Binomial Fibonacci Sums from Chebyshev Polynomials

Kunle Adegoke

Department of Physics and Engineering Physics

Obafemi Awolowo University

220005 Ile-Ife

Nigeria

adegoke00@gmail.com

Robert Frontczak Reutlingen Germany

robert.frontczak@web.de

Taras Goy

Faculty of Mathematics and Computer Science Vasyl Stefanyk Precarpathian National University

Shevchenka Str. 57 76018 Ivano-Frankivsk

Ukraine

taras.goy@pnu.edu.ua

Abstract

We explore new types of binomial sums with Fibonacci and Lucas numbers. The binomial coefficients under consideration are \frac{n}{n+k} \binom{n+k}{n−k} and \frac{k}{n+k} \binom{n+k}{n−k}. We derive the identities by relating the underlying sums to Chebyshev polynomials. Finally, we study some combinatorial sums and derive a connection with a recent paper by Chu and Guo from 2022.

1 Preliminaries

As usual, the Fibonacci numbers F_{n} and the Lucas numbers L_{n} are defined, for n ∈ ℤ, through the recurrence relations F_{n} = F_{n−1} + F_{n−2}, n ≥ 2, with initial values F_{0} = 0, F_{1} = 1 and L_{n} = L_{n−1} + L_{n−2} with L_{0} = 2, L_{1} = 1. For negative subscripts we have F_{−n} = (−1)^{n−1}F_{n} and L_{−n} = (−1)^{n}L_{n}. They possess the explicit formulas (Binet forms)

F_{n} = \frac{α^{n} − β^{n}}{α − β} , L_{n} = α^{n} + β^{n}, n ∈ ℤ,  (1)

with α = (1 + \sqrt{5})/2 being the golden ratio and β = −1/α. The sequences (F_{n})_{n≥0} and (L_{n})_{n≥0} are indexed in the On-Line Encyclopedia of Integer Sequences [19] as entries \underline{A000045} and \underline{A000032}, respectively. For more information we refer to Koshy [15] and Vajda [21], who have written excellent books dealing with Fibonacci and Lucas numbers.

For an integer n ≥ 0, the Chebyshev polynomials (T_{n}(x))_{n≥0} of the first kind are defined by the second-order recurrence relation [18]

T_{0}(x) = 1, T_{1}(x) = x, T_{n+1}(x) = 2xT_{n}(x) − T_{n−1}(x),

while the Chebyshev polynomials (U_{n}(x))_{n≥0} of the second kind are defined by

U_{0}(x) = 1, U_{1}(x) = 2x, U_{n+1}(x) = 2xU_{n}(x) − U_{n−1}(x).

The Chebyshev polynomials possess the representations

T_{n}(x) = ∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}(x^{2} − 1)^{k}x^{n−2k},  (2)

U_{n}(x) = ∑_{k=0}^{⌊n/2⌋} \binom{n + 1}{2k + 1}(x^{2} − 1)^{k}x^{n−2k}.  (3)

Also, the sequences T_{n}(x) and U_{n}(x) have the Binet-like formulas

T_{n}(x) = \frac{1}{2} ((x + \sqrt{x^{2} − 1})^{n} + (x − \sqrt{x^{2} − 1})^{n}),  (4)
U_{n}(x) = \frac{1}{2\sqrt{x^{2} − 1}} ((x + \sqrt{x^{2} − 1})^{n+1} − (x − \sqrt{x^{2} − 1})^{n+1}).  (5)

The Chebyshev polynomials of the first and second kind are connected by the equations

T_{n+1}(x) = xT_{n}(x) − (1 − x^{2})U_{n−1}(x),
U_{n+1}(x) = xU_{n}(x) + T_{n+1}(x),  (6)

from which we also get

T_{n}(x) = \frac{1}{2}(U_{n}(x) − U_{n−2}(x)).

The properties of Chebyshev polynomials have been studied extensively in the literature. In the recent papers [1, 7, 8, 9, 13, 14, 16, 17, 22] the reader can find additional information about them, especially about their products, convolutions, and power sums, as well as their connections with Fibonacci numbers and polynomials.

There exists a countless number of binomial sum identities involving Fibonacci and Lucas numbers. For some new papers in this field, we refer to [2, 3, 4, 5]. Motivated by these studies, the goal of this paper is to extend existing results to other types of binomial coefficients. More precisely, in this paper we deal with sums of the following forms

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}x_{k}  and  ∑_{k=0}^{n} \frac{k}{n + k} \binom{n + k}{n − k}x_{k},

where the x_{k} are some weighted Fibonacci (Lucas) entries. Among the huge number of Fibonacci (Lucas) sums in the literature, we could not find references treating these forms, although the binomial coefficients remind us of coefficients of Girard–Waring type [10], Jennings [11], and also Kilic and Ioanescu [12]. It may be also of interest that we can write the sums under consideration equivalently involving three binomial coefficients. For instance, the first type of sum equals

∑_{k=0}^{n} \begin{matrix}\binom{n+k−1}{k} & \binom{n}{k} \\ ( & {2k \atop k} & ) \end{matrix} x_{k},

and the second is similar.

2 Binomial Fibonacci and Lucas sums from identities involving T_{n}(x)

We start by deriving two identities involving T_{n}(x), which we prove for the readers’ convenience.

Theorem 1. For all x ∈ ℂ and non-negative integers n and m, we have the following identities:

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}(1 ∓ x)^{k} = (±1)^{n}T_{n}(x),  (7)

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}(1 ∓ T_{m}(x))^{k} = (±1)^{n}T_{nm}(x).  (8)

Proof. It suffices to prove (7). Let P_{n}^{(a,b)}(x), a, b > −1, be the Jacobi polynomial. Then P_{n}^{(a,b)}(x) has the truncated series representation

P_{n}^{(a,b)}(x) = \frac{Γ(a + n + 1)}{n!Γ(a + b + n + 1)} ∑_{k=0}^{n} \binom{n}{k}\frac{Γ(a + b + n + k + 1)}{2^{k}Γ(a + k + 1)}(x − 1)^{k}.

Now we use the fact that T_{n}(x) = \frac{4^{n}(n!)^{2}}{(2n)!} P_{n}^{(−1/2,−1/2)}(x). The statement follows after some

steps of simplifications using Γ(n + \frac{1}{2}) = \frac{(2n)!\sqrt{π}}{4^{n}n!}.

Identity (8) follows immediately if x → T_{m}(x), because T_{n}(T_{m}(x)) = T_{nm}(x).

