Binomial sum relations involving Fibonacci and Lucas numbers

In this paper, we introduce relations between binomial sums involving (generalized) Fibonacci and Lucas numbers, and different kinds of binomial coefficients. We also present some relations between sums with two and three binomial coefficients. In the course of exploration we rediscover a few relations presented as problem proposals.


Introduction and motivation
The literature on Fibonacci numbers is immensely rich.There exist dozens of articles and problem proposals dealing with binomial sums involving these sequences as (weighted) summands.We attempt to give a short survey, not claiming completeness.The following binomial sums have been studied (X n stands for a (weighted) Fibonacci or Lucas number, alternating or non-alternating, or a product of them): • Standard form and variants of it [2,6,12,14,19,24,26,27] n k=0 n k X k ; • Forms coming from the Waring formula and studied by Gould [17], for instance, • Forms introduced by Filipponi [15] n k=0 • Forms introduced by Jennings [22] n k=0 n + k 2k X k ; • Forms introduced by Kilic and Ionascu [23] n k=0 2n n + k X k ; • Forms studied recently by Bai, Chu and Guo [9] ⌊n/2⌋ k=0 2n n − 2k X k and ⌊n/2⌋ k=0 2n + 2 n − 2k X k ; • Forms studied by the authors in the recent paper [4] ⌊n/2⌋ k=0 n 2k X k ; • Forms studied by the authors in the recent paper [3] We note that Let (W j (a, b; p, q)) j≥0 be the Horadam sequence [21] defined for all non-negative integers j by the recurrence where a, b, p and q are arbitrary complex numbers, with p = 0 and q = 0. Extension of the definition of (W j ) to negative subscripts is provided by writing the recurrence relation as Two important cases of (W j ) are the Lucas sequences of the first kind, (U j (p, q)) = (W j (0, 1; p, q)), and of the second kind, (V j (p, q)) = (W j (2, p; p, q)), so that U 0 = 0, U 1 = 1, U j = pU j−1 − qU j−2 , j ≥ 2; and The most well-known Lucas sequences are the Fibonacci sequence (F j ) = (U j (1, −1)) and the sequence of Lucas numbers (L j ) = (V j (1, −1)).
In this paper, we introduce relations between binomial sums involving (generalized) Fibonacci and Lucas numbers, and different kinds of binomial coefficients.We also present some relations between sums with two and three binomial coefficients.In the course of exploration we rediscover a few relations presented as problem proposals.
In particular, Lemma 3. Let r and d be any integers.Then In particular, [19], Lemma 4. For any integer j, Proof.See [5, Lemma 1] for a proof of (14).Identity ( 15) is a consequence of the Binet formula.

Relations from a classical polynomial identity
The first binomial sum relations follow from the next classical polynomial identity which we state in the next lemma.
Lemma 5 ([29, Identity 6.21]).If x is a complex variable and m and n are non-negative integers, then According to Gould [29], identity ( 16) is due to Laplace.In addition, we note that the binomial theorem is a special case of ( 16) which occurs at m = n − 1.
Using −n k = (−1) k n + k − 1 k and replacing m by m − 1 we have the equivalent and useful form of Lemma 5: Theorem 1.If r, s and t are any integers and m and n are non-negative integers, then Proof.Set x = −U r+s /(U r σ s ) in ( 16), use (11) and multiply through by τ t , obtaining Similarly, setting x = −U r+s /(U r τ s ) in ( 16), using (10) and multiplying through by σ t , yields The results follow by combining these identities according to the Binet formulas (2) and Lemma 4.
In particular, and with the special cases Corollary 2. If r, s and t are any integers and n is a non-negative integer, then Proof.Set m = n in Theorem 1.
In particular, Corollary 3. If r, s and t are any integers and n is a non-negative integer, then In particular, with the special cases We mention that identities (19) and (20) exhibit strong similarities to those derived by Hoggatt, Phillips and Leonard in [20].
Proof.Make the substitutions r → 2r, s → −r and t → −rn in Theorem 1 and simplify.
In particular, n k=0 with the special cases By making appropriate substitutions in Theorem 1 many new sum relations can be established.For example, setting r = 1, s = −2, and t = 1 (or r = −1, s = 2 and t = −1) in (17) gives which at m = 2n gives The corresponding Fibonacci sums from (18) are of exactly the same structure with the special case Another example is the relation and its Fibonacci counterparts: Theorem 5.If m and n are non-negative integers and r, s are any integers, then Proof.Set x = −U r σ s /U r+s in (16), use (11) and multiply through by τ t , obtaining Similarly, setting x = −U r τ s /U r+s in (16), using (13) and multiplying through by σ t , yields Now, the result follows immediately upon combining according to the Binet formulas (2).
