Fibonacci sums modulo 5

We develop closed form expressions for various finite binomial Fibonacci and Lucas sums depending on the modulo 5 nature of the upper summation limit. Our expressions are inferred from some trigonometric identities.


Preliminaries
As usual, the Fibonacci numbers F n and the Lucas numbers L n are defined, for n ∈ Z, by the following recurrence relations for n ≥ 2: For negative subscripts we have F −n = (−1) n−1 F n and L −n = (−1) n L n .
Throughout this paper, we denote the golden ratio by α = 1+ The sequences {F n } n≥0 and {L n } n≥0 are indexed in the On-Line Encyclopedia of Integer Sequences [17] as entries A000045 and A000032, respectively.For more information we refer to Koshy [12] and Vajda [18] who have written excellent books dealing with Fibonacci and Lucas numbers.
There exists a countless number of binomial sums involving Fibonacci and Lucas numbers.For some new articles in this field we refer to the papers [1,2,4,6].
In this paper, we introduce closed form expressions for finite Fibonacci and Lucas sums involving different kinds of binomial coefficients and depending on the the modulo 5 nature of the upper summation limit.Our expressions are derived from various trigonometric identities, particularly utilizing Waring formulas and Chebyshev polynomials of the first and second kinds.We also present some series involving Bernoulli polynomials.
We note that some of our results were announced without proofs in [5].
Lemma 2. If n is an integer, then Relations stated in (4) follow directly from (3).
In our first main results we state Lucas (Fibonacci) identities involving binomial coefficient and additional parameter.
Theorem 1.If n is a positive integer and t is any integer, then Proof.Set x = π/5 in (1) and use (3) and the fact that for any integer r.
We proceed with some corollaries.
Corollary 2. If n is a positive integer, then if n ≡ 2 or 3 (mod 5).
A variant of the Lucas and Fibonacci sums with even subscripts is stated as the next corollary.

Fibonacci sums modulo 5 from Waring formulas
This section is based on utilizing the following trigonometric identities with the use of Waring formulas.
Proof.Similar to the proof of Lemma 4. We use the dual to the Waring formula If n is a positive integer and t is any integer, then if n ≡ 2 or 3 (mod 5).
Proof.We apply equation (10).Inserting x = π/5 and x = 3π/5, respectively, and keeping in mind the trigonometric identity cos 3x = 4 cos 3 x − 3 cos x we end with if n ≡ 2 or 3 (mod 5); To complete the proof simplify the terms in brackets and combine according the Binet form.
From Theorem 7 we can immediately obtain the following finite binomial sums.

Fibonacci sums modulo 5 from Chebyshev polynomials
For any integer n ≥ 0, the Chebyshev polynomials {T n (x)} n≥0 of the first kind are defined by the second-order recurrence relation [16] T while the Chebyshev polynomials {U n (x)} n≥0 of the second kind are defined by The Chebyshev polynomials possess the representations and have the exact Binet-like formulas The properties of Chebyshev polynomials of the first and second kinds have been studied extensively in the literature.The reader can find in the recent papers [7,8,11,14,15,19] additional information about them, especially about their products, convolutions, power sums as well as their connections to Fibonacci numbers and polynomials.Lemma 6.For all x ∈ C and a positive integer n, we have the following identities: Proof.Identities ( 14) and ( 15) are consequences of the identity derived in [3].

Lemma 7.
If n is a non-negative integer, then Proof.Evaluate the identity T n (cos x) = cos nx at x = 4π/5 and x = 2π/5, in turn.Theorem 9.If n is a positive integer and t is any integer, then Proof.Using x = −α/2 and x = −β/2, in turn, in (16) with the upper sign gives, in view of Lemma 7, if n ≡ 1 or 4 (mod 5); −(λα t−1 + β t−1 )/2, if n ≡ 2 or 3 (mod 5); from which ( 17) and ( 18) now follow upon setting λ = 1 and λ = −1, in turn, and using the Binet formulas and the summation identity We observe the following special cases of the prior result.
Using Theorem 11, we have the following binomial Fibonacci identities modulo 5.
To get Theorem 13 simplify the terms in brackets and replace t by t + 1.
Applying Theorem 13 yields the following two corollaries.Corollary 15.If n is a positive integer and t is any integer, then we have: If n ≡ 0 (mod 5), then if n ≡ 1 or 4 (mod 5), then if n ≡ 2 or 3 (mod 5), then Proof.Compare Theorem 13 with Theorem 4.

Lemma 9.
If n is a non-negative integer, then Proof.Evaluate the identity [3] where B n is the nth Bernoulli number, defined by the power series We have B n (1) = B n (0) = B n for all n ≥ 2 and B 2n+1 = 0 for all n ≥ 1.
Theorem 19.Let m be a non-negative integer.Then and Fibonacci and Lucas numbers possess the explicit formulas (Binet forms)

( 4 )
Proof.Relations stated in (3) can be proved easily by elementary methods.For instance, they follow by applying the addition theorem for the cosine function cos(a + b) = cos a cos b − sin a sin b combined with the special values cos

( 7 )
From Lemma 3 we can deduce the following Lucas and Fibonacci binomial identities modulo 5.

Theorem 4 .
If n is a positive integer and t is any integer, then ⌊n/2⌋ k=0