We note that the relation T_{n}(2x^{2} − 1) = T_{2n}(x) in conjunction with (7) yields

∑_{k=0}^{n} 4^{k} \frac{n}{n + k} \binom{n + k}{n − k}(x^{2} − 1)^{k} = T_{2n}(x)  (9)

and

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}x^{2k} = (−1)^{n}T_{2n}(x).  (10)

Example 2. Setting x = 0 and x = −1 in (7) we get

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k} = \begin{cases}0, & if n is odd; \\ (−1)^{n/2}, & otherwise; \end{cases}

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k} = (−1)^{n}.

Example 3. Directly from (7) we have the following interesting sums:

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{k} = T_{n}(α) + T_{n}(β),

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{k} = \frac{T_{n}(α) − T_{n}(β)}{−\sqrt{5}}.

It follows from these formulas that T_{n}(α) + T_{n}(β) and \frac{1}{\sqrt{5}}(T_{n}(α) − T_{n}(β)) are integers. The sequence (\frac{1}{\sqrt{5}}(T_{n}(α) − T_{n}(β)))_{n≥0} = \{0, 1, 2, 5, 16, 45, 130, 377, 1088, 3145, … \} is sequence \underline{A138573} in the OEIS [19]. But more is true as we show in the next theorem.

Theorem 4. For an integer s, the sequences (T_{n}(α^{s}) + T_{n}(β^{s}))_{n≥0} and (\frac{T_{n}(α^{s})−T_{n}(β^{s})}{\sqrt{5}})_{n≥0} are integers. Moreover, the sequences (U_{n}(α^{s}) + U_{n}(β^{s}))_{n≥0} and (\frac{U_{n}(α^{s})−U_{n}(β^{s})}{\sqrt{5}})_{n≥0} are also integer sequences for each s.

Proof. For a complex variable x, from (2) we have

T_{n}(α^{s}x) = ∑_{k=0}^{⌊n/2⌋} ∑_{j=0}^{k}(−1)^{j}\binom{n}{2k}\binom{k}{j}(α^{s}x)^{n−2j},

T_{n}(β^{s}x) = ∑_{k=0}^{⌊n/2⌋} ∑_{j=0}^{k}(−1)^{j}\binom{n}{2k}\binom{k}{j}(β^{s}x)^{n−2j}.

In particular,

T_{n}(α^{s}) = \begin{cases}∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}L_{s}^{k}α^{s(n−k)}, & if s is odd; \\ ∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}(\sqrt{5}F_{s})^{k}α^{s(n−k)}, & otherwise, \end{cases}

T_{n}(β^{s}) = \begin{cases}∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}L_{s}^{k}β^{s(n−k)}, & if s is odd; \\ ∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}(−\sqrt{5}F_{s})^{k}β^{s(n−k)}, & otherwise. \end{cases}

Thus,

T_{n}(α^{s}) + T_{n}(β^{s}) = ∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}L_{s}^{k}L_{s(n−k)},  s odd,

\frac{T_{n}(α^{s}) − T_{n}(β^{s})}{\sqrt{5}} = ∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}L_{s}^{k}F_{s(n−k)},  s odd,

and

T_{n}(α^{s}) + T_{n}(β^{s}) = ∑_{k=0}^{⌊n/4⌋} 5^{k}F_{s}^{2k} (\binom{n}{4k}L_{s(n−2k)} + 5\binom{n}{4k + 2}F_{s}F_{s(n−2k−1)}),  s even,

\frac{T_{n}(α^{s}) − T_{n}(β^{s})}{\sqrt{5}} = ∑_{k=0}^{⌊n/4⌋} 5^{k}F_{s}^{2k} (\binom{n}{4k}F_{s(n−2k)} + \binom{n}{4k + 2}F_{s}L_{s(n−2k−1)}),  s even.

For Chebyshev polynomials of the second kind similar formulas hold. From (3) using Binet’s formulas (1) we have

U_{n}(α^{s}) + U_{n}(β^{s}) = ∑_{k=0}^{⌊n/2⌋} \binom{n + 1}{2k + 1}L_{s}^{k}L_{s(n−k)},  s odd,

\frac{U_{n}(α^{s}) − U_{n}(β^{s})}{\sqrt{5}} = ∑_{k=0}^{⌊n/2⌋} \binom{n + 1}{2k + 1}L_{s}^{k}F_{s(n−k)},  s odd,

and

U_{n}(α^{s}) + U_{n}(β^{s}) = ∑_{k=0}^{⌊n/4⌋} 5^{k}F_{s}^{2k} (\binom{n + 1}{4k + 1}L_{s(n−2k)} + 5\binom{n + 1}{4k + 3}F_{s}F_{s(n−2k−1)}),  s even,

\frac{U_{n}(α^{s}) − U_{n}(β^{s})}{\sqrt{5}} = ∑_{k=0}^{⌊n/4⌋} 5^{k}F_{s}^{2k} (\binom{n + 1}{4k + 1}F_{s(n−2k)} + \binom{n + 1}{4k + 3}F_{s}L_{s(n−2k−1)}),  s even.

Example 5. Here we proceed with additional interesting sums of the same kind as in Example 3:

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{2k} = (−1)^{n}(T_{n}(α) + T_{n}(β)),

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{2k} = \frac{(−1)^{n}}{\sqrt{5}} (T_{n}(α) − T_{n}(β)),

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{3k} = (−1)^{n}(T_{n}(2α) + T_{n}(2β)),

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{3k} = \frac{(−1)^{n}}{\sqrt{5}} (T_{n}(2α) − T_{n}(2β)),

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}L_{k} = T_{n}(\frac{\sqrt{5}α}{2}) + T_{n}(\frac{\sqrt{5}β}{2} ),

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}F_{k} = \frac{1}{\sqrt{5}}(T_{n}(\frac{\sqrt{5}α}{2}) − T_{n}(\frac{\sqrt{5}β}{2} )),

∑_{k=0}^{n} 4^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{k} = T_{2n}(α) + T_{2n}(β),

∑_{k=0}^{n} 4^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{k} = \frac{1}{\sqrt{5}}(T_{2n}(α) − T_{2n}(β)).

By substituting x = L_{p}/2 and x = \sqrt{5}F_{p}/2 in (4), in turn, we obtain the results stated in Lemma 6.

