + =

sdot =

=

+ + + + minus =

U drugačijim situacijama ćemo koristiti neko trikče i neki od ova tri rezultata

Recimo 2 2 2 21 2 3 n+ + + + =

Podjimo od formulice za kub binoma

3 3 2

3 3 2

( 1) 3 3 1 a odavde je

( 1) 3 3 1

k k k k

k k k k

+ = + + +

+ minus = + +



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Ovde ćemo redom stavljati umesto k vrednosti k=1k=2k=3k=n

3 3 2

3 3 2

3 3 2

3 3 2

3 3 2

(2) 1 3 1 3 1 1

(3) 2 3 2 3 2 1

(4) 3 3 3 3 3 1

( ) ( 1) 3 ( 1) 3 ( 1) 1

( 1) 3 3 1

n n n n

n n n n

minus = sdot + sdot +



minus minus = sdot minus + sdot minus +

+ minus = sdot + sdot +

Sad saberemo sve ove jednakostiNa levoj strani se potiru svi sem

3 3 2 2 2

3 2 2 2 2

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

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

n n n n

n n n n n n

+ minus = sdot + + + + sdot + + + + sdot

+ sdot + = sdot + + + + sdot + + + + sdot

Iskoristimo da je( 1)

1 2 3 2

n nn

sdot ++ + + + = i izrazimo ono što nam treba

3 2 2 2 2

3 2 2 2 2

2 2 2 3 2 2

2 2 2 3 2

2 2 2 2

2 2 2

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

2

2 6 6 6 (1 2 ) 3 ( 1) 2

6 (1 2 ) 2 6 6 3 3 2

6 (1 2 ) 2 3

6 (1 2 ) (2 3 1)

6 (1 2 )

n nn n n n n

n n n n n n n

n n n n n n n

n n n n

n n n n

n

++ sdot + = sdot + + + + sdot +

+ sdot + = sdot + + + + sdot + +

sdot + + + = + sdot + minus sdot minus sdot minus

sdot + + + = minus sdot +

sdot + + + = + sdot +

sdot + + + =

2 2 2

( 1)(2 1)

( 1)(2 1)1 2

6

n n n

n n nn

+ +

+ ++ + + =

U našim primerima se javlja i

1 1 2 3 3 5 (2 1) n nsdot + sdot + sdot + + sdot minus =

Ovaj zadnji član nam daje ideju šta treba raditi 2(2 1) 2n n n nsdot minus = minus

Ako primenimo ovo na svaki sabirak niza dobijamo

2 2 2

2 2 2

2 2 2

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

2 1 2 2 2 3 2 1 2 3

2 (1 2 3 ) (1 2 3 )

Iskoristimo formulice od malopre

( 1)(2 1) ( 1) ( 1)(2(2 1) 3) ( 1)(4 1)2

6 2 6 6

n n

n n

n n

n n n n n n n n n n n

sdot minus + sdot minus + sdot minus + + sdot minus =

sdot + sdot + sdot + + sdot minus minus minus minus minus =sdot + + + + minus + + + + =

+ + sdot + sdot + + minus sdot + minussdot minus = =

( 1)(4 1)1 1 2 3 3 5 (2 1)

6

Dakle

n n nn n

sdot + minussdot + sdot + sdot + + sdot minus =



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Sad sa lakoćom pošto znamo formule možemo rešiti zadate primere

a)

2 2 2 2 3 2

3 3 3

( 1)(2 1)1 2 3 2 3 2 16lim lim lim

6 6 3n n n

n n n

n n n n

n n nrarrinfin rarrinfin rarrinfin

+ ++ + + + + +

= = = =

b)

2 2

2

1 3 5 (2 1) 1lim lim lim 1

2 4 6 (2 ) ( 1) 1n n n

n n n

n n n n nrarrinfin rarrinfin rarrinfin

+ + + + minus= = = =

+ + + + + +

c)

2 2 2 2

( 1)( 1)(4 1)1 1 2 3 3 5 (2 1) 6lim lim lim

( 1)(2 1)1 2 3

6

n n n

n nn n n

n n

n n nnrarrinfin rarrinfin rarrinfin

sdot +sdot + minussdot + sdot + sdot + + sdot minus

= =+ ++ + + +

(4 1)

6

n minus

( 1)n n + (2 1)

