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A Proof of the Curious Binomial Coefficient Identity Which Is Connected with the Fibonacci Numbers

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A Proof of the Curious Binomial Coefficient Identity Which Is Connected with the Fibonacci Numbers Open Access Journal of Mathematical and Theoretical Physics Review Article Open Access A proof of the curious binomial coefficient identity which is connected with the fibonacci numbers Abstract Volume 1 Issue 1 - 2017 We give an elementary proof of the curious binomial coefficient identity, which is Jovan Mikic connected with the Fibonacci numbers, by using system of auxiliary sums and the induction principle. We discover some interesting relations between main sum and Jovan Cvijić High School Center, Balkans auxiliary sums, where appear the Fibonacci numbers. With help of these relations, Correspondence: Jovan Mikic, Jovan Cvijić High School we found a second order linear recurrence with main sums only. We easily solve this Center (JU SŠC), Modric a 74480, Bosnia and Herzegovina, recurrence by using the Binet formula for the Fibonacci numbers and then prove the Balkans, Email [email protected] desired identity. Received: October 31, 2017 | Published: December 04, 2017 Keywords: binomial coefficient, fibonacci number, recurrence equation, auxiliary sum, combinatorial identity Introduction Theorem 1 We consider the following sum with binomial coefficients: Let n be a natural number; and let k be a natural number greater than 1. Then the following relations hold between main and auxiliary nn nni− ni−− ni sum: Sn( )= ∑∑ ∑ 1 2 k (1) k n ii=0 =0 i =0i2 i3 i1 Sn( ) = F⋅ Sn ( −+ 1) FPn ⋅ ( −+ 1) F (4) 12 k kkkkkk+−11 Where n is a non negative integer and k is a natural number greater Pnk( ) = F kk⋅ S ( n −+ 1) F k−1 ⋅ Pn k ( − 1) (5) than 1. We call this sum Snk () a main sum. Let us Fn denote the n From our main theorem, we derive a second order linear recurrence -th Fibonacci number. Namely, F =0, F =1, and FF= + F 0 1 nn−−12 n between main sums only. We have: ; if n ≥ 2 . Our main goal is to prove the following identity: Corollary 1 Fkn(+ 1) Let n be a non negative integer; and let k be a natural number Sn( )= (2) k greater than 1. Then the following relation holds: Fk 2 Snk(+ 2) = ( F k+1 + F k − 1 ) ⋅ Sn k ( ++ 1) ( F kkk − F−+11 ⋅ F)() ⋅ Sn k The Identity (2) can be found1 as the Identity 142; and it arises as (6) a generalization of the Identity 5 (when k =2) from the same book. The same variant of the Identity (2), also, can be found in paper2 as the With initial conditions: Identity 3 (with small error). The Identity (2) is interesting, mainly, S (0) = 1 (7) because of its connection with the Fibonacci numbers. k There are no many proofs of the Identity (2). It is known that S(1) = FF+ (8) k kk+−11 exists1 a purely combinatorial proof of the Identity (2). We give an elementary proof of the Identity (2) by using system of auxiliary sums Recall that the Cassini identity states that and the induction principle. 21k − FFFkkk−⋅−+11= ( − 1) . By the Cassini identity, the Relation (6) Method of auxiliary sums is a new method in proving binomial simplifies to coefficient identities. This method is introduced by the mathematician k −1 Jovan Mikic in papers.3,4 In this paper, we show how the same method Sn(+ 2) = ( F + F ) ⋅ Sn ( + 1) +− ( 1) ⋅Sn ( ) (9) k kkk+−11 k works on harder example; such is the Identity (2). In that sense, this proof of the Identity (2) is interesting, particularly, because of a choice More definitions of auxiliary sums. In order to prove main theorem (1), beside the auxiliary sum Pnk () First, we introduce the auxiliary sum Pnk (), as follows: , we need to define a whole system of auxiliary sums. nn nni− ni−− ni Definition 1 ∑∑ ∑ 1 2 k Pnk ( )= (3) ii=0 =0 i =0 i i i − 1 12 k 2 3 1 Let n be a natural number; and k be a natural number greater than1. Let l be a natural number such that lk≤+2 . We define auxiliary We establish our main theorem: sums Snkl, (), as follows: Submit Manuscript | http://medcraveonline.com Open Acc J Math Theor Phy. 2017;1(1):1‒7. 