Lemma 6. If p is an integer, then we have

T_{n}(\frac{L_{p}}{2} ) = \frac{1}{2}L_{pn},  p even,  (11)

and

T_{n}(\frac{\sqrt{5}F_{p}}{2} ) = \begin{cases}\frac{1}{2}L_{pn}, & if p is odd and n is even; \\ \frac{\sqrt{5}}{2}F_{pn}, & if p is odd and n is odd. \end{cases}  (12)

In particular, we have

T_{n}(\frac{3}{2}) = \frac{1}{2}L_{2n},
T_{n}(\frac{\sqrt{5}}{2}) = \begin{cases}\frac{1}{2}L_{n}, & if n is even; \\ \frac{\sqrt{5}}{2}F_{n}, & otherwise; \end{cases}  (13)

T_{n}(\sqrt{5}) = \begin{cases}\frac{1}{2}L_{3n}, & if n is even; \\ \frac{\sqrt{5}}{2}F_{3n}, & otherwise. \end{cases}  (14)

Theorem 7. If n is a positive integer and p is an integer, then the following identities hold:

∑_{k=0}^{n}(−1)^{(p−1)(n−k)} \frac{n}{n + k} \binom{n + k}{n − k}L_{p}^{2k} = \frac{L_{2pn}}{2} ,
∑_{k=0}^{n}(−1)^{p(n−k)} \frac{n}{n + k} \binom{n + k}{n − k}5^{k}F_{p}^{2k} = \frac{L_{2pn}}{2} .

Proof. Set x = L_{2p}/2 in (7), use (11) and the fact that

L_{2p} − 2 = \begin{cases}5F_{p}^{2}, & if p is even; \\ L_{p}^{2}, & otherwise; \end{cases}  L_{2p} + 2 = \begin{cases}5F_{p}^{2}, & if p is odd; \\ L_{p}^{2}, & otherwise; \end{cases}  (15)

to get

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}L_{p}^{2k} = \frac{1}{2}L_{2pn},  p odd,
∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}(−1)^{k}L_{p}^{2k} = \frac{(−1)^{n}}{2} L_{2pn},  p even,

and

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}(−5)^{k}F_{p}^{2k} = \frac{(−1)^{n}}{2} L_{2pn},  p odd,
∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}5^{k}F_{p}^{2k} = \frac{1}{2}L_{2pn},  p even,

from which the stated identities follow.

Theorem 7 can be generalized in the following way. Theorem 8. If n is a positive integer and p is an integer, then we have

∑_{k=0}^{n}(−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}(2 ± \sqrt{5}F_{pm})^{k} = \frac{±\sqrt{5}}{2}F_{pmn},  p, m, n odd,
∑_{k=0}^{n}(−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}(2 ± L_{pm})^{k} = \frac{(±1)^{n}}{2} L_{pmn},  p odd, m even,
∑_{k=0}^{n}(−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}(2 ± L_{pm})^{k} = \frac{(±1)^{n}}{2} L_{pmn},  p even.

Proof. Combine (8) with (11) and (12).

Some particular cases of Theorems 7 and 8 stated in the next Example.

Example 9. We have

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}α^{−3k} = ∑_{k=0}^{n}(−1)^{k+1} \frac{n}{n + k} \binom{n + k}{n − k}α^{3k} = \frac{\sqrt{5}F_{n}}{2} ,  n odd,

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}4^{k}α^{−k} = ∑_{k=0}^{n}(−1)^{k+1} \frac{n}{n + k} \binom{n + k}{n − k}4^{k}α^{k} = \frac{\sqrt{5}F_{3n}}{2} ,  n odd,

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k} = \frac{L_{2n}}{2} ,  ∑_{k=0}^{n}(−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}5^{k} = \frac{L_{2n}}{2} ,

∑_{k=0}^{n} \frac{n}{n + k} \binom{n + k}{n − k}5^{k} = \frac{L_{4n}}{2} ,  ∑_{k=0}^{n}(−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}9^{k} = \frac{L_{4n}}{2} .

Lemma 10. If p is a non-zero integer, then

T_{n}(\frac{\sqrt{5}F_{p}}{L_{p}} ) = cosh(n arctanh (\frac{2}{\sqrt{5}F_{p}} )),  p odd,  (16)
T_{n}(\frac{L_{p}}{\sqrt{5}F_{p}} ) = cosh(n arctanh (\frac{2}{L_{p}} )),  p even.  (17)

Proof. Setting x = \sqrt{5}F_{p}/L_{p} in (4) and making use of 5F_{p}^{2} − 4(−1)^{p+1} = L_{p}^{2} with p odd produces

T_{n}(\frac{\sqrt{5}F_{p}}{L_{p}} ) = \frac{(\sqrt{5}F_{p} − 2)^{n} + (\sqrt{5}F_{p} + 2)^{n}}{2L_{p}^{n}}  (18)

from which (16) follows upon using the identity

(x − y)^{n} + (x + y)^{n} = 2(\sqrt{x^{2} − y^{2}})^{n} cosh(n arctanh (\frac{y}{x})).

For low values of p, it is easier to use (18) directly for evaluation. Thus, at p = 1 we recover (14) while p = 3 gives (13). On account of (15), Formula (17) also implies

T_{n}(\frac{L_{2p}}{\sqrt{5}F_{2p}} ) = \frac{(5F_{p}^{2})^{n} + (L_{p}^{2})^{n}}{2(\sqrt{5}F_{2p})^{n}}  (19)

for every non-zero integer p.

We also note that

∑_{k=0}^{2n} (\frac{4}{5})^{k}\binom{2n + k}{2n − k}\frac{4^{k} − (−1)^{k}L_{2p}^{2k}}{(2n + k)F_{2p}^{2k}} = 0,  p ≠ 0,

∑_{k=0}^{2n} (\frac{4}{5})^{k}\binom{2n + k}{2n − k}\frac{4^{k} + (−1)^{k}L_{2p}^{2k}}{(2n + k)F_{2p}^{2k}} = \frac{625^{n}F_{p}^{8n} + L_{p}^{8n}}{2n 25^{n}F_{2p}^{4n}} ,  p ≠ 0.

Theorem 11. If p is a non-zero integer and n is a positive integer, then we have

∑_{k=0}^{n} (\frac{16}{5})^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{2p}^{2(n−k)} = \frac{25^{n}F_{p}^{4n} + L_{p}^{4n}}{2 · 5^{n}} ,
∑_{k=0}^{n}(−1)^{n−k}(\frac{4}{5})^{k} \frac{n}{n + k} \binom{n + k}{n − k}(\frac{F_{2p}}{L_{2p}} )^{2(n−k)} = \frac{25^{n}F_{p}^{4n} + L_{p}^{4n}}{2 · 5^{n}L_{2p}^{2n}} .

In particular, we obtain

∑_{k=0}^{n} (\frac{16}{5})^{k} \frac{n}{n + k} \binom{n + k}{n − k} = \frac{25^{n} + 1}{2 · 5^{n}},
∑_{k=0}^{n}(−1)^{n−k}(\frac{36}{5})^{k} \frac{n}{n + k} \binom{n + k}{n − k} = \frac{25^{n} + 1}{2 · 5^{n}}.