6

n +

4 1lim 2

2 1n

n

nrarrinfin

minus= =

+

2

Izračunati graniču vrednost niza

a)1 1 1

lim 1 2 2 3 ( 1)n n nrarrinfin

+ + +

sdot sdot sdot +

b)2 2

lim 1 2 3 ( 1)( 2)n n n nrarrinfin

+ +

sdot sdot sdot + +

Rešenje

a)

Ideja kod ovog tipa zadatka je da pokušamo da rastavimo sabirke( Pogledajte fajl o integraciji

racionalne funkcije vezan za integrale)



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1Ovde su A i B konstante koje tražimo

( 1) 1

1 ( 1)

( 1) 1

1 ( 1)

1

1 ( ) Sad vršimo uporedjivanje uz n pa slobodne članove0

1 1

1

(

A B

n n n n

A Bn n

n n n n

A n Bn

An A Bn

n A B A A B

A B

Dakle

n n

= + rarrsdot + +

= + sdot +sdot + +

= + +

= + +

= + + rarr+ =

= rarr = minus

sdot

1 1

1) 1n n= minus

+ +

Evo nama ideje kako razbijamo svaki član

1 1 1 1 1 1 1 1 1 1 1 1 2 2 3 ( 1) 1 2 2 3 3 4 1

Svi unutrašnji članovi se potiru

1 1

1 2

n n n n+ + + = minus + minus + minus + + minussdot sdot sdot + +

= minus1

2+

1

3minus

1

3+

1

4minus

1

n+ +

1 11

1 1n nminus = minus

+ +

1 1 1 1lim lim 1 1 0 1

1 2 2 3 ( 1) 1n nn n nrarrinfin rarrinfin

+ + + = minus = minus = sdot sdot sdot + +

b)

2 2 2

2 2 2

2

2 ( 1)( 2)

( 1)( 2) 1 2

2 ( 1)( 2) ( 2) ( 1)

2 ( 3 2) 2

2 3 2 2

2 ( ) (3 2 ) 2

0

3 2 0

2 2 1

1

2

A B C n n n

n n n n n n

A n n Bn n Cn n

A n n Bn Bn Cn Cn

An An A Bn Bn Cn Cn

n A B C n A B C A

A B C

A B C

A A

B C

B

= + + sdot + +sdot + + + +

= + + + + + +

= + + + + + +

= + + + + + +

= + + + + + +

+ + =

+ + =

= rarr =

+ = minus

+ 3

2 1

2 1 2 1

( 1)( 2) 1 2

C

B C

n n n n n n

= minus

= minus rarr =

= minus +sdot + + + +



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E sad je zgodno zadatak napisati preko sume

1

2 2 2lim lim

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

n

n nk n n n k k k rarrinfin rarrinfin

=

+ + =

sdot sdot sdot + + sdot + + sum

Primenimo razlaganje koje smo našli

1 1 1 1

1

1 2

2

1 3

1

1 1 1 1 1 2

2 1 1 12

( 1)( 2) 1 2

1Doteramo sve tri sume da budu oblika

1 1

1

1 1

2

Sada je

2 1 1 1 1 12 2

( 1)( 2) 1 2

n n n n

k k k k

n n

k k

n n

k k

n n n n n n

k k k k k k

k k k k k k

k

k k

k k

k k k k k k k k

= = = =

+

= =

+

= =

+

= = = = = =


=+

=+

= minus + = minussdot + + + +

sum sum sum sum

sum sum

sum sum

sum sum sum sum sum sum2

3

1 3

1

2 3

2

3 3

1 2

1 2 3

1

Dalje doteramo sve sume da kreću od k=3 a da se završavaju sa n

1 1 1 1

1 2

1 1 1 1

2 1

1 1 1 11 2

Vratimo se na zadatak

1 1 1 12

1

n

k

n n

k k

n n

k k

n n

k k

n n n

k k k

k

k k

k k n

k k n n

k k k

+

=

= =

+

= =

+

= =

+ +

= = =

+

= + +

= + ++

= + ++ +

minus + =

sum

sum sum

sum sum

sum sum

sum sum sum3 3 3

3 3 3

1 1 1 1 1 1 1 12

2 2 1 1 2

Ovo malo prisredimo

1 1 1 2 1 1 11 1 2

2 1 1 2

Primetimo da se sume potiru

1 1 1

2 1 2

Sad pustimo limes

1 1 1lim

2 1 2

n n n

k k k

n n n

k k k

n

k k n k n n

k k n k n n

n n

n n

= = =

= = =

rarrinfin

+ + minus + + + + + + + +

= + + minus minus minus + + ++ + +

= minus ++ +

minus + + +

sum sum sum

sum sum sum

1 wwwmatematiranjeinrs

2=



983090

Sličan trik možemo primeniti i za zbir prvih n parnih ili neparnih brojeva

2 4 6 2 n+ + + + =

2 4 6 2n S + + + + =

2 (2 2) (2 4) 2n n n S + minus + minus + + =

Odavde je

(2 2) 2 2 ( 1) 22 ( 1)