In particular, with the special cases If n is a non-negative integer and r and s are any integers, then Proof.Set m = n in ( 24) and (25).
In particular, We mention that setting m = n − 1 in Theorem 5 gives again Corollary 3.
Corollary 7. If m and n are non-negative integers and r is any integer, then Proof.Make the substitutions r → 2r, s → −r, t → −rn in (24), (25) and simplify.
In particular, Theorem 8.If m and n are non-negative integers and s, t are integers, then Proof.Set x = V s /(∆U s ) in ( 16) and multiply through by τ t to obtain Similarly, set x = −V s /(∆U s ) in ( 16) and multiply through by σ t to obtain Combine ( 26) and ( 27) according to the Binet formula while making use also of (14).Consider the cases n → 2n and n → 2n − 1, in turn.
In particular, with the special cases Note that in ( 28)-( 31), we used (see, for example, [1, Identities (1.16), (1.17)]) Lemma 6.If x is a complex variable and m, n are non-negative integers, then Proof.Use the transformation x 1+x → x in ( 16).Theorem 9.If m and n are non-negative integers and s, t are integers, then Writing 2n for n in (38) (after multiplying through by β t ) and comparing the coefficients of √ 5 produces (34) and (35).Writing 2n − 1 for n gives (36) and (37).
Corollary 10.If m and n are non-negative integers and s is an integer, then Corollary 11.If n is a non-negative integer and s is any integer, then 3 Relations from a recent identity by Alzer In 2015 Alzer [8], building on the work of Aharonov and Elias [7], studied the polynomial Among other things he showed that Such a polynomial identity immediately offers many appealing Fibonacci and Lucas sum relations as can been seen from the next series of theorems.
Theorem 12.For each non-negative integer n we have the relations Proof.Set x = α and x = β in (39) and (40), respectively, and combine according to the Binet formulas (3).
The next theorem generalizes Theorem 12.
Theorem 14.For non-negative integers n and m we have the relations Proof.Set x = α m /L m and x = β m /L m in (39) and (40), respectively, and combine according to the Binet formulas.
When m = 1 then Theorem 14 reduces to Theorem 12.As additional examples we state the next relations: which also appears in Alzer's paper [8] as Eq.(1.4), and Theorem 15.For non-negative integer n and any integers m and t, we have the relations 39) and (40), respectively, and combine according to the Binet formulas, while making use of Lemma 4.
In particular, with the special cases Theorem 16.For each non-negative integer n we have the relations Proof.Set x = α/2 and x = β/2 in (39) and (40), respectively, and combine according to the Binet formulas.
Theorem 17.For each non-negative integer n we have the relations Proof.Set x = 1/α and x = 1/β in (39) and (40), respectively, and combine according to the Binet formulas.
Remark.Combining Theorem 13 with Theorem 17 gives the relations Theorem 18.For each non-negative integer n we have the relations Proof.Set x = −α 4 and x = −β 4 in (39) and (40), respectively, and combine according to the Binet formulas.
The last Theorem in this set involves mixed identities.
Theorem 19.For each non-negative integer n we have the relations Proof.Set x = −α 3 and x = −β 3 in (39) and (40), respectively, and combine according to the Binet formulas.
As a final remark in this section we note that some of the identities presented in this section follow also from the following lemma.
4 Relations involving two central binomial coefficients Lemma 8. Let x be a complex variable.Then Proof.From Riordan's book [30] it is known that for the polynomial we have the relation Set x = 2t − 1 and simplify.
Theorem 20.For each integer r and each non-negative integer n we have the relations Proof.Set x = α and x = β in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas (3).
Theorem 21.For each integer r and each non-negative integer n we have the relations Proof.Set x = α 2 and x = β 2 in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
We note the following particular results: Theorem 22.For each integer r and each non-negative integer n we have the relations Proof.Set x = α/2 and x = β/2 in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
Theorem 23.For each integer r and each non-negative integer n we have the relations Proof.Set x = √ 5/2 and x = − √ 5/2 in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
Theorem 24.For each integer r and each non-negative integer n we have the relations Proof.Set x = α 3 and x = β 3 in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
Theorem 25.For each integer r and each non-negative integer n we have the relations Proof.Set x = α 4 and x = β 4 in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
In particular, We proceed with some identities involving an additional parameter.