Proof. Combine (19) with (9) and (10), respectively, while setting x = L_{2p}/(\sqrt{5}F_{2p}). Remark 12. We note the following relations:

∑_{k=0}^{n} (\frac{5}{16})^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}F_{2p}^{2(n−k)} = ∑_{k=0}^{n} \binom{2n}{2k}\frac{L_{2p}^{2(n−k)}}{4^{2n−k}} ,
∑_{k=0}^{n} (− \frac{5}{4})^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}(\frac{F_{2p}}{L_{2p}} )^{2(n−k)} = ∑_{k=0}^{n} \binom{2n}{2k}\frac{4^{k−n}}{L_{2p}^{2k}} .

Theorem 13. If p is an odd integer and n is a positive integer, then

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}\frac{L_{pk+t}}{L_{p}^{k}} = cosh(n arctanh (\frac{2}{\sqrt{5}F_{p}} )) · \begin{cases}L_{t}, & if n is even; \\ −\sqrt{5}F_{t}, & otherwise; \end{cases}
∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}\frac{F_{pk+t}}{L_{p}^{k}} = cosh(n arctanh (\frac{2}{\sqrt{5}F_{p}} )) · \begin{cases}F_{t}, & if n is even; \\ −\frac{L_{t}}{\sqrt{5}}, & otherwise. \end{cases}

Proof. Set x = \sqrt{5}F_{p}/L_{p} in (7), taking the upper signs, and then use (16). Corollary 14. If p is an odd integer and n is a positive integer, then

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}\frac{L_{pk}}{L_{p}^{k}} = 0,  n odd,
∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}\frac{F_{pk}}{L_{p}^{k}} = 0,  n even.

Corollary 15. If n is a positive integer, then we have the following combinatorial results:

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{k+t} = \begin{cases}\frac{1}{2}L_{t}L_{3n}, & if n is even; \\ −\frac{5}{2}F_{t}F_{3n}, & otherwise; \end{cases}

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{k+t} = \begin{cases}\frac{1}{2}F_{t}L_{3n}, & if n is even; \\ −\frac{1}{2}L_{t}F_{3n}, & otherwise. \end{cases}

The corresponding identities to Theorem 13 for p even are stated next. Theorem 16. If p is a non-zero even integer, n is a positive integer and t is an integer, then we have the following sum relations:

∑_{k=0}^{⌊n/2⌋} (\frac{16}{5F_{p}^{2}} )^{k}(\binom{n + 2k}{n − 2k}\frac{L_{2pk+t}}{n + 2k} − \frac{4}{F_{p}} \binom{n + 2k + 1}{n − 2k − 1}\frac{F_{p(2k+1)+t}}{n + 2k + 1})
= \frac{1}{n} cosh(n arctanh (\frac{2}{L_{p}} )) · \begin{cases}L_{t}, & if n is even; \\ −\sqrt{5}F_{t}, & otherwise; \end{cases}
∑_{k=0}^{⌊n/2⌋} (\frac{16}{5F_{p}^{2}} )^{k}(\binom{n + 2k}{n − 2k}\frac{F_{2pk+t}}{n + 2k} − \frac{4}{5F_{p}} \binom{n + 2k + 1}{n − 2k − 1}\frac{L_{p(2k+1)+t}}{n + 2k + 1})
= \frac{1}{\sqrt{5} n} cosh(n arctanh (\frac{2}{L_{p}} )) · \begin{cases}\sqrt{5}F_{t}, & if n is even; \\ −L_{t}, & otherwise. \end{cases}

The non-alternating versions of the identities in Theorem 13 are derived from (16) and (7) by considering the bottom sign combinations. This is left to the interested reader. Example 17. Noting that

∑_{k=0}^{n}(−2)^{k} \frac{n}{n + k} \binom{n + k}{n − k}((1 + x)^{k} ∓ (1 − x)^{k}) = ((−1)^{n} ∓ 1)T_{n}(x),  (20)

with x = \sqrt{5}/2 and (13) we get

∑_{k=1}^{n}(−1)^{k−1} \frac{n}{n + k} \binom{n + k}{n − k}F_{3k} = \frac{1 − (−1)^{n}}{2} F_{n},  (21)

as well as

∑_{k=0}^{n}(−1)^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{3k} = \frac{1 + (−1)^{n}}{2} L_{n}.  (22)

As we will see, identities (21) and (22) are special cases of Theorem 20 below.

Example 18. We derive the inverse relation of (21) and (22). Starting with (20) we set x = −\sqrt{5}, and make use of T_{n}(−\sqrt{5}) = (−1)^{n}T_{n}(\sqrt{5}) together with (14) to derive the following identities:

∑_{k=1}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{k} = \frac{(−1)^{n} − 1}{2}F_{3n},

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{k} = \frac{1 + (−1)^{n}}{2} L_{3n}.

Example 19. It is obvious that from (20) more appealing relations can be derived. We give just four examples:

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}(L_{2k} ± (−1)^{k}L_{k}) = ((−1)^{n} ± 1)(T_{n}(α^{3}) + T_{n}(β^{3}))

= ((−1)^{n} ± 1) ∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}4^{k}L_{3(n−k)},

∑_{k=0}^{n}(−4)^{k} \frac{n}{n + k} \binom{n + k}{n − k}(F_{2k} ± (−1)^{k}F_{k}) = \frac{(−1)^{n} ± 1}{\sqrt{5}}(T_{n}(α^{3}) − T_{n}(β^{3}))

= \frac{(−1)^{n} ± 1}{\sqrt{5}} ∑_{k=0}^{⌊n/2⌋} \binom{n}{2k}4^{k}F_{3(n−k)},

∑_{k=0}^{n} (− \frac{4}{3})^{k} \frac{n}{n + k} \binom{n + k}{n − k}L_{2k} = ((−1)^{n} + 1)T_{n}(\frac{\sqrt{5}}{3}),

∑_{k=0}^{n} (− \frac{4}{3})^{k} \frac{n}{n + k} \binom{n + k}{n − k}F_{2k} = \frac{(−1)^{n} − 1}{\sqrt{5}} T_{n}(\frac{\sqrt{5}}{3}).