( 1)2

2 4 6 2 ( 1)

n n S n n S

n nS n n

n n n

sdot + = rarr sdot + =sdot += = sdot +

+ + + + = sdot +

Za zbir prvih n neparnih brojeva bi bilo

( )

( )

( )

2

2

1 3 5 2 1

1 3 5 2 1

(2 1) (2 3) 1

2 2 2 2

2 2

1 3 5 2 1

n

n S

n n S

n n n S

n n S

S n

n n

+ + + + minus =

+ + + + minus =

minus + minus + + =

+ + + =

sdot =

=

+ + + + minus =

U drugačijim situacijama ćemo koristiti neko trikče i neki od ova tri rezultata

Recimo 2 2 2 21 2 3 n+ + + + =

Podjimo od formulice za kub binoma

3 3 2

3 3 2

( 1) 3 3 1 a odavde je

( 1) 3 3 1

k k k k

k k k k

+ = + + +

+ minus = + +



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3 3 2

3 3 2

3 3 2

3 3 2

3 3 2

(2) 1 3 1 3 1 1

(3) 2 3 2 3 2 1

(4) 3 3 3 3 3 1

( ) ( 1) 3 ( 1) 3 ( 1) 1

( 1) 3 3 1

n n n n

n n n n







3 3 2 2 2

3 2 2 2 2

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

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

n n n n

n n n n n n




1 2 3 2

n nn


3 2 2 2 2

3 2 2 2 2

2 2 2 3 2 2

2 2 2 3 2

2 2 2 2

2 2 2

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

2

2 6 6 6 (1 2 ) 3 ( 1) 2

6 (1 2 ) 2 6 6 3 3 2

6 (1 2 ) 2 3

6 (1 2 ) (2 3 1)

6 (1 2 )

n nn n n n n

n n n n n n n

n n n n n n n

n n n n

n n n n

n

++ sdot + = sdot + + + + sdot +

+ sdot + = sdot + + + + sdot + +



sdot + + + = + sdot +

sdot + + + =

2 2 2

( 1)(2 1)

( 1)(2 1)1 2

6

n n n

n n nn

+ +

+ ++ + + =





2 2 2

2 2 2

2 2 2

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

2 1 2 2 2 3 2 1 2 3

2 (1 2 3 ) (1 2 3 )


( 1)(2 1) ( 1) ( 1)(2(2 1) 3) ( 1)(4 1)2

6 2 6 6

n n

n n

n n





( 1)(4 1)1 1 2 3 3 5 (2 1)

6

Dakle

n n nn n




983092


a)

2 2 2 2 3 2

3 3 3

( 1)(2 1)1 2 3 2 3 2 16lim lim lim

6 6 3n n n

n n n

n n n n


+ ++ + + + + +

= = = =

b)

2 2

2

1 3 5 (2 1) 1lim lim lim 1

2 4 6 (2 ) ( 1) 1n n n

n n n


+ + + + minus= = = =

+ + + + + +

c)

2 2 2 2

( 1)( 1)(4 1)1 1 2 3 3 5 (2 1) 6lim lim lim

( 1)(2 1)1 2 3

6

n n n

n nn n n

n n



= =+ ++ + + +

(4 1)

6

n minus

( 1)n n + (2 1)

6

n +

4 1lim 2

2 1n

n

nrarrinfin

minus= =

+

2


a)1 1 1


+ + +

sdot sdot sdot +

b)2 2


+ +

sdot sdot sdot + +

Rešenje

a)





983093


( 1) 1

1 ( 1)

( 1) 1

1 ( 1)

