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Int. J. Contemp. Math. Sciences, Vol. 7, 2012, no. 38, 1871 - 1876 Fibonacci Identities as Binomial Sums Mohammad K. Azarian Department of Mathematics, University of Evansville 1800 Lincoln Avenue, Evansville, IN 47722, USA E-mail: [email protected] Abstract To facilitate rapid numerical calculations of identities pertaining to Fibonacci numbers, we present each identity as a binomial sum. Mathematics Subject Classification: 05A10,11B39 Keywords: Fibonacci numbers, Fibonacci identities, Fibonacci sequence, Pascal’s identity, Pascal’s (Khayyām-Pascal’s) triangle 1. Preliminaries The most prominent linear homogeneous recurrence relation of order two with constant coefficients is the one that defines Fibonacci numbers (or Fibonacci sequence). It is defined recursively as Fn+2 = Fn+1 + Fn , where F0 = 0, F1 = 1, and n ≥ 0. 1872 M. K. Azarian It is well-known that the function g(x) = generates Fibonacci sequence. can be verified by long division. x 1 − x − x2 (1) Bicknell and Hoggatt in [2] stated that (1) But, since the method of long division is a long process, especially for a large n, the author in [1] used the method of generating functions to verify (1) which is quicker, regardless of the value of n, and obtained √ √ 1+ 5 n 1 1− 5 n ) −( ) , n ≥ 0. Fn = √ ( 2 2 5 (2) Finding the exact value of Fn from (2) requires multiple steps of busy and messy algebraic calculations which is not desirable. So, our goal in this note is to present Fn as a binomial sum for quick numerical calculations. Likewise, we use this binomial sum to write some well-known and fundamental identities concerning Fibonacci numbers as binomial sums as well. It is known that Fibonacci numbers are the sum of the numbers along the rising diagonals of Pascal’s (Khayyam-Pascal’s) triangle, and if we write the elements of Pascal’s triangle as binomial terms we have 0 =1 S1 = 0 1 =1 S2 = 0 2 1 S3 = + =2 0 1 3 2 + =3 S4 = 0 1 4 3 2 + + =5 S5 = 0 1 2 5 4 3 + + =8 S6 = 0 1 2 1873 Fibonacci identities as binomial sums . . . n n−1 n−2 Sn+1 = + + + ..., n ≥ 0. 0 1 2 n+1 n n Now, using Pascal’s identity = + , (n ≥ r ≥ 0) r r r−1 we can easily verify that Sn+2 = Sn+1 + Sn , and hence the binomial sum Sn satisfies the Fibonacci relation. In practice, we need to know all terms in the binomial sum of Sn . By inspection we can see that for n ≥ 0, Sn+1 n 2 n − n2 + 1 n − n2 n n−1 n−2 n−i = + + + ... + , + = 0 1 2 n2 − 1 n2 i i=0 where n represents the floor function. Again, if we use Pascal’s identity, we can easily show that the above Sn does indeed satisfy the Fibonacci relation. 2. Identities It is well-known that the left-hand side of each identity in Corollary 1 can be written as a (power of a ) single Fibonacci number. For example, n n as early as 1876 Lucas has shown that 1 + Fi = Fn+2 , 1 + F2i = F2n+1 , and n i=1 i=1 F2i−1 = F2n . One could use the principle of mathematical induction, i=1 combinatorial argument, or just simple algebra to verify the validity of these identities. These Fibonacci identities have been developed over the centuries by mathematicians and number enthusiasts alike, and their proofs can be found in various sources. So, as we stated earlier, the goal of this note is to write some of these fundamental identities as binomial sums for quick numerical calculations. 1874 M. K. Azarian Theorem 1. If Fn is any Fibonacci number, then Fn+1 n − n2 n − n2 + 1 n n−1 n−2 + = + + + ... + 0 1 2 n2 − 1 n2 n 2 n−i = , n ≥ 0. i i=0 Proof follows from our above discussion. Also, Theorem 1 can be proven by using the principle of mathematical induction or combinatorial methods. As a direct consequence of Theorem 1 and the definition of Fibonacci numbers we obtain the following corollary. Corollary 1. If n is any nonnegative integer, then n+1 n 2 n+1−i Fi = (i) 1 + i i=0 i=0 n n 2n − i F2i = (ii) 1 + i i=0 i=0 2n+1 2 n 2n + 1 − i (iii) F2i+1 = i i=0 i=0 3n+3 n 2 3n + 3 − i (iv) 1 + 2 F3i+2 = i i=0 i=0 3n+2 2 n 3n + 2 − i (v) 2 F3i+1 = i i=0 i=0 3n+1 2 n 3n + 1 − i (vi) 1 + 2 F3i = i i=0 i=0 2 n n 2n − i (vii) 1 + F4i = i i=0 i=0 2n+1 n+1 2 n+1 2n + 1 − i Fn+1−i = (viii) i i i=0 i=0 n n+1 2n + 2 − i (ix) 1 + Fn+1 Fn+2 + 2 Fi Fi+1 = i i=0 i=0 1875 Fibonacci identities as binomial sums n+3−i (x) 2 + (n + 1)Fn+2 − iFi = i i=0 i=0 n 2 n−i n+1 2 3 (xi) Fn Fn+3 − Fn+2 = (−1) i i=0 n+1 2 n + 1 − i n+1 2 3 (xii) Fn+3 Fn − Fn+1 = (−1) i i=0 n n+3 2 3. Conclusion We presented just a few widely-known Fibonacci identities as binomial sums. We are hoping that this article would motivate the curious reader to write her/his favorite Fibonacci identities as binomial sums too. The author himself will be working on writing some other Fibonacci and Lucas identities as binomial sums as well. References [1] Mohammad K. Azarian, The Generating Function for the Fibonacci Sequence, Missouri Journal of Mathematical Sciences, Vol. 2, No. 2, Spring 1990, pp. 78-79. [2] Marjorie Bicknel and Verner E. Hoggatt, Fibonacci’s Problem Book, The Fibonacci Association, 1974. [3] Verner E. Hoggatt and Gerald E. Bergum, A Problem of Fermat and the Fibonacci Sequence, The Fibonacci Quarterly, Vol. 15, No. 4, Oct. 1977, pp. 323-330. [4] K. Subba Rao, Some Properties of Fibonacci Numbers, The American Mathematical Monthly, Vol. 60, No. 10, Dec. 1953, pp. 680-684. [5] Neil. J. Sloan, http://oeis.org/A000045&A007318. 1876 Received: April, 2012 M. K. Azarian