Theorem 20. If n is a positive integer and t is an integer, then we have

∑_{k=0}^{n} (−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}L_{3k+t} = \begin{cases}\frac{5}{2}F_{t}F_{n}, & if n is odd; \\ \frac{1}{2}L_{t}L_{n}, & otherwise; \end{cases}

∑_{k=0}^{n} (−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}F_{3k+t} = \begin{cases}\frac{1}{2}L_{t}F_{n}, & if n is odd; \\ \frac{1}{2}F_{t}L_{n}, & otherwise. \end{cases}

Proof. Set x = \sqrt{α^{3}}/2 and x = \sqrt{β^{3}}/2, in turn, in (10) and use

T_{2n}(\frac{\sqrt{α^{3}}}{2} ) = \begin{cases}\frac{\sqrt{5}}{2}F_{n}, & if n is odd; \\ \frac{1}{2}L_{n}, & otherwise; \end{cases} T_{2n}(\frac{\sqrt{β^{3}}}{2} ) = \begin{cases}−\frac{\sqrt{5}}{2}F_{n}, & if n is odd; \\ \frac{1}{2}L_{n}, & otherwise. \end{cases}

to obtain

∑_{k=0}^{n} (−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}(α^{3k+t} + λβ^{3k+t}) = \begin{cases}\frac{\sqrt{5}}{2}F_{n}(α^{t} − λβ^{t}), & if n is odd; \\ \frac{1}{2}L_{n}(α^{t} − λβ^{t}), & otherwise; \end{cases}

from which the stated results follow upon setting λ = 1 and λ = −1.

Note that Example 17 is the special case (t = 0) of Theorem 20. Theorem 21. If n is a positive integer and p is a non-zero integer, then we have

∑_{k=0}^{n}(−1)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}(\frac{2L_{2p}}{\sqrt{5}F_{2p}} )^{2k} = \frac{1}{2}((\frac{\sqrt{5}F_{p}}{L_{p}} )^{2n} + (\frac{L_{p}}{\sqrt{5}F_{p}} )^{2n}).

Proof. Set x = L_{2p}/(\sqrt{5}F_{2p}) in (10) and then use (19). Theorem 22. If n is a non-negative integer and p and q are non-zero integers, then we have

∑_{k=0}^{n}(−1)^{(p−q)(n−k)} \frac{n}{n + k} \binom{n + k}{n − k} (\frac{F_{p+q}F_{p−q}}{F_{p}F_{q}} )^{2k} = \frac{F_{p}^{4n} + F_{q}^{4n}}{2(F_{q}F_{p})^{2n}} .

Proof. Set x = \frac{(−i)^{p−q+1}F_{p+q}F_{p−q}}{2F_{q}F_{p}}, where i = \sqrt{−1}, in (10) and use

T_{n} (\frac{i^{p−q+1}F_{p+q}F_{p−q}}{2F_{q}F_{p}} ) = i^{n(p−q+1)} \frac{F_{p}^{2n} + (−1)^{n(p−q+1)}F_{q}^{2n}}{2(F_{q}F_{p})^{n}} .

Theorem 23. If n is a non-negative integer and p is a non-zero integer, then we have

∑_{k=0}^{n}(−1)^{p(n−k)} \frac{n}{n + k} \binom{n + k}{n − k} (\frac{F_{3p}}{F_{2p}} )^{2k} = \frac{L_{p}^{4n} + 1}{2L_{p}^{2n}},
∑_{k=0}^{n}(−1)^{(p+1)(n−k)} \frac{n}{n + k} \binom{n + k}{n − k}5^{n−k} (\frac{L_{3p}}{F_{2p}} )^{2k} = \frac{5^{2n}F_{p}^{4n} + 1}{2F_{p}^{2n}}.

Proof. Set x = \frac{(−i)^{p+1}F_{3p}}{2F_{2p}} and x = \frac{(−i)^{p}L_{3p}}{2\sqrt{5}F_{2p}}, in turn, in (10) and use

T_{n} (\frac{(−i)^{p+1}F_{3p}}{2F_{2p}} ) = (−i)^{n(p+1)} \frac{L_{p}^{2n} + (−1)^{n(p+1)}}{2L_{p}^{n}} ,
T_{n} (\frac{(−i)^{p}L_{3p}}{2\sqrt{5}F_{2p}} ) = (−i)^{np} \frac{5^{n}F_{p}^{2n} + (−1)^{np}}{2\sqrt{5^{n}}F_{p}^{n}} .

Lemma 24. If p and q are integers, then the following identities involving Chebyshev polynomials T_{n}(x) hold:

T_{2n} (\sqrt{\frac{(−1)^{q+1}F_{q}^{2}}{4F_{p}F_{p+q}} α^{2p+q}}) = \frac{(−1)^{n}}{2} ((−1)^{nq} \frac{F_{p+q}^{n}}{F_{p}^{n}} α^{qn} + \frac{F_{p}^{n}}{F_{p+q}^{n}} β^{qn}), p ≠ 0, p ≠ −q, (23)

T_{2n} (\sqrt{\frac{(−1)^{q+1}F_{q}^{2}}{4F_{p}F_{p+q}} β^{2p+q}}) = \frac{(−1)^{n}}{2} ((−1)^{nq} \frac{F_{p+q}^{n}}{F_{p}^{n}} β^{qn} + \frac{F_{p}^{n}}{F_{p+q}^{n}} α^{qn}), p ≠ 0, p ≠ −q,

T_{2n} (\sqrt{\frac{5(−1)^{q+1}F_{q}^{2}}{4L_{p}L_{p+q}} α^{2p+q}}) = \frac{(−1)^{n}}{2} ((−1)^{nq} \frac{L_{p+q}^{n}}{L_{p}^{n}} α^{qn} + \frac{L_{p}^{n}}{L_{p+q}^{n}} β^{qn}),

T_{2n} (\sqrt{\frac{5(−1)^{q+1}F_{q}^{2}}{4L_{p}L_{p+q}} β^{2p+q}}) = \frac{(−1)^{n}}{2} ((−1)^{nq} \frac{L_{p+q}^{n}}{L_{p}^{n}} β^{qn} + \frac{L_{p}^{n}}{L_{p+q}^{n}} α^{qn}).

Theorem 25. If n is a positive integer and p, q and t are integers with p ≠ 0 and q ≠ 0, then we have the following combinatorial results:

∑_{k=0}^{n} \frac{(−1)^{(n−k)q} n}{n + k}\binom{n + k}{n − k}F_{p+q}^{n−k}F_{p}^{n−k}F_{q}^{2k}L_{k(2p+q)+t} = \frac{1}{2} (F_{p+q}^{2n}L_{t+qn} + F_{p}^{2n}L_{t−qn}), (24)

∑_{k=0}^{n} \frac{(−1)^{(n−k)q} n}{n + k}\binom{n + k}{n − k}F_{p+q}^{n−k}F_{p}^{n−k}F_{q}^{2k}F_{k(2p+q)+t} = \frac{1}{2} (F_{p+q}^{2n}F_{t+qn} + F_{p}^{2n}F_{t−qn}), (25)

∑_{k=0}^{n} \frac{(−1)^{(n−k)q} n}{n + k}\binom{n + k}{n − k}5^{k}L_{p+q}^{n−k}L_{p}^{n−k}F_{q}^{2k}L_{k(2p+q)+t} = \frac{1}{2} (L_{p+q}^{2n}L_{t+qn} + L_{p}^{2n}L_{t−qn}), (26)