Theorem 26.If n is a non-negative integer and r, s are any integers, then Proof.Set x = σ s /τ s and x = τ s /σ s , in turn, in Lemma 8, multiply through by τ r and σ r , respectively, and combine according to the Binet formulas.
In particular, with the special cases Remark.Note that Theorems 21 and 25 are particular cases of (47) and (48) at s = 1 and s = 2, respectively.
Theorem 27.For integers r and s ≥ 1, and each non-negative integer n we have the relations Proof.Set x = √ 5F s /L s and x = − √ 5F s /L s in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
Theorem 28.For integers r and s ≥ 1, and each non-negative integer n we have the relations Proof.Set x = L s /( √ 5F s ) and x = −L s /( √ 5F s ) in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
Theorem 29.For each integer r and each non-negative integer n we have the relations Proof.Set x = α 3 / √ 5 and x = β 3 / √ 5, in turn, in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas, using also the fact that √ 5 − α 3 = −2 and √ 5 + α 3 = 4α.
Theorem 30.For each integer r, and each non-negative integer n we have the relations Proof.Set x = α 3 /3 and x = β 3 /3, in turn, in Lemma 8, multiply through by α r and β r , respectively, and combine according to the Binet formulas, using also the fact that 3−α 3 = 2β and 3 + α 3 = 2 √ 5α.
Theorem 31.If n is a non-negative integer and r, s are any integers, then Proof.Set x = ∆U s /V s in Lemma 8 and multiply through by σ r .Repeat for x = −∆U s /V s and multiply through by τ r .Now combine the resulting equations using the Binet formula.
In particular, with the special cases 5 Another class of identities with squared binomial coefficients Lemma 9 ([28]).If n is a non-negative integer and x is any complex variable, then Theorem 32.Let r, s and m be arbitrary integers with r = 0. Then for each non-negative integer n we have the relations Proof.Set x = −F s β r /(F r α s ) and x = −F s α r /(F r β s ), respectively, in Lemma 9, and use Lemma 3. Multiply through by α m and β m , respectively, and combine according to the Binet formulas.
Corollary 33.For each integer m and each non-negative integer n we have Proof.Set r = 2 and s = −1 in Theorem 32.
The case m = 0 in Corollary 33 was proposed by Carlitz as a problem in the Fibonacci Quarterly [10] (with a typo).
Corollary 34.For each integer m and each non-negative integer n we have Proof.Set r = s = 1 in Theorem 32.
The case m = 0 in Corollary 34 was proposed by Carlitz as another problem in the Fibonacci Quarterly [11].
Corollary 35.For each integer m and each non-negative integer n we have Proof.Set r = 2 and s = 1 in Theorem 32.
Corollary 36.For each integer m and each non-negative integer n we have Proof.Set r = −1 and s = 2 in Theorem 32.
Corollary 37.For each integer r and each non-negative integer n we have Proof.Set r = 1 and s = 2 in Theorem 32.
Corollary 38.For each integer r and each non-negative integer n we have Proof.Set s = r in Theorem 32.
Corollary 39.For each integer r and each non-negative integer n we have Proof.Set s = 2r in Theorem 32 and make use of F 3r /F r = L 2r + (−1) r .
Theorem 40.For each non-negative integers r, m and n we have the relations Proof.Set x = α m /L m and x = β m /L m in Lemma 9, multiply through by α r and β r , respectively, and combine according to the Binet formulas.
When m = 1 then Theorem 40 reduces to Corollary 33.When m = 2 then Theorem 41.For each non-negative integer n, any odd integer m and any integer r we have the relations Proof.Set x = α 2m and x = β 2m , m odd, in Lemma 9, and use the facts that Multiply through by α r and β r , respectively, and combine according to the Binet formulas.
When m = 1 then Theorem 41 reduces to Corollary 34.
We conclude this section with a sort of inverse relations compared to those from Theorem 32.
Theorem 42.For each non-negative integer n and any integers m, r and s, we have the relations Proof.Set x = F r+s /(α s F r ) and x = F r+s /(β s F r ), respectively, in Lemma 9, and use Lemma 3. Multiply through by α m and β m , respectively, and combine according to the Binet formulas.
