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 nni− ni−−  ni  sum: Sn( )= ∑∑ ∑ 1 2 k (1) k    n ii=0 =0 i =0i2 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 nni− 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) ii=0 =0 i =0 i i 12 k 2 1 Sk,2 ( n ) = F kk⋅ S ( n −+ 1) F k−1 ⋅ Pn k ( − 1) (20) n−−11 nn nni−−1 n−−1 i  ni −  ni− ∑ ∑∑ ∑ 1 ll−1 k Snkl, ( )=      i=0 ii =0 =0 i =0 i ii i 1

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 S( n ) = Pn (− 1) (13) kk,2+ k 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 nn0 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  nnnj− 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 nni− 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 nni−−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 223 1  Corollary 2 nn−1 nni−−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 nni−−1 ni−−  ni  following equations hold ∑∑ ∑ 1 2 k (26) −    ii=0 =0 i =0i2 1 i3 i1  S()= n F⋅ S () n +⋅ FS() n (18) 12 k 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 nni−−1 ni−−  ni  k −1 ∑∑ ∑1 2 k  Pascal formula on the binomial coefficient . We have i i ik ii12=0 =0 ik =0i2 3 1  gradually:

nn−−11 n ni−−1 ni−− ni 1 2 k  n−−11 nnni−−1 ni− ni− = ∑∑ ∑   ∑ ∑∑ 1 k −1 k i i i Snkk,1− ( )=  ii12=0 =0 ik =02 3 1  ii i11=0 iikk− =0 =02 k i1 =Sn ( ). (27) n−−11 nnni−−1 ni−−1 ni− k ,2 = ∑ ∑∑1 k −1 k + ii The Equation (26) becomes gradually: i11=0 iikk− =0 =02 k i1 n−−11 nnni−−1 ni−−1 ni− nn−1 n −− −− 1 k −1 k ni1 1 ni2  nik  ∑ ∑∑. ∑∑ ∑   i=0 ii =0 =0 ii− 1 i i − 1 i 11kk− 2 k 1 ii12=0 =0 ik =02 3 i1  n−1 nn −− 11ni−−1 ni−−1 ni− nn−1 n −− −− 1 k −1 k ni1 1 ni2  nik  = ∑ ∑∑+ = ∑∑ ∑   i=0 ii =0 =0 iii i − 1 i 11kk− 2 k 1 ii12=0 =1 ik =02 3 i1  n−−11 nn ni−−1 ni−−1 k −1 ni− nn−−11 nni−−1 n−−1 t  ni −  ∑ ∑∑ 1 k ∑∑ ∑ 1 2 k (28)  =   (ti22 =− 1) i=0 ii =0 =1 ii− 1 i it=0 =0 i =0 t i i 11kk− 2 k 1 12 k 2 3 1 

=Snkk, ()+ If k =2, then the above sum in the Equation (28) is Snk (− 1) according to the Equation (1). Due to the Equation (12) from the n−1 nn −− 11ni−−1 ni−−1 nt−−1 ∑ ∑∑1 k −1 k . Defination (1), we have that Sn(− 1)= S ( n )= S ( n ). it k kk,+ 1 k,3 i11=0 itkk− =0 =02 k i1 If k >2, then the above sum in the Equation (28) is, as follows: In the last equation above, we used substitution tikk=1− . Therefore, from the last equation above, we obtain that nn−−11 n n ni−−1 n−−1 t ni − ni − 1 23   k  S()= n S () n+− Sn ( 1) ∑∑∑ ∑     kk,1− kk , k t i it123=0 =0 i =0 ik =02 3 ii41   

=SnS ()+ + (); n(30) (by the Equation (12)). The Equation =Sn ()According to the Defination (1) and the Equation kk, kk ,1 k ,3 (30) proves the third case. (11). In both cases, the sum in the Equation (28) is equal to Snk ,3 () The fourth case:13. We . Now, from the Equation (25), (26), (27), and (28), it follows that give a short sketch of the proof. The proof of this case is very similar