∑_{k=0}^{n} \frac{(−1)^{(n−k)q} n}{n + k}\binom{n + k}{n − k}5^{k}L_{p+q}^{n−k}L_{p}^{n−k}F_{q}^{2k}F_{k(2p+q)+t} = \frac{1}{2} (L_{p+q}^{2n}F_{t+qn} + L_{p}^{2n}F_{t−qn}). (27)

Proof. In (10) set x = \sqrt{\frac{(−1)^{q+1}F_{q}^{2}}{4F_{p}F_{p+q}} α^{2p+q}} and use (23) to obtain

∑_{k=0}^{n} (−1)^{kq} \frac{n}{n + k} \binom{n + k}{n − k}F_{p+q}^{n−k}F_{p}^{n−k}F_{q}^{2k}α^{k(2p+q)+t} = \frac{(−1)^{nq}}{2} F_{p+q}^{2n}α^{qn+t} + \frac{(−1)^{t}}{2} F_{p}^{2n}β^{qn−t},

from which (24) and (25) follow. The proof of (26) and (27) is similar; in (10), set x =

\sqrt{\frac{5(−1)^{q+1}F_{q}^{2}}{4L_{p}L_{p+q}} α^{2p+q}}.

Example 26. If n is a positive integer and t is an integer, then we get

∑_{k=0}^{n}(−2)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}L_{5k+t} = \frac{1}{2}(L_{t−n} + 4^{n}L_{n+t}),

∑_{k=0}^{n}(−2)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}F_{5k+t} = \frac{1}{2}(F_{t−n} + 4^{n}F_{n+t}),

∑_{k=0}^{n}(−2)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}5^{k}L_{k+t} = \frac{1}{2}(4^{n}L_{t−n} + L_{n+t}),

∑_{k=0}^{n}(−2)^{n−k} \frac{n}{n + k} \binom{n + k}{n − k}5^{k}F_{k+t} = \frac{1}{2}(4^{n}F_{t−n} + F_{n+t}).

3 Binomial Fibonacci and Lucas sums from identities involving U_{n}(x)

Using the fact that \frac{d}{dx}T_{n}(x) = nU_{n−1}(x), from (7) we get

∑_{k=1}^{n}(−2)^{k} \frac{k}{n + k} \binom{n + k}{n − k}(1 ∓ x)^{k−1} = ∓U_{n−1}(x),  (28)

and from (9) and (10)

∑_{k=1}^{n} 4^{k} \frac{k}{n + k} \binom{n + k}{n − k}(x^{2} − 1)^{k−1} = \frac{U_{2n−1}(x)}{x}

and

∑_{k=1}^{n}(−4)^{k} \frac{k}{n + k} \binom{n + k}{n − k}x^{2k} = (−1)^{n}xU_{2n−1}(x).  (29)

From here, setting x = 3/2 in (28) and using U_{n−1}(3/2) = F_{2n}, we get

∑_{k=1}^{n} \frac{k}{n + k} \binom{n + k}{n − k} = \frac{F_{2n}}{2}

and

∑_{k=1}^{n}(−5)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k} = \frac{(−1)^{n−1}}{2} F_{2n}.

Also, with x = \sqrt{5}/2 upon combining we produce

∑_{k=1}^{n}(−1)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}L_{3(k−1)} = \frac{1 − (−1)^{n}}{2} L_{n},

where we used the fact that

U_{n}(\frac{\sqrt{5}}{2}) = \begin{cases}L_{n+1}, & if n is even; \\ \sqrt{5}F_{n+1}, & otherwise. \end{cases}

In general, working with (8) we get the relations

∑_{k=1}^{n}(−2)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}(1 ∓ T_{m}(x))^{k−1} = (±1)^{n−1} \frac{U_{nm−1}(x)}{2U_{m−1}(x)},

where we have used that

U_{m−1}(T_{n}(x)) = \frac{U_{mn−1}(x)}{U_{n−1}(x)}.

Lemma 27. If n is a positive integer and p an integer, then we have the following identities involving Chebyshev polynomials of the second kind:

U_{n}(\frac{L_{p}}{2} ) = \frac{F_{p(n+1)}}{F_{p}} , p even, p ≠ 0,
U_{n}(\frac{iL_{p}}{2} ) = \frac{i^{n}F_{p(n+1)}}{F_{p}} , p odd,

U_{n}(\frac{\sqrt{5}F_{p}}{2} ) = \begin{cases}L_{p(n+1)}/L_{p}, & if p is odd and n is even; \\ \sqrt{5}F_{p(n+1)}/L_{p}, & if p and n are odd; \end{cases}

U_{n}(\frac{i\sqrt{5}F_{p}}{2} ) = \begin{cases}i^{n}L_{p(n+1)}/L_{p}, & if p and n are even; \\ i^{n}\sqrt{5}F_{p(n+1)}/L_{p}, & if p is even and n is odd. \end{cases}

Theorem 28. If n is a positive integer and p is an integer, then we have

∑_{k=1}^{n}(−1)^{(p−1)(n−k)} \frac{k}{n + k} \binom{n + k}{n − k}L_{p}^{2k−1} = \frac{F_{2np}}{2F_{p}} ,  p ≠ 0,

∑_{k=1}^{n}(−1)^{p(n−k)} \frac{k}{n + k} \binom{n + k}{n − k}5^{k−1}F_{p}^{2k−1} = \frac{F_{2np}}{2L_{p}} .

Proof. Evaluate (29) at x = L_{p}/2, x = iL_{p}/2, x = \sqrt{5}F_{p}/2, and x = i\sqrt{5}F_{p}/2, in turn, using Lemma 27.

Lemma 29. If n is a positive integer, then

\sqrt{\frac{5}{2α}} U_{2n−1}(\sqrt{\frac{α^{5}}{8}} ) = 2^{n}α^{n} − (−1)^{n} \frac{β^{n}}{2^{n}} ,
\sqrt{\frac{5}{2β}} U_{2n−1}(\sqrt{\frac{β^{5}}{8}} ) = 2^{n}β^{n} − (−1)^{n} \frac{α^{n}}{2^{n}} .  (30)

Theorem 30. If n is a positive integer and t is an integer, then we have

∑_{k=1}^{n}(−2)^{n−k} \frac{k}{n + k} \binom{n + k}{n − k}L_{5k+t} = \frac{1}{2} (4^{n}F_{t+n+3} − F_{t−n+3}),
∑_{k=1}^{n}(−2)^{n−k} \frac{k}{n + k} \binom{n + k}{n − k}F_{5k+t} = \frac{1}{10} (4^{n}L_{t+n+3} − L_{n−t−3}).