1


1 1

1

(

A B

n n n n

A Bn n

n n n n

A n Bn

An A Bn

n A B A A B

A B

Dakle

n n

= + rarrsdot + +

= + sdot +sdot + +

= + +

= + +

= + + rarr+ =

= rarr = minus

sdot

1 1

1) 1n n= minus

+ +


1 1 1 1 1 1 1 1 1 1 1 1 2 2 3 ( 1) 1 2 2 3 3 4 1


1 1

1 2


= minus1

2+

1

3minus

1

3+

1

4minus

1

n+ +

1 11

1 1n nminus = minus

+ +

1 1 1 1lim lim 1 1 0 1



b)

2 2 2

2 2 2

2

2 ( 1)( 2)

( 1)( 2) 1 2

2 ( 1)( 2) ( 2) ( 1)

2 ( 3 2) 2

2 3 2 2

2 ( ) (3 2 ) 2

0

3 2 0

2 2 1

1

2

A B C n n n

n n n n n n

A n n Bn n Cn n

A n n Bn Bn Cn Cn

An An A Bn Bn Cn Cn

n A B C n A B C A

A B C

A B C

A A

B C

B

= + + sdot + +sdot + + + +

= + + + + + +

= + + + + + +

= + + + + + +

= + + + + + +

+ + =

+ + =

= rarr =

+ = minus

+ 3

2 1

2 1 2 1

( 1)( 2) 1 2

C

B C

n n n n n n

= minus

= minus rarr =




983094


1

2 2 2lim lim

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

n


=

+ + =



1 1 1 1

1

1 2

2

1 3

1

1 1 1 1 1 2

2 1 1 12

( 1)( 2) 1 2


1 1

1

1 1

2

Sada je

2 1 1 1 1 12 2

( 1)( 2) 1 2

n n n n

k k k k

n n

k k

n n

k k

n n n n n n

k k k k k k

k k k k k k

k

k k

k k

k k k k k k k k

= = = =

+

= =

+

= =

+

= = = = = =


=+

=+


sum sum sum sum

sum sum

sum sum


3

1 3

1

2 3

2

3 3

1 2

1 2 3

1


1 1 1 1

1 2

1 1 1 1

2 1

1 1 1 11 2


1 1 1 12

1

n

k

n n

k k

n n

k k

n n

k k

n n n

k k k

k

k k

k k n

k k n n

k k k

+

=

= =

+

= =

+

= =

+ +

= = =

+

= + +

= + ++

= + ++ +

minus + =

sum

sum sum

sum sum

sum sum

sum sum sum3 3 3

3 3 3

1 1 1 1 1 1 1 12

2 2 1 1 2

Ovo malo prisredimo

1 1 1 2 1 1 11 1 2

2 1 1 2


1 1 1

2 1 2

Sad pustimo limes

1 1 1lim

2 1 2

n n n

k k k

n n n

k k k

n

k k n k n n

k k n k n n

n n

n n

= = =

= = =

rarrinfin

+ + minus + + + + + + + +


= minus ++ +

minus + + +

sum sum sum

sum sum sum


2=



983091


3 3 2

3 3 2

3 3 2

3 3 2

3 3 2

(2) 1 3 1 3 1 1

(3) 2 3 2 3 2 1

(4) 3 3 3 3 3 1

( ) ( 1) 3 ( 1) 3 ( 1) 1

( 1) 3 3 1

n n n n

n n n n







3 3 2 2 2

3 2 2 2 2

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

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

n n n n

n n n n n n




1 2 3 2

n nn


3 2 2 2 2

3 2 2 2 2

2 2 2 3 2 2

2 2 2 3 2

2 2 2 2

2 2 2

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

2

2 6 6 6 (1 2 ) 3 ( 1) 2

6 (1 2 ) 2 6 6 3 3 2

6 (1 2 ) 2 3

6 (1 2 ) (2 3 1)

6 (1 2 )

n nn n n n n

n n n n n n n

n n n n n n n

n n n n

n n n n

n

++ sdot + = sdot + + + + sdot +

+ sdot + = sdot + + + + sdot + +



sdot + + + = + sdot +

sdot + + + =

2 2 2

( 1)(2 1)

( 1)(2 1)1 2

6

n n n

n n nn

+ +

+ ++ + + =





2 2 2

2 2 2

2 2 2

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

2 1 2 2 2 3 2 1 2 3

2 (1 2 3 ) (1 2 3 )


( 1)(2 1) ( 1) ( 1)(2(2 1) 3) ( 1)(4 1)2

6 2 6 6

n n

n n

n n





( 1)(4 1)1 1 2 3 3 5 (2 1)