Snkkk,1 ()= SnSn,2 ()+ ,3 (). This proves the first case. to the proof of the first case. We use the Pascal formula on the binomial The second case: lk= . Since in≤−1 , we can apply the Pascal ni− nini− −−11 ni −− k coefficient l and we get l= ll+ i i ii− 1 ni− l+1 l+1 ll ++ 11 formula on the binomial coefficient k . We have gradually:  . Then the sum Sn( ) splits on two sums. It is easy to see that the i1 kl, − −− n1 nn 11ni−−1 ni−−1 ni− first sum is Snkl, (). Sn( )= ∑ ∑∑ 1 k −1 k kk, ( i )  2 i i11=0 iikk− =0 =0 k i1 For the second sum, we need to introduce the substitution ti=1− . Then the second sum becomes Sn(). Therefore, n−1 nn −− 11ni−−1 ni−−1 ni−−1 ll++11 kl,2+ = ∑ ∑∑1 k −1 k + it follows that Sn()= S++ () nS+ () n. This proves the fourth i=0 ii− =0 =0ii2 k i1 kl, kl ,1kl ,2 11kk case. n−1 nn −− 11 −− ni−−1 1 ni1 k −1 ni−−1 k ∑ ∑∑. ii− A proof of the corollary (2) i11=0 iikk− =0 =02 k i1 1 We need to prove Equations. (18), (19), and (20). All these

=Skk ( n−+ 1) Pn ( − 1) (29) According to the Equation (1) and equations are direct consequences of the Equation (17). the Equation (3). A proof of the equation (18) According to the Equation (12) and the Equation (13) from the Defination (1), the Equation (29) becomes Proof: We assume that integers n, k, and l are fixed. We give a proof by using the induction principle on m. If m =0, then the Equation Snkk,()= S kk ,1++ () nS+ kk ,2 () n. This proves the second case. (18) is satisfied, because F0 =0. Thus, we confirm the base of the

induction. Let us suppose that the Equation (18) holds for some m such that 0≤mk <1 +− l. In other words, our induction hypothesis nn n ni− 1 ni−−2  nik  is S()= n F⋅ S () n +⋅ FS() n. Then lmk+≤. By = ∑∑ ∑   kl, m+ 1, kl + m m kl,1++ m i i − ii12=1 =0 ik =02 3 i1 1  the Lemma (1) and the Equation (17), we know that Now, we introduce the substitution ti=1− . Then the last S()= n S () nS+ () n (31) 11 kl,+ m kl ,1++ m kl ,2++ m equation above becomes From our induction hypothesis and the Equation (31), we have that nn−1 nnt−−1 ni−−  ni  ∑∑ ∑ 1 2 k ⋅ +⋅ Pnk ( )=    Skl,()= n F m+ 1, S kl + m () n FS m kl,1++ m () n i i ti12=0 =0 ik =02 3 t1  nn−−11 n nt−−1 ni−− ni =FSnSnFSn⋅ ( () + ()) +⋅ () 1 2 k  m+1 kl ,1 ++ m kl ,2++ m m kl,1++ m = ∑∑ ∑   i i ti12=0 =0 ik =02 3 t1 

=(Fm+1+⋅ FS m ) kl ,1++ m () nF +m + 1,2 ⋅ S kl ++ m () n =Sn () (34) k ,2 According to the Equation (11) from the Defination (1). The =Fm+2⋅ S kl ,1 ++ m () nF +⋅m + 1,2 S kl ++ m (). n Equation (34) completes the proof of the Lemma (2). From the last equation above, it follows that Equation (18) is satisfied for m + 1 ; where 0

=F⋅ S ( n −+ 1) F ⋅ Pn ( − 1) (32) nj− k+1 k kk = 2 (bythebinomialtheorem) According to the Equations (12) and (13) from the Defination (1). j nj− The Equation (32) is our desired the Equation (19). This proves the = FF23⋅ Equation (19) j nj− =.FF(4−− 2)⋅ (4 1) A proof of the equation (20) The last equation above proves the base of induction. Let us Proof: Again, this proof is straightforward. Just set l =2 and suppose that the Equation (22) holds for some k ≥ 4 . This is our mk=1− in the Equation (2). Since, mk≤ +−1 l, this is allowed. induction hypothesis. Let us consider ∆kj+1, . From the Equation (18), we get nn nnj− ni− ni− ∑∑ ∑ 3 k −1 S()= n FS⋅ () n +⋅ F S() n ∆kj+1, =  k,2 k kk,+ 1 k −+1 kk , 2 i i ii34=0 =0 ik =03 i4 k ⋅ −+ ⋅ − =Fkk S ( n 1) F k−1 Pn k ( 1) (33) n− nn− − nj ni3 nik −1 = ∑ ∑∑ According to the Equations (12) and (13) from the Defination (1). i i i34=03 ii =0k =0i4 k The Equation (33) is our desired the Equation (20). This proves the n nj− Equation (20) and completes the proof of the Corollary (2). ∑ = ⋅∆ki, (n ). (35) 3 i =0i3 A proof of the lemma (2) 3 By the induction hypothesis, it follows that Proof i ni− ∆⋅33 This proof immediately follows from the Equation (3) and the ki,(nF )= k−− 21 F k (36) 3 Defination (1). We need to introduce the substitution ti=1− . We 11 By the Equation (36), the Equation (35) becomes have:

