Evaluation of various partial sums of Gaussian q-binomial sums

Transcription

1 Arab J Math (018) 7: Arabian Journal of Matheatics Erah Kılıç Evaluation of various partial sus of Gaussian -binoial sus Received: 3 February 016 / Accepted: Noveber 017 / Published online: 8 Deceber 017 The Author(s) 017 This article is an open access publication Abstract We present three new sets of weighted partial sus of the Gaussian -binoial coefficients To prove the claied results we will use -analysis Rothe s forula and a -version of the celebrated algorith of Zeilberger Finally we give soe applications of our results to generalized Fibonoial sus Matheatics Subject Classification 11B65 05A30 1 Introduction The authors of [ present soe results about the partial sus of rows of Pascal s triangle as well as its alternating analogues As alternating partial su of Pascal s triangle fro [ wehave ( ) ( ) r r 1 ( 1) ( 1) integer The authors ) note that there is no closed for for the partial su of a row of Pascal s triangle S ( n) However they give a curious exaple for the partial su of the row eleents ultiplied by their ( n distance fro the center: ( ) r (r ) 1 ( ) r integer 1 Ollerton [11 considered the sae partial su of a row of Pascal s triangle S ( n) and developed forulae for the sus by use of generating functions E Kılıç (B) Matheatics Departent TOBB Econoics and Technology University Anara Turey E-ail: eilic@etuedutr

2 10 Arab J Math (018) 7: We recall a different result fro [6: for any nonnegative integer t ( )( n 1 ) ( ) n 1 n t t These inds of partial sus as well as certain weighted sus (both alternating and non alternating) are very rare in the current literature Calin [1 proved the following curious identity: j0 n ( ) n 3 n 3n 1 3n 3n j j0 j0 ( n n ) n Hirschhorn [5 established the following two identities on sus of powers of binoial partial sus: n ( ) n n ( ) n n 1 n and n n n 1 n n ( ) n j j n Recently He [4 gave-analogues of the results of Hirschhorn as well as a new version of another result For exaple fro [4 we recall that n j0 n ( j ( ) ; ) n (n1) ( 1; ) n j 1 where the -Pochhaer sybol is (x; ) n (1 x)(1 x)(1 x n 1 ) and the Gaussian -binoial coefficients are n (; ) n (; ) (; ) n Note that li 1 n ( ) n Meanwhile Guo et al [3 gave soe partial sus including -binoial coefficients for exaple n ( n )( n ) ( 1) n n n j j0 For later use we recall that one version of the Cauchy binoial theore is given by n ( 1 )[ n x n (1 x ) 1 and Rothe s forula is n )[ n 1 n ( 1) ( x (x; ) n (1 x ) Nowadays there is increasing interest in deriving -analogues of certain cobinatorial stateents This includes for exaple -analogues of certain identities involving haronic nubers we refer to [910

3 Arab J Math (018) 7: RecentlyKılıç and Prodinger [7 coputed half Gaussian -binoial sus with certain weight functions They also gave applications of their results as the generalized Fibonoial su forulas For exaple they showed that for n 1 such that n 1 r n 1 n ( 1) 1 ( (r1) ) n (1 ) r1 r n n and n n 1 n 3 ( 3) (1 ) 3 n n 1 (1 ) (1 n )(1 n 1 ) More recently Kılıç and Prodinger [8 coputed three types of sus involving products of the Gaussian -binoial coefficients They are of the following fors: for any real nuber a n n n n n n ( 1) n( ) (a ) ( 1) 1 n( ) a and n [ n n 1 ( 1) n(1 ) a b They also presented interesting applications of their results to generalized Fibonoial and Lucanoial sus In this paper inspired by the partial sus given in [ we present three sets of weighted partial sus of the Gaussian binoial -binoial coefficients Before each set we first give a lea and then present our ain results To prove the claied results we will use Rothe s forula -analysis and a-version of the celebrated algorith of Zeilberger For the sae of siplicity we will only prove one result fro each set The ain results In this section we will exaine each set in separate subsections We start with the first set of weighted partial sus 1 First ind of weighted partial sus We start with the following lea to use in proving our first set of results Lea 1 (i) For odd n (ii) For even n n n n 1 ( n) i (n ) ( 1) 0 n 1 ( n 1) i (n) (1 ) 0