6

Dakle

n n nn n




983092


a)

2 2 2 2 3 2

3 3 3

( 1)(2 1)1 2 3 2 3 2 16lim lim lim

6 6 3n n n

n n n

n n n n


+ ++ + + + + +

= = = =

b)

2 2

2

1 3 5 (2 1) 1lim lim lim 1

2 4 6 (2 ) ( 1) 1n n n

n n n


+ + + + minus= = = =

+ + + + + +

c)

2 2 2 2

( 1)( 1)(4 1)1 1 2 3 3 5 (2 1) 6lim lim lim

( 1)(2 1)1 2 3

6

n n n

n nn n n

n n



= =+ ++ + + +

(4 1)

6

n minus

( 1)n n + (2 1)

6

n +

4 1lim 2

2 1n

n

nrarrinfin

minus= =

+

2


a)1 1 1


+ + +

sdot sdot sdot +

b)2 2


+ +

sdot sdot sdot + +

Rešenje

a)





983093


( 1) 1

1 ( 1)

( 1) 1

1 ( 1)

1


1 1

1

(

A B

n n n n

A Bn n

n n n n

A n Bn

An A Bn

n A B A A B

A B

Dakle

n n

= + rarrsdot + +

= + sdot +sdot + +

= + +

= + +

= + + rarr+ =

= rarr = minus

sdot

1 1

1) 1n n= minus

+ +


1 1 1 1 1 1 1 1 1 1 1 1 2 2 3 ( 1) 1 2 2 3 3 4 1


1 1

1 2


= minus1

2+

1

3minus

1

3+

1

4minus

1

n+ +

1 11

1 1n nminus = minus

+ +

1 1 1 1lim lim 1 1 0 1



b)

2 2 2

2 2 2

2

2 ( 1)( 2)

( 1)( 2) 1 2

2 ( 1)( 2) ( 2) ( 1)

2 ( 3 2) 2

2 3 2 2

2 ( ) (3 2 ) 2

0

3 2 0

2 2 1

1

2

A B C n n n

n n n n n n

A n n Bn n Cn n

A n n Bn Bn Cn Cn

An An A Bn Bn Cn Cn

n A B C n A B C A

A B C

A B C

A A

B C

B

= + + sdot + +sdot + + + +

= + + + + + +

= + + + + + +

= + + + + + +

= + + + + + +

+ + =

+ + =

= rarr =

+ = minus

+ 3

2 1

2 1 2 1

( 1)( 2) 1 2

C

B C

n n n n n n

= minus

= minus rarr =




983094


1

2 2 2lim lim

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

n


=

+ + =



1 1 1 1

1

1 2

2

1 3

1

1 1 1 1 1 2

2 1 1 12

( 1)( 2) 1 2


1 1

1

1 1

2

Sada je

2 1 1 1 1 12 2

( 1)( 2) 1 2

n n n n

k k k k

n n

k k

n n

k k

n n n n n n

k k k k k k

k k k k k k

k

k k

k k

k k k k k k k k

= = = =

+

= =

+

= =

+

= = = = = =


=+

=+


sum sum sum sum

sum sum

sum sum


3

1 3

1

2 3

2

3 3

1 2

1 2 3

1


1 1 1 1

1 2

1 1 1 1

2 1

1 1 1 11 2


1 1 1 12

1

n

k

n n

k k

n n

k k

n n

k k

n n n

k k k

k

k k

k k n

k k n n

k k k

+

=

= =

+

= =

+

= =

+ +

= = =

+

= + +

= + ++

= + ++ +

minus + =

sum

sum sum

sum sum

sum sum

sum sum sum3 3 3

3 3 3

1 1 1 1 1 1 1 12

2 2 1 1 2

Ovo malo prisredimo

1 1 1 2 1 1 11 1 2

2 1 1 2


1 1 1

2 1 2

Sad pustimo limes

1 1 1lim

2 1 2

n n n

k k k

n n n

k k k

n

k k n k n n

k k n k n n

n n

n n

= = =

= = =

rarrinfin

+ + minus + + + + + + + +


= minus ++ +

minus + + +

sum sum sum

sum sum sum


2=



983092


a)

2 2 2 2 3 2

3 3 3

( 1)(2 1)1 2 3 2 3 2 16lim lim lim

6 6 3n n n

n n n

n n n n


+ ++ + + + + +

= = = =

b)