6 More identities with two binomial coefficients Lemma 10 ([18, Identity (3.17)]).If n is a non-negative integer, m is any real number and x is any complex variable, then In particular, Proof.The first particular case is obvious.The second follows upon setting m = n − 1/2 in (52) and using Remark.Comparing (49) with (53) we immediately get an "identity" of the form Such an identity does not contain any new information as the identities can be trivially transformed into each other by reindexing n k=0 a k = n k=0 a n−k .This shows that the binomial sum relations from the previous section are actually special instances of those derived now.
Theorem 43.Let r, s and p be arbitrary integers with r = 0, and let m is any real number.Then for all non-negative integer n we have the relations In particular, Proof.Set x = −F s β r /(F r α s ) and x = −F s α r /(F r β s ), respectively, in (52), and use Lemma 3. Multiply through by α p and β p , respectively, and combine according to the Binet formulas.
Corollary 44.If n is a non-negative integer, m is any real number and p is any integer, then In particular, Corollary 45.If n is a non-negative integer, m is any real number and p is any integer, then In particular, Corollary 46.If n is a non-negative integer, m is any real number and p is any integer, then In particular, Corollary 47.If n is a non-negative integer, m is any real number and p is any integer, then In particular, n k=0 Theorem 48.If n is a non-negative integer, m is any real number and s, r are any integers, then In particular, Proof.Set x = α s /L s and x = β s /L s in (52), multiply through by α r and β r , respectively, and combine according to the Binet formulas.
Theorem 49.If n is a non-negative integer, m is any real number, r is any integer and s is an odd integer, then In particular, Proof.Set x = α 2s in (52) and use the fact that α 2s − 1 = α s L s if s is an odd integer.
Theorem 50.Let n be a non-negative integer and let m be a real number.Let r, s, p and t be any integers.Then 52) and use ( 12) to obtain from which the results follow.
Corollary 51.Let n be a non-negative integer; let m be a real number and let s and p be any integers.Then Proof.Set t = −sp and r = −s in (59) and (60).
Corollary 52.Let n be a non-negative integer and let s and p be any integers.Then Theorem 53.If n is a non-negative integer, m is any real number and r, s and t are any integers, then Proof.Set x = U r+s and y = τ r U s in (61) and multiply by σ t to obtain Similarly obtain Combine ( 62) and (63) using the Binet formula.
In particular, with the special cases Lemma 12 ([18, Identity (3.84)]).If n is a non-negative integer and x is any complex variable, then Theorem 54.If n is a non-negative integer and r, s, t and m are any integers, then Proof.Set x = U r+s /(τ r U s ) in (64); use (10) and multiply through the resulting equation by τ t to obtain Similarly obtain Combine (65) and (66) according to the Binet formula.
In particular, with the special cases Lemma 13 ([16, 31]).If n is a non-negative integer and x is any complex variable, then (−1) Proof.To prove (68), set x = q s /τ 2s and x = q s /σ s , in turn, in (67) and use (4).Combine the resulting equations according to the Binet formula.For (69), use x = τ 2s /q s and x = σ s /q s , in turn, in (67).
In particular, Next we present an obvious extension of (33) and some associated Fibonacci-Lucas sums.
Lemma 14.Let x and y be complex variables.Let m and n be non-negative integers and let r be any integer.Then The result follows from (71), (72) and the Binet formula.
8 Identities with three binomial coefficients Concerning identities with three binomial coefficients some classical Fibonacci (Lucas) examples exist.For instance, Carlitz [13] presented the identities In addition, Zeitlin [35,37] derived In his solution to Carlitz' proposal from above Zeitlin [36] proved mutatis mutandis the identity In this section we provide more examples of this kind using "Zeitlin's identity" in its equivalent form given in the next lemma.

Lemma 1 .
If a, b, c and d are rational numbers and λ is an irrational number, then a + b λ = c + d λ ⇐⇒ a = c, b = d.

Corollary 4 .
If m and n are non-negative integers and r is any integer, then n k=0 L t+r(m−k)−sk .
Theorem 55.If n is a non-negative integer and s, t and m are any integers, then W 2s(m−k)+t ,

Theorem 59 .n k 3 q
If n is a non-negative integer and r and t are any integers, then n k=0 rk W r(n−2k)+t = W Still other classes of identities with two binomial coefficients Lemma 11 ([18, Identity (3.18)]).If n is a non-negative integer, m is any real number and x, y are any complex variables, then