4 104 Arab J Math (018) 7: Proof We only give a proof for the first case (i) The second case (ii) is siilarly done For odd n we write n1 [ n 1 i 1 n1 ( ) n ( 1) (1 [ n 1 )n n1 [ n 1 ( ) n ( 1) n i i 1 ( i 1 () n ( 1) n i i 1 ) ( i) n1 [ n 1 which by Rothe forula euals i 1 ( ( 1)n i n ; ) n1 i 1 (( 1)n i n ; ) n1 i 1 n (1 ( 1) n i n ) i 1 n ( ) n ( 1) n ( i) (1 ( 1) n i n ) Here we consider two subcases: first if n is even ( i 1 n (1 i n ) i 1 n n (1 i n ) i (i (1 )) i 1 n (n1 ) n (1 )[( 1) n 1 1 which clearly euals 0 for even n The other subcase n is odd can be done in a siilar way Theore For n 0 n1 n 4 n 1 [ n 4 1 Proof Consider the LHS of the clai n1 n 4 1 (1 n4 ) ( 1) ( )n 1 ( 3) n ) n ( i (1 )) 1 { (1 n1 )(1 n ) if n is even (1 n1 )(1 n ) if n is odd (1 n4 ) ( 1) ( )n 1 ( 3) n ( 1) n (1 1 ) (3)n n1 n 4 (1 n4 ) ( 1) (n) ( 1 ) ( (n) 3) By Lea 1 (ii) we have n4 n 4 (1 n4 ) ( 1) (n) ( 1 ) ( (n) 3) 0 Apart fro the constant factor we consider the LHS of the clai and write n1 n 4 (1 n4 ) ( 1) (n) ( 1 ) ( (n) 3) 1 n 4 (1 n4 ) ( 1) (n) ( 1 ) ( (n) 3)

5 Arab J Math (018) 7: n4 n 4 (1 n4 ) n ( 1) (n) ( 1 ) ( (n) 3) which after soe rearrangeents euals 1 n 4 (1 n4 ) ( 1) (n) ( 1 ) ( (n) 3) n 4 (1 ) ( 1) nn(1) ( )( 1 )1 1 ( 1)(n) 1 n 4 (1 n4 ) ( 1) (n) ( 1 ) ( (n) 3) n 4 ( 1) (1)(n) (1 ) ( 1 ( 1)( n4) 1 ) After further rearrangeents this euals ( 1 n4 ) ( 1 n 3 n 3 1 (1 n4 ) ( 1 ( 1) (n) 1 ( 1) 1 ( n 3) ( 1) (1)(n) (1 ) 1 ( 1)( n4) ) n 3 1 n 3 which by telescoping euals ( 1) (n) ( ) 1 ( n 3) ( 1) (n) ( )1 1 ( n 3) ) ( ( 1 n4) 1 n 3 ( 1) (n) ( 1 ) ( n 3) 1 ) n 3 ( 1) (n) ( 1 ) ( n 3) ( ( 1 n4) n 3 ( 1) 1 (n)(1) 1 (1)(n) 1 ) n 3 ( 1) 1 (n1) 1 (n3) ( 1 n4) ( 1) 1 (n1) 1 (n3) ( [n ) 3 n 3 ( 1) n (n1) 1

6 106 Arab J Math (018) 7: Thus we get ( 1) 1 (n1) 1 n 4 (1)(n) 1 { ( 1 n1 )( 1 n) if n is even ( 1 n1)( 1 n) if n is odd n1 n 4 (n1) [ n 4 1 ( 1 n4 ) ( 1) ( )n 1 ( 3) n { ( 1 n1 )( 1 n) if n is even ( 1 n1)( 1 n) if n is odd as claied We present the following results without proof Theore 3 For n 0 and 1 n n n 1 n Theore 4 For n 0 n n1 (1 n 1 ) n ( 1) n ( ) ( 1) n n 1 ( )n 1 ) n 1 (1 n ) 1 () n ( 1) (1 n 1 1 ( n 1) ( 1) n (1 ) n 1 n 3 1 ( n 3) ( 1) ( )n ( 1) n n 1 [ n 4 1 { (1 n1 )/(1 n ) if n is odd ( 1 n1 ) / ( 1 n) if n is even Second ind of weighted partial sus Now we continue with our second set of sus Before that we need the following lea which could be proven siilar to Lea 1 Lea 5 (i) If n is odd (ii) If n is even n n n ( 1) 1 ( n) 0 n 1 ( n1) ( 1) (1 n ) 0 We start with the following result

7 Arab J Math (018) 7: Theore 6 For n 0 4n1 ( 1) 1 ( 4n 3) n 1 (1 n1 )(1 n ) (1 1 ) Proof Consider the LHS of the clai by taing instead of wenote 4n1 ( 1) 1 (4n 3) 4n1 ( 1) 1 (4n3) ( 1) 1 ( 4n 3) and by Lea 5 (i) Thus we obtain 4n3 4n1 1 4n3 4n ( 1) 1 ( 4n 3) 0 ( 1) 1 ( 4n 3) ( 1) 1 ( 4n 3) ( 1) 1 ( 4n 3) which by soe rearrangeents euals 1 By telescoping this euals which euals 1 [ 4n 1 1 ( 1) 1 ( 4n 3) ( 1) ( 1 (1)(4n) 1 ( 1) 1 ( 1)(4n) ( 1) 1 ( 4n 3) () 4n3 ( 1) 1 ( 4n 3) ) n1 ( 1) 1 (1)(4n) ( (1 4n3 ) n1 (1 1 )) () 1 () 4n3