n nj− i ni− nn nni− ni−−  ni  ∆= ∑ ⋅⋅FF33 ∑∑ ∑ 1 2 k kj+1, k−−2 k 1 Pnk ( )=    i =0 i i i − 3 3 ii12=0 =0 ik =02 3 i1 1  nj− i ni− nj− 33 = ∑ ⋅⋅FF−− i kk21 i3 =03

A proof of the corollary (1)

nj− i n−− ji Proof j nj− 33 = F−⋅∑  ⋅⋅FF−− k1 i kk21 i3 =03 The Corollary (1) directly follows from the Theorem (1) and from its Equations (4) and (5). Let us suppose that n is a natural j nj− =FF⋅+ ( F ) (bythebinomialtheorem) kk−−121 k − number. First, we prove the Equation (6). We express sums Pnk () and Pn(− 1) by main sums with help of the Equation (4). Then we j nj− k =.FF⋅ (37) kk−1 use the Equation (5), in order to get relation between main sums only. The Equation (37) proves the step of the induction principle. This First, from the Equation (4), we obtain that: completes the proof of the Lemma (3). n Snk( )− F kk+−11 ⋅ Sn ( −− 1) F k Pnk (− 1) = . (42) A proof of the lemma (4) F k It is easily verified, by direct calculation, that If we use the substitution nt− 1= , the Equation (42) becomes Sn( ) = 1; 2,in = t+1 1 St(+− 1) F ⋅ St ( ) − F Pt()= k kk+−11 k; (43) Sn( ) = 1. k 3,in = F 1 k So, we can conclude, from the above equations, that Where t is a non negative integer. Therefore, when n is a natural n S2,in = ( nF )=1 ; (38) number, the Equation (43) implies that 1 n+1 n Sn(+− 1) F ⋅ Sn ( ) − F S( nF )= . (39) k kk+−11 k 3,in = 2 Pnk ( )= . (44) 1 F k Therefore, the Equation (22) is true, when k =2 and k =3. Now, we use the Equation (5). By Equations (42) and (44), the Now, let us suppose that k ≥ 4 . According to the Equations (16) Equation (5) becomes gradually: and (22), we have that nn+1 Snk(+−⋅ 1) F kk+−11 Sn ( ) − F k Snk( )−⋅ F kk+1 Sn ( −− 1) F k − 1 =FSnkk⋅ ( −+ 1) F k−1 ⋅ Snki, = n( ) =∆ k,0 ( n )(bytheEquation (16)) 1 FFkk

00n− nn+12 Sn(+− 1) FSn+− ( ) − F = FSn (−+ 1) F− ( Sn ( ) − FSn + ( −− 1) F − ) =FFkk−−21⋅ (bytheEquation (16)) k kk11 k kk k1 k kk 1 k 1 2 n Sn(+− 1) FSn ( ) = FSn (−+ 1) F ( Sn ( ) − FSn ( − 1)) =.F (40) k kk+1 kk k−+11 k kk k −1 2 The Equations (38), (39), and (40) prove the Equation (23). This Sn(+ 1) = ( F + F ) ⋅ Sn ( ) +−⋅ ( F F F) ⋅ Sn ( − 1). completes the proof of the Lemma (4). k k+1 k − 1 k kkk−+11 k If we use the substitution nt− 1= , the last equation above A proof of the theorem (1) becomes 2 St(+ 2) = ( F + F ) ⋅ St ( ++ 1) ( F − F ⋅ F) ⋅ St ( ); Proof k k+1 k − 1 k kkk−+11 k (45) Let n be a natural number. First, we prove the Equation (4). Due to the Equation (10) from the Defination (1) and the Defination (3), Where t is a non negative integer. The Equation (45) is same as the we know that: Equation (6). Therefore, the Equation (6) is proved. Now, we prove Equations (7) and (8). The Equation (7) directly follows from the Snk()= S k,1 () n+ S ki, = n (). n (41) 1 Equation (1). Obviously, from the Equation (3), it follows that By Equations (19) and (23), the Equation (41) becomes Pk (0) = 0 (46) n Sn( ) = ( F⋅ Sn ( −+ 1) FPn ⋅ ( − 1)) + F . The last equation k k+−11 k kk k Therefore, the Equation (8) follows from Equations. (4), (7), and above proves the Equation (4). Now, we prove the Equation (5). The (46). We set n =1 in the Equation (4) proof of the Equation (5) immediately follows from the Equations ⋅++1 (20) and (21). Sk(1) = FS kk+−11 (0) P k (0) F k Pn( ) = S ( n )(bytheEquation (21)) kk =FFkk+−11⋅+ 10 +