Proof. Set x = \sqrt{β^{5}/8} in (29) and use (30) to obtain

\sqrt{5} ∑_{k=1}^{n} \frac{(−1)^{k−1}}{2^{k}} \frac{k}{n + k} \binom{n + k}{n − k}β^{5k+t} = (−2)^{n−1}β^{n+t+3} + \frac{(−1)^{t}}{2^{n+1}} α^{n−t−3},

from which the results follow. Theorem 31. If n is a non-negative integer and t is an integer, then we have

∑_{k=1}^{n}(−1)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}L_{3k+t} = \begin{cases}\frac{1}{2}L_{t+3}L_{n}, & if n is odd; \\ −\frac{5}{2}F_{t+3}F_{n}, & otherwise; \end{cases}
∑_{k=1}^{n}(−1)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}F_{3k+t} = \begin{cases}\frac{1}{2}F_{t+3}L_{n}, & if n is odd; \\ −\frac{1}{2}L_{t+3}F_{n}, & otherwise. \end{cases}

Proof. Set x = \sqrt{α^{3}}/2 in (29) and use

U_{2n−1}(\frac{\sqrt{α^{3}}}{2} ) = \begin{cases}\sqrt{α^{3}}L_{n}, & if n is odd; \\ \sqrt{5α^{3}}F_{n}, & otherwise; \end{cases}

to obtain

∑_{k=1}^{n}(−1)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}β^{3k+t} = \frac{β^{t+3}}{2} · \begin{cases}L_{n}, & if n is odd; \\ \sqrt{5}F_{n}, & otherwise; \end{cases}

from which the results follow. Theorem 32. If n is a non-negative integer and t is an integer, then we have

∑_{k=1}^{n}(−4)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}L_{k+t} = \begin{cases}−\frac{1}{2}L_{t+1}L_{3n}, & if n is odd; \\ \frac{5}{2}F_{t+1}F_{3n}, & otherwise; \end{cases}
∑_{k=1}^{n}(−4)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}F_{k+t} = \begin{cases}−\frac{1}{2}F_{t+1}L_{3n}, & if n is odd; \\ \frac{1}{2}L_{t+1}F_{3n}, & otherwise. \end{cases}

Proof. Set x = \sqrt{α} in (29) and use

U_{2n−1}(\sqrt{α}) = \begin{cases}\frac{\sqrt{α}}{2}L_{3n}, & if n is odd; \\ \frac{\sqrt{5α}}{2}F_{3n}, & otherwise, \end{cases}

to obtain

∑_{k=1}^{n}(−4)^{k−1} \frac{k}{n + k} \binom{n + k}{n − k}β^{k+t} = −\frac{β^{t+1}}{2} · \begin{cases}L_{3n}, & if n is odd; \\ \sqrt{5}F_{3n}, & otherwise, \end{cases}

from which the results follow. Theorem 33. If n is a non-negative integer and p is a non-zero integer, then we have

∑_{k=1}^{n} (−1)^{n−k} \frac{k}{n + k} \binom{n + k}{n − k} (\frac{2L_{2p}}{\sqrt{5}F_{2p}} )^{2k} = (−1)^{p} \frac{L_{2p}(L_{p}^{4n} − 5^{2n}F_{p}^{4n})}{4(\sqrt{5}F_{2p})^{2n}}.

Proof. Set x = L_{2p}/(\sqrt{5}F_{2p}) in (29) and use

U_{n−1}(\frac{L_{2p}}{\sqrt{5}F_{2p}} ) = \frac{(−1)^{p}(L_{p}^{2n} − 5^{n}F_{p}^{2n})}{4(\sqrt{5}F_{2p})^{n−1}}.  (31)

Theorem 34. If n is a non-negative integer and p and q are non-zero integers, then we have

∑_{k=1}^{n}(−1)^{(p−q)(n−k)} \frac{k}{n + k} \binom{n + k}{n − k} (\frac{F_{p+q}F_{p−q}}{F_{p}F_{q}} )^{2k} = \frac{(F_{p}^{4n} − F_{q}^{4n})F_{p+q}F_{p−q}}{2(F_{p}^{2} + (−1)^{p−q}F_{q}^{2})(F_{q}F_{p})^{2n}} .

Proof. Set x = \frac{(−i)^{p−q+1}F_{p+q}F_{p−q}}{2F_{q}F_{p}} in (29) and use

U_{n−1} (\frac{i^{p−q+1}F_{p+q}F_{p−q}}{2F_{q}F_{p}} ) = \frac{F_{p}^{2n} + (−1)^{p−q}F_{q}^{2n}}{(F_{p}^{2} + (−1)^{p−q}F_{q}^{2})(F_{q}F_{p})^{n−1}} .  (32)

Lemma 35. We have

(−1)^{(q+1)(n+3/2)+⌊q/2⌋+1}α^{p+q/2} U_{2n−1}(\sqrt{\frac{(−1)^{q+1}F_{q}^{2}}{4F_{p}F_{p+q}} α^{2p+q}})

= \frac{F_{p+q}^{2n+1}α^{qn+p} − F_{p}^{2n}F_{p+q}(−β)^{qn−p} + F_{p+q}^{2n}F_{p}α^{qn+p+q} − F_{p}^{2n+1}(−β)^{qn−p−q}}{(F_{p}F_{p+q})^{n−1/2}(F_{p+q}^{2} + F_{p}F_{p+q}L_{q} + (−1)^{q}F_{p}^{2})} , (33)

(−1)^{(q+1)(n+5/2)+⌊q/2⌋+p}β^{p+q/2} U_{2n−1}(\sqrt{\frac{(−1)^{q+1}F_{q}^{2}}{4F_{p}F_{p+q}} β^{2p+q}}) ,

= \frac{F_{p+q}^{2n+1}β^{qn+p} − F_{p}^{2n}F_{p+q}(−α)^{qn−p} + F_{p+q}^{2n}F_{p}β^{qn+p+q} − F_{p}^{2n+1}(−α)^{qn−p−q}}{(F_{p}F_{p+q})^{n−1/2}(F_{p+q}^{2} + F_{p}F_{p+q}L_{q} + (−1)^{q}F_{p}^{2})} ,

(−1)^{(q+1)(n+3/2)+⌊q/2⌋+1}α^{p+q/2} U_{2n−1}(\sqrt{\frac{(−1)^{q+1}5F_{q}^{2}}{4L_{p}L_{p+q}} α^{2p+q}})

= \frac{L_{p+q}^{2n+1}α^{qn+p} − L_{p}^{2n}L_{p+q}(−β)^{qn−p} + L_{p+q}^{2n}L_{p}α^{qn+p+q} − L_{p}^{2n+1}(−β)^{qn−p−q}}{(L_{p}L_{p+q})^{n−1/2}(L_{p+q}^{2} + L_{p}L_{p+q}L_{q} + (−1)^{q}L_{p}^{2})} ,

(−1)^{(q+1)(n+3/2)+p}β^{p+q/2} U_{2n−1}(\sqrt{\frac{(−1)^{q+1}5F_{q}^{2}}{4L_{p}L_{p+q}} β^{2p+q}})

= \frac{L_{p}^{2n+1}β^{qn+p} − L_{p}^{2n}L_{p+q}(−α)^{qn−p} + L_{p+q}^{2n}L_{p}β^{qn+p+q} − L_{p}^{2n+1}(−α)^{qn−p−q}}{(L_{p}L_{p+q})^{n−1/2}(L_{p+q}^{2} + L_{p}L_{p+q}L_{q} + (−1)^{q}L_{p}^{2})} .