2 2

2

1 3 5 (2 1) 1lim lim lim 1

2 4 6 (2 ) ( 1) 1n n n

n n n


+ + + + minus= = = =

+ + + + + +

c)

2 2 2 2

( 1)( 1)(4 1)1 1 2 3 3 5 (2 1) 6lim lim lim

( 1)(2 1)1 2 3

6

n n n

n nn n n

n n



= =+ ++ + + +

(4 1)

6

n minus

( 1)n n + (2 1)

6

n +

4 1lim 2

2 1n

n

nrarrinfin

minus= =

+

2


a)1 1 1


+ + +

sdot sdot sdot +

b)2 2


+ +

sdot sdot sdot + +

Rešenje

a)





983093


( 1) 1

1 ( 1)

( 1) 1

1 ( 1)

1


1 1

1

(

A B

n n n n

A Bn n

n n n n

A n Bn

An A Bn

n A B A A B

A B

Dakle

n n

= + rarrsdot + +

= + sdot +sdot + +

= + +

= + +

= + + rarr+ =

= rarr = minus

sdot

1 1

1) 1n n= minus

+ +


1 1 1 1 1 1 1 1 1 1 1 1 2 2 3 ( 1) 1 2 2 3 3 4 1


1 1

1 2


= minus1

2+

1

3minus

1

3+

1

4minus

1

n+ +

1 11

1 1n nminus = minus

+ +

1 1 1 1lim lim 1 1 0 1



b)

2 2 2

2 2 2

2

2 ( 1)( 2)

( 1)( 2) 1 2

2 ( 1)( 2) ( 2) ( 1)

2 ( 3 2) 2

2 3 2 2

2 ( ) (3 2 ) 2

0

3 2 0

2 2 1

1

2

A B C n n n

n n n n n n

A n n Bn n Cn n

A n n Bn Bn Cn Cn

An An A Bn Bn Cn Cn

n A B C n A B C A

A B C

A B C

A A

B C

B

= + + sdot + +sdot + + + +

= + + + + + +

= + + + + + +

= + + + + + +

= + + + + + +

+ + =

+ + =

= rarr =

+ = minus

+ 3

2 1

2 1 2 1

( 1)( 2) 1 2

C

B C

n n n n n n

= minus

= minus rarr =




983094


1

2 2 2lim lim

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

n


=

+ + =



1 1 1 1

1

1 2

2

1 3

1

1 1 1 1 1 2

2 1 1 12

( 1)( 2) 1 2


1 1

1

1 1

2

Sada je

2 1 1 1 1 12 2

( 1)( 2) 1 2

n n n n

k k k k

n n

k k

n n

k k

n n n n n n

k k k k k k

k k k k k k

k

k k

k k

k k k k k k k k

= = = =

+

= =

+

= =

+

= = = = = =


=+

=+


sum sum sum sum

sum sum

sum sum


3

1 3

1

2 3

2

3 3

1 2

1 2 3

1


1 1 1 1

1 2

1 1 1 1

2 1

1 1 1 11 2


1 1 1 12

1

n

k

n n

k k

n n

k k

n n

k k

n n n

k k k

k

k k

k k n

k k n n

k k k

+

=

= =

+

= =

+

= =

+ +

= = =

+

= + +

= + ++

= + ++ +

minus + =

sum

sum sum

sum sum

sum sum

sum sum sum3 3 3

3 3 3

1 1 1 1 1 1 1 12

2 2 1 1 2

Ovo malo prisredimo

1 1 1 2 1 1 11 1 2

2 1 1 2


1 1 1

2 1 2

Sad pustimo limes

1 1 1lim

2 1 2

n n n

k k k

n n n

k k k

n

k k n k n n

k k n k n n

n n

n n

= = =

= = =

rarrinfin

+ + minus + + + + + + + +


= minus ++ +

minus + + +

sum sum sum

sum sum sum


2=



983093


( 1) 1

1 ( 1)

( 1) 1

1 ( 1)