=F⋅ S ( n −+ 1) F− ⋅ Pn ( − 1).(bytheEquation (22)) kk k1 k =.FFkk+−11+ The last equation above proves the Equation (5). This completes The last equation above proves the Equation (8). This proves the the proof of the Theorem (1). Corollary (1).

A proof of the identity (2) By using Equations (48) and (49). The general solution of the Finally, we prove our main Identity (2). We use the Corollary (1) Equation (6) is, as follows: and the Binet formula for the Fibonacci numbers. Recall that the Binet nn formula states that Snk ( )= C11⋅+⋅λλ C 2 2

1 15+−nn 15 15+−kn 15kn F =⋅− [( ) ( ) ] (47) =CC12⋅ (( ) ) +⋅ (( ) ) (Eqns(54)and(55) n 22 5 22

Further, we use facts that 15+−kn 15kn =CC12⋅ ( ) +⋅ ( ). (56) 22 15+ k FFFkkk+−11+ +⋅5 ( )= ; (48) By Equations (7) and (8), from the Equation (56), we evaluate 22 constants C1 and C2 . We get a system of equations, as follows: 15− k FFF+−+ −⋅5 ( )=kkk11 . (49) CC+ = 1; (57) 12 22 15+− 15 Namely, Equations (48) and (49) are consequences of following kk C1⋅( ) +⋅ C2 ( )= FFkk+−11+ . (58) relations: 22 After short calculation, from Equations (57) and (58), we obtain 15++k 15 ()=();FFkk⋅+−1 that 22

1 15+ k 15−−k 15 C1 =⋅ ( ); (59) ()=();FFkk⋅+−1 5 ⋅ F 2 22 k −−1 15 Which can be easily proved by the induction principle. Proof: The k C2 =⋅ ( ). (60) characteristic equation of the recurrence (6) is 5 ⋅ F 2 k 22 λλ−+⋅+⋅−()(F F FF F) = 0. (50) k+1 k − 1 kk −+ 11 k By using Equations (59) and (60), the Equation (56) becomes The discriminant of the Equation (50) is, as follows gradually 22 DF=(k+1+ F k − 1 ) −⋅ 4 ( FF kk −+ 11 ⋅ − F k) 1 15++k 15kn 15 −−k 15kn Snk ()=⋅ [( )( ⋅ ) −⋅ ( )( )] 22 5 ⋅ Fk 22 22 =(FFkk+−11− ) +⋅ 4 F k 22 =FF+⋅ 4 (bythedefinitionoftheFibonaccinumbers) 1 15+−kn++ k 15kn k kk =⋅− [( ) ( ) ] 2 5 ⋅ Fk 22 =5⋅ Fk .

1 1 15+−kn(++ 1) 15kn( 1) From the last equation above, it follows that =⋅⋅ ( [( ) − ( ) ]) 2 DF=5⋅ . (51) Fk 5 22 k 1 By using the Equation (51), it follows that roots of the Equation =⋅ F (theEquation (47)) (50) are kn(+ 1) Fk

FFF+ +⋅5 Fkn(+ 1) kkk+−11 =. λ1 =; (52) 2 Fk The last equation above proves the Equation (2). This completes FFFkkk+−11+ −⋅5 λ2 =. (53) the proof of the Equation (2). 2 Acknowledgements Equations (52) and (53) can be written, as follows: I want to thank to my professor Duško Jojić. I am also grateful to 15+ k Stamatina Theofilou. λ1 =( ) ; (54) 2 Conflict of interest 15− k The author declares no conflict of interest. λ2 =( ) ; (55) 2

References 2. http://math.sun.ac.za/%20hproding/pdffiles/Double 1. Benjamin AT, Quinn JJ. Proof that Really Count. The Art of Combinato- 3. Miki J. A proof of Dixon’s identity. J Integer Seq. 2016;19:1–5. rial Proof. The Dolciani Mathematical Expositions, Mathematical Asso- 4. Miki J. A proof of a famous identity concerning the convolution of the ciation of America; 2003. 27:209. central binomial coefficients.J Integer Seq. 2016;19:1–10.

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