8 108 Arab J Math (018) 7: () 4n3 ( 1) 1 (1)(4n) (1 n1 )(1 n ) () () 4n3 1 1 ( 1) 1 (1)(4n) (1 n1 )(1 n ) (1 1 ) Thus we derive 4n1 ( 1) 1 ( 4n 3) n 1 (1 n1 )(1 n ) (1 1 ) as claied Using siilar techniues as in the above proof we obtain the following identities for all n 0 4n 4 ( 1) 1 ( 4n 3) (1 4n4 ) n1 (1 4n4 ) (1 n1 )(1 n ) (1 1 ) ( 1) 1 ( 4n 3) 4n1 4n 4n1 4n 4n 4n 4n 1 4n 1 4n 4 4n n (1 4n ) ( 1) (1 ) n n (1 n1 )(1 n ) 1 1 4n 1 ) n ( 1) n 1 (1 (1) 4n 1 1 (4n 3) ( 1) (1 4n4 ) (1 4n4 ) 1 ( 4n1) ( 1) (1 4n ) 4n Third ind of weighted partial sus We start with the following lea for later use Lea 7 For > 0 and integer r r ( 1) 1 ( 4n 1) (1 n1 ) ( 1) r 1 4n 1 (r)(4n r1) (1 1n ) r

9 Arab J Math (018) 7: Proof Denote the left-hand side of this identity by S[ r Using the celebrated algorith of Zeilberger in Matheatica we obtain S [ r ( 1) r 1 (r)(4n r1) 1 () 4n (1 1n ) () 4n1 r () r which is easily seen to be the sae as the right-hand side of the desired identity Theore 8 For n 0 and > 0 4n3 n1 (1 n )(1 ) (1 n1 ) 1 ( 4n 3) ( 1) (1 ) 1 Proof Note that and also Thus we write 4n3 1 (4n3) ( 1) 4n3 4n 4n3 1 ( 4n 3) ( 1) (1 ) 1 ( 4n 3) ( 1) (1 ) 1 ( 4n 3) ( 1) (1 ) 0 1 ( 4n 3) ( 1) (1 ) 1 1 ( 4n 3) ( 1) (1 ) ( 1) 1 (1)( 4n ) (1 4n ) 1 ( 1)(4n4) ( 1) 1 (1 1 ) 1 ( 1) [ 1 ( 4n 3) (1 ) 1 (1)( 4n ) (1 4n ) 1 ( 1)(4n4) ( 1) 1 (1 1 ) 1 (n1) (1 n1 ) ( 1) 1 ( 4n 1) (1 n1 )

10 110 Arab J Math (018) 7: which by Lea 7 with the case r euals Thus 4n3 as claied ( 1) 1 1 ( 1)(4n4) ( [4n ) 4n 1 (1 1 ) (1 n1 )(1 1n ) 1 ( 1) 1 ( 1)(4n4) 1 () 4n1 (1 n )(1 n1 )(1 ) () 1 () 4n3 1 ( 4n 3) ( 1) (1 ) n1 (1 n )(1 ) (1 n1 ) 1 We present the following results without proof The proofs use Lea 7 andaresiilartothatof Theore 8 Theore 9 For n 0 and 1 4n Theore 10 For n 0 4n1 4n 1 ( 4n 3) ( 1) (1 ) (1 n1 )(1 1n ) 4n 1 1 ( 1) 1 4n 1 ( 4n 3) (1 ) n 1 (1 n1 )(1 n1 ) 1 ( 4n 3) ( 1) (1 ) 4n 1 (1 4n )(1 n ) (1 n1 ) 1 3 Applications In this section we will give soe applications of our results to the generalized Fibonoial coefficients Define the non-degenerate second-order linear seuences {U n } and {V n } by for n > 1 U n pu n 1 U n U 0 0 U 1 1 V n pv n 1 V n V 0 V 1 p The Binet forulas of {U n } and {V n } are U n αn β n α β and V n α n β n where α β (p ± p 4)/ For n 1 define the generalized Fibonoial coefficient by { } n U 1 U U n : (U 1 U U ) (U 1 U U n ) U

11 Arab J Math (018) 7: with { n 0 }U { n} n U 1 When p 1 we obtain the usual Fibonoial coefficients denoted by { n} F The lin between the generalized Fibonoial and Gaussian -binoial coefficients is { } n n α n U with α By taing β/α the Binet forulae are reduced to the following fors: n 1 1 n U n α 1 and V n α n (1 n ) where i 1 α Fro each set we now give one application Fro Theore 3 E(1) and Theore 9 wehavethe following identities by taing β/α respectively: for n 0and 1 4n 4n n { } n { } { } U U U { } n 1 ( 1) U U n 1 U { } ( 1) 1 ( 1) U ( 1) 1 ( 1) U U n1 V 1n { 4n 1 1 As a showcase we will prove the second identity just above First we convert the second identity into -for Thus it taes the for or or 4n 4n 4n α ()(4n3 ()) ( 1) 1 ( 1) α ()( 4n3) ( 1) 1 ( 1) which was already given as the identity (1) ( 1) 1 (4n 3) } U α (4n3 ) α (4n3) Open Access This article is distributed under the ters of the Creative Coons Attribution 40 International License ( creativecoonsorg/licenses/by/40/) which perits unrestricted use distribution and reproduction in any ediu provided you give appropriate credit to the original author(s) and the source provide a lin to the Creative Coons license and indicate if changes were ade

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