1


1 1

1

(

A B

n n n n

A Bn n

n n n n

A n Bn

An A Bn

n A B A A B

A B

Dakle

n n

= + rarrsdot + +

= + sdot +sdot + +

= + +

= + +

= + + rarr+ =

= rarr = minus

sdot

1 1

1) 1n n= minus

+ +


1 1 1 1 1 1 1 1 1 1 1 1 2 2 3 ( 1) 1 2 2 3 3 4 1


1 1

1 2


= minus1

2+

1

3minus

1

3+

1

4minus

1

n+ +

1 11

1 1n nminus = minus

+ +

1 1 1 1lim lim 1 1 0 1



b)

2 2 2

2 2 2

2

2 ( 1)( 2)

( 1)( 2) 1 2

2 ( 1)( 2) ( 2) ( 1)

2 ( 3 2) 2

2 3 2 2

2 ( ) (3 2 ) 2

0

3 2 0

2 2 1

1

2

A B C n n n

n n n n n n

A n n Bn n Cn n

A n n Bn Bn Cn Cn

An An A Bn Bn Cn Cn

n A B C n A B C A

A B C

A B C

A A

B C

B

= + + sdot + +sdot + + + +

= + + + + + +

= + + + + + +

= + + + + + +

= + + + + + +

+ + =

+ + =

= rarr =

+ = minus

+ 3

2 1

2 1 2 1

( 1)( 2) 1 2

C

B C

n n n n n n

= minus

= minus rarr =




983094


1

2 2 2lim lim

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

n


=

+ + =



1 1 1 1

1

1 2

2

1 3

1

1 1 1 1 1 2

2 1 1 12

( 1)( 2) 1 2


1 1

1

1 1

2

Sada je

2 1 1 1 1 12 2

( 1)( 2) 1 2

n n n n

k k k k

n n

k k

n n

k k

n n n n n n

k k k k k k

k k k k k k

k

k k

k k

k k k k k k k k

= = = =

+

= =

+

= =

+

= = = = = =


=+

=+


sum sum sum sum

sum sum

sum sum


3

1 3

1

2 3

2

3 3

1 2

1 2 3

1


1 1 1 1

1 2

1 1 1 1

2 1

1 1 1 11 2


1 1 1 12

1

n

k

n n

k k

n n

k k

n n

k k

n n n

k k k

k

k k

k k n

k k n n

k k k

+

=

= =

+

= =

+

= =

+ +

= = =

+

= + +

= + ++

= + ++ +

minus + =

sum

sum sum

sum sum

sum sum

sum sum sum3 3 3

3 3 3

1 1 1 1 1 1 1 12

2 2 1 1 2

Ovo malo prisredimo

1 1 1 2 1 1 11 1 2

2 1 1 2


1 1 1

2 1 2

Sad pustimo limes

1 1 1lim

2 1 2

n n n

k k k

n n n

k k k

n

k k n k n n

k k n k n n

n n

n n

= = =

= = =

rarrinfin

+ + minus + + + + + + + +


= minus ++ +

minus + + +

sum sum sum

sum sum sum


2=



983094


1

2 2 2lim lim

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

n


=

+ + =



1 1 1 1

1

1 2

2

1 3

1

1 1 1 1 1 2

2 1 1 12

( 1)( 2) 1 2


1 1

1

1 1

2

Sada je

2 1 1 1 1 12 2

( 1)( 2) 1 2

n n n n

k k k k

n n

k k

n n

k k

n n n n n n

k k k k k k

k k k k k k

k

k k

k k

k k k k k k k k

= = = =

+

= =

+

= =

+

= = = = = =


=+

=+


sum sum sum sum

sum sum

sum sum


3

1 3

1

2 3

2

3 3

1 2

1 2 3

1


1 1 1 1

1 2

1 1 1 1

2 1

1 1 1 11 2


1 1 1 12

1

n

k

n n

k k

n n

k k

n n

k k

n n n

k k k

k

k k

k k n

k k n n

k k k

+

=

= =

+

= =

+

= =

+ +

= = =

+

= + +

= + ++

= + ++ +

minus + =

sum

sum sum

sum sum

sum sum

sum sum sum3 3 3

3 3 3

1 1 1 1 1 1 1 12

2 2 1 1 2

Ovo malo prisredimo

1 1 1 2 1 1 11 1 2

2 1 1 2


1 1 1

2 1 2

Sad pustimo limes

1 1 1lim

2 1 2

n n n

k k k

n n n

k k k

n

k k n k n n

k k n k n n

n n

n n

= = =

= = =

rarrinfin

+ + minus + + + + + + + +


= minus ++ +

minus + + +

sum sum sum

sum sum sum


2=

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