1 © 2017 Mikic. This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and build upon your work non-commercially. Copyright: A proof of the curious binomial coefficient identity which is connected with the fibonacci numbers ©2017 Mikic 2 nn−1 n − ni− 1 nik ⋅ −+ ⋅ − (19) ∑∑ ∑ Sk,1 ( n ) = F k+1 S k ( n 1) F kk Pn ( 1) Snk ,1 ( )= (10) i ii12=0 =0 ik =02 i1 Sk,2 ( n ) = F kk⋅ S ( n −+ 1) F k−1 ⋅ Pn k ( − 1) (20) n−−11 nn nni−−1 n−−1 i ni − ni− ∑ ∑∑ ∑ 1 ll−1 k Snkl, ( )= i=0 ii =0 =0 i =0 i ii i 1<lk≤ (11) 1 lk l+ 1 21 ll + 1 Lemma 2 S( n ) = Sn (− 1) (12) kk,1+ k Let n be a non negative integer; and k be a natural integer greater than 1. The following relation holds Skk,2+ ( n ) = Pnk (− 1) (13) Pnkk()= S,2 () n (21) Then we define the following sum: Lemma 3 Definition 2 Let n, k, and j be from the Defination (3). Then the following Let n be a non negative integer; and k be a natural integer greater equation holds: than 1. We define the sum Snki,= n(), as follows: 1 j nj− ∆⋅(nF )= F (22) kj, k−− 21 k nn0 ni− ni− ∑∑ 2 k Snki,= n( )= (14) 1 ii=0 =0 i i n Lemma 4 2 k 2 3 Let n and k be from the Defination (2). Then the following equation In other words, if we set in1 = in the main sum Snk (), we get the holds: sum Sn(). In order to calculate the sum Sn(), we need to n ki,= n ki,= n Ski,= n( nF )= k−1 (23) 1 1 1 define one more sum: Definition 3 A proof of the lemma (1) Our proof of the Equation (17) relies on the well-known Pascal’s Let n be a non negative integer; and k be a natural integer greater formula for binomial coefficients: than 3. Let j be a non negative integer such that jn≤ . We define the sum ∆ ()n , as follows: nn −−11 n kj, = + (24) k kk − 1 nnnj− ni− ni− ∑∑ 3 k −2 ∆kj, (n )= (15) ii=0 =0 i i i Where n is a natural number and k may be an arbitrary integer. 31k − 3 4 k −1 Due to the Definition (1), our proof consists of four parts. All There is a connection between these two sums from the Definition proofs of these four parts are very similar. We prove three cases and (2) and Definition (3). Namely, when ≥ , the following equation k 4 give a sketch of a proof for the fourth case. holds Snki, = n()=∆ k,0 () n (16) Proof 1 The first case: l =1. Since in≤−1 , we can apply the Pascal The Equation (16) is true because of the fact that in= implies 1 1 formula on the binomial coefficient that both i2 and ik must be equal to zero. ni− 1 Main lemmas . i2 Before we prove main theorem (1), we need to prove several We have gradually: lemmas. Here, we give a list of all lemmas which are important for us. nn−1 nni− ni−− ni Lemma 1 ∑∑ ∑ 1 2 k Snk ,1 ( )= ii=0 =0 i =0 i i i Let integers n, k, and l be from the Definition (1); with condition 12 k 2 3 1 lk≤ . Then the following equation holds nn−1 nni−−11 ni −− ni−− ni =(∑∑ ∑11+ )2 k Sn()= S () nS+ () n (17) ii=0 =0 i =0 ii− 1 i i kl, kl ,1++kl ,2 12 k 223 1 Corollary 2 nn−1 nni−−1 ni−− ni = ∑∑ ∑1 2 k + (25) ii=0 =0 i =0 i i i Let integers n, k, and l be from the Defination (1); with condition 12 k 2 3 1 lk≤ . Let m be a non negative integer such that mk≤ +−1 l. Then nn−1 nni−−1 ni−− ni following equations hold ∑∑ ∑1 2 k (26) i − 1 i ii12=0 =0 ik =02 3 i1 S()= n F+⋅ S + () n +⋅ FS++() n (18) kl, m 1, kl m m kl,1 m Citation: Mikic J. A proof of the curious binomial coefficient identity which is connected with the fibonacci numbers. Open Acc J Math Theor Phy. 2017;1(1):1‒7. DOI: 10.15406/oajmtp.2017.01.00001 Copyright: A proof of the curious binomial coefficient identity which is connected with the fibonacci numbers ©2017 Mikic 3 Note that the first binomial coefficient in the Equation (25) perishes if in= . Therefore, we have 2 The third case: lk=1− . Since ink −1 ≤−1 , we can apply the ni− nn−1 nni−−1 ni−− ni k −1 ∑∑ ∑1 2 k Pascal formula on the binomial coefficient . We have i i ik ii12=0 =0 ik =0i2 3 1 gradually: nn−−11 n ni−−1 ni−− ni 1 2 k n−−11 nnni−−1 ni− ni− = ∑∑ ∑ ∑ ∑∑ 1 k −1 k i i i Snkk,1− ( )= ii12=0 =0 ik =02 3 1 ii i11=0 iikk− =0 =02 k i1 =Sn ( ). (27) n−−11 nnni−−1 ni−−1 ni− k ,2 = ∑ ∑∑1 k −1 k + ii The Equation (26) becomes gradually: i11=0 iikk− =0 =02 k i1 n−−11 nnni−−1 ni−−1 ni− nn−1 n −− −− 1 k −1 k ni1 1 ni2 nik ∑ ∑∑.
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