A Relation Between Binomial Coefficients and Fibonacci Numbers to the Higher Power Yuhong Che Weinan Normal University, Weinan Shaanxi, 714000, China
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Advances in Engineering Research (AER), volume 107 2nd International Conference on Materials Engineering and Information Technology Applications (MEITA 2016) A Relation Between Binomial Coefficients and Fibonacci Numbers to the Higher Power Yuhong Che Weinan Normal University, Weinan Shaanxi, 714000, China Key words: Binomial coefficient, Fibonacci numbers, Convolution. Abstract. In this paper, we calculatate high power of Fibonacci numbers by elementary mathematical methods and prove an interesting identity between the binomial coefficients and the high power of Fibonacci numbers. Introduction n n n ni− i ()ab+=∑ ab Let us begin by recalling the binomial theorem i=0 i ,for arbitrary integers n and i n with ni≥≥0. The coefficient i are known as binomial coefficient and is given by n n! = i in!(− i )! . (1) n It is clear from the above formular (1) that i must be a positive integer.The binomial nin (1+=xx ) theorem,for arbitrary complex number x is given by i . F Let l (for abitrary non-negative l ) be the l− th Fibonacci number given by αβll− F = , l αβ− L and l be the l− th Lucas number defined by L =αβnn + l . For a long time,properties of binomial coefficients and Fibonacci numbers had been very interesting topics in ‘combinatorics’ and in ‘number’,and the identities relating binomial coefficients and Fibonacci numbers attracted many experts.We refer the reader to [1],[2],[3],[4],[5],[6] and [7] for more exhaustive details. For abitrary fixed nonnegative integer k and abitrary fixed positive integer n ,we denote the n n m n i {}Fki+ f(, kmn ,) convolution of sequence i=0 and i=0 by and define that as follows: n nnmm n m n m f(,,) kmn= Fk + Fk+1 ++... Fkn ++ =∑ Fki 01 nii=0 (2) The question of calculating f(, kmn ,) is being considered in [8],[9],[11],[12] and in [13].For example when m = 1, 2, 3, it had been proved that: fk( ,1, n ) = F kn+2 Copyright © 2017, the Authors. Published by Atlantis Press. 281 This is an open access article under the CC BY-NC license (http://creativecommons.org/licenses/by-nc/4.0/). Advances in Engineering Research (AER), volume 107 n 2 5Ln+ ,2| = 2kn fk( , 2, n ) n−1 2 5Fn2kn+ ,2 | 1 (2nF−− ( 1)kn+ 3 F ) kn≥ 5 32k+− n kn fk( , 3, n ) = 1 (2n F+< 3 F ) kn. 32k+− n nk 5 FFF,, , F (k+− 2) n th (2k+− n ) th ,(3 k + 2 n ) − th ()k−− n th Here kn+2 2 kn ++− 32 kn kn denotes the , and Fibonacci L (2k+− n ) th numbers respectively.Also, 2kn+ denotes the Lucas numbers.Inspired by the above work,a natural question would be whether there exist a similar conclusion to the higher powers of the Fibonacci numbers as well.We try to explore this here in this work.To this end,let fk( , 4, n ) be the in= n 4 n i {}Fki+ convolution of sequence i=0 and i=0 ,i.e. in= nn44 n 4 4 fk( ,4, n ) = Fk + Fk+1 ++... Fkn ++ =∑ F ki. 01 n i=0 (3) In this paper,we consider this question and prove an identity between the binomial coefficients and the quartic of Fibonacci numbers. In other words,we prove the following result. Theorem For abitrary fixed nonnegative integer k and positive integer n ,let fk( , 4, n ) be the n n 4 n i {}Fki+ convolution of sequence i=0 and i=0 ,then n+1 1 n ()ki+ fk( ,4, n )= [3 L42k++ n −− 4( 1) 5F2kn+ 6] 25 . (4) Proof of the theorem Proof.It is known that, αβll− Fl = , αβ− (5) where 15+− 15 αβ=,, = 22 (6) Now using (5)、(6) we get 41 4l 3 ll 22 l l l 3 l 4 l Fl =(4α −+ α β 6 α β −+ 4 αβ β ) 25 1 =(α4l −− 4( 1) l α 2 l +− 6( 1) 2 l − 4αβ ll 34 + β l ) 25 (7) According (2) and (7) we see that nn nn4 1 fk( , 4, n ) =∑∑ Fki+ = X ii=00ii25 = (8) 282 Advances in Engineering Research (AER), volume 107 where 4()ki+++++++ ()2()ki ki 2()ki ()2()4()ki ki ki X ==α −−4( 1) α +−6( 1) −− 4( 1) ββ + We know from (2) that n n 44in ∑αα=(1 + ) i=0 i (9) and n n 44in ∑ββ=(1 + ) . i=0 i (10) Now if we use the relations (6) and (7) we find that 42 42 2 1−=−=α β ,1 β α , αβ =−+= 1,1 α 3 α ,1 += β 3 β ,1 += α 5 α (11) Substituting these into (11) we have, n 1 n 44ki+ ( ki + ) 22 ki + 22 ki + ki ++22 ki 44) ki + fk( , 4, n ) = ∑(α−−4( 1) α +6(− 1)−− 4( 1) ββ + ) 25 i=0 i nn 1 44()kiknn++ikik222 2i = [αα∑∑ −−4( 1) αα +−6( 1) 25 ii=00ii= nn k+ikiki2244nn −−4( 1) ββββ∑∑+ ] ii ii=00= 1 =[αα44k (1 +−− ) n 4( 1) ()22ki+ αα k (1 ++−−− ) n 6( 1) 22k++ i 4( 1) kiββββ 2244 k (1 +++ ) n k (1 ) n ] 25 1 = [αα4k (3 2 ) n−− 4( 1)(ki+ ) α 2 k (1 + α 2 ) n +− 6( 1)22k++ i −− 4( 1) kiβ2 k (1 + β 2 ) n + ββ 4 k (3 2 ) n ] 25 1 =[3nα42 k++ n −− 4( 1) ki αα2 k (1 + 2 ) n +−− 6 4( 1)ki+ ββ2 k (1 + 2 ) n + 3 n β 42 k+ n ] 25 1 n =[3n (α42 k+ n + β 42 k + n) −−4( 1)ki + {52 (αβ2 kn ++ − 2kn)] + 6} 25 . Thus finally we have n+1 1 n ()ki+ fk( ,4, n )= [3 L42k++ n −− 4( 1) 5F2kn+ 6] 25 . The question of calculating f(, kmn ,) for m = 5 or for even higher powers could be solved in the same way but the form may be too complicated. Acknowledgments This research was supported by the Education Department of Shaanxi Province (Grant NO. 16JK1265). 283 Advances in Engineering Research (AER), volume 107 References [1] W. Andrzez,Some identities for the generalized Fibonacci numbers and the generalized Lucas numbers, Applied Mathematics and Computation,219(2013),5564--5568. [2] J J Chao, The relation between generalized Yang Hui triangle and Fibonacci sequence, Journal of Capital Normal University:Natural Science Edition,32(2) (2011),11--13. [3] X F Chen,Another relation between Fibonacci number and Yang Hui triangle, Journal of Capital Normal University:Natural Science Edition, 35(3) (2014),1--2. [4] L K Hua, Begin from Yang Hui triangle[M].Beijing, China Youth Press,1962. 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