Theorem 36. If n is a positive integer, p, q, and t are integers, then we have the following sum evaluations

\begin{matrix}∑_{k=1}^{n} & (−1)^{q(n−k)} \frac{k}{n + k} \binom{n + k}{n − k}F_{p}^{n−k}F_{p+q}^{n−k}F_{q}^{2k−1}L_{(2p+q)k+t} \\ = \frac{F_{p+q}^{2n}(F_{p+q}L_{p+t+qn} + F_{p}L_{q+p+t+qn}) − F_{p}^{2n}(F_{p+q}L_{p+t−qn} + F_{p}L_{q+p+t−qn})}{2(F_{p+q}^{2} + F_{p}F_{p+q}L_{q} + (−1)^{q}F_{p}^{2})}, \end{matrix}  (34)

\begin{matrix}∑_{k=1}^{n} & (−1)^{q(n−k)} \frac{k}{n + k} \binom{n + k}{n − k}F_{p}^{n−k}F_{p+q}^{n−k}F_{q}^{2k−1}F_{(2p+q)k+t} \\ = \frac{F_{p+q}^{2n}(F_{p+q}F_{p+t+qn} + F_{p}F_{q+p+t+qn}) − F_{p}^{2n}(F_{p+q}F_{p+t−qn} + F_{p}F_{p+q+t−qn})}{2(F_{p+q}^{2} + F_{p}F_{p+q}L_{q} + (−1)^{q}F_{p}^{2})}, \end{matrix}  (35)

\begin{matrix}∑_{k=1}^{n} & (−1)^{q(n−k)} \frac{k}{n + k} \binom{n + k}{n − k}L_{p}^{n−k}L_{p+q}^{n−k}F_{q}^{2k−1}5^{k}F_{(2p+q)k+t} \\ = \frac{L_{p+q}^{2n}(L_{p+q}L_{p+t+qn} + L_{p}L_{q+p+t+qn}) − L_{p}^{2n}(L_{p+q}L_{p+t−qn} + L_{p}L_{p+q+t−qn})}{2(L_{p+q}^{2} + L_{p}L_{p+q}L_{q} + (−1)^{q}L_{p}^{2})}, \end{matrix}  (36)

and

\begin{matrix}∑_{k=1}^{n} & (−1)^{q(n−k)} \frac{k}{n + k} \binom{n + k}{n − k}L_{p}^{n−k}L_{p+q}^{n−k}F_{q}^{2k−1}5^{k−1}L_{(2p+q)k+t} \\ = \frac{L_{p+q}^{2n}(L_{p+q}F_{qn+p+t} + L_{p}F_{q+p+t+qn}) − L_{p}^{2n}(L_{p+q}F_{p+t−qn} + L_{p}F_{p+q+t−qn})}{2(L_{p+q}^{2} + L_{p}L_{p+q}L_{q} + (−1)^{q}L_{p}^{2})}. \end{matrix}  (37)

Proof. In (29) set x = \sqrt{\frac{(−1)^{q+1}F_{q}^{2}}{4F_{p}F_{p+q}} α^{2p+q}} and use (33) to obtain

2 ∑_{k=1}^{n} (−1)^{qk} \frac{k}{n + k} \binom{n + k}{n − k}F_{p}^{n−k}F_{p+q}^{n−k}F_{q}^{2k} α^{(2p+q)k+t} = \frac{F_{q}}{2(F_{p+q}^{2} + F_{p}F_{p+q}L_{q} + (−1)^{q}F_{p}^{2})}

×((−1)^{nq}F_{p+q}^{2n+1} 2α^{qn+p+t} − (−1)^{p+t}F_{p}^{2n}F_{p+q} 2β^{qn−p−t}
+ (−1)^{nq}F_{p+q}^{2n}F_{p} 2α^{qn+q+p+t} − (−1)^{p+q+t}F_{p}^{2n+1} 2β^{qn−p−q−t}),

from which (34) and (35) follow. The proof of (36) and (37) is similar; in (29), set x =

\sqrt{\frac{(−1)^{q+1}5F_{q}^{2}}{4L_{p}L_{p+q}} α^{2p+q}}.

4 Some binomial-coefficient weighted Fibonacci and Lucas sums

In this section we present some Fibonacci and Lucas sums having binomial coefficients as weights.

Theorem 37. If n is a positive integer and p is an integer, then we have the following identities

∑_{k=0}^{n}(−1)^{(p−1)(n−k)}\binom{n + k}{n − k}L_{p}^{2k} = \frac{F_{(2n+1)p}}{F_{p}} ,

∑_{k=0}^{n}(−1)^{p(n−k)}\binom{n + k}{n − k}5^{k}F_{p}^{2k} = \frac{L_{(2n+1)p}}{L_{p}} .

Proof. Addition of each identity stated in Theorem 7 and its counterpart in Theorem 28 while making use of L_{r}F_{s} + L_{s}F_{r} = 2F_{r+s} and L_{r}L_{s} + 5F_{r}F_{s} = 2L_{r+s} gives the proof.

The identities stated below in Theorem 38 follow immediately upon addition of each identity in Theorem 25 to the respective corresponding identity in Theorem 36.

Theorem 38. If n is a non-negative integer and p, q and t are integers, then we have

∑_{k=0}^{n}(−1)^{q(n−k)}\binom{n + k}{n − k}F_{p}^{n−k}F_{p+q}^{n−k}F_{q}^{2k}L_{(2p+q)k+t} = \frac{1}{2} (F_{p+q}^{2n}L_{t+qn} + F_{p}^{2n}L_{t−qn})

+ \frac{F_{q}(F_{p+q}^{2n}(F_{p+q}L_{p+t+qn} + F_{p}L_{q+p+t+qn}) − F_{p}^{2n}(F_{p+q}L_{p+t−qn} + F_{p}L_{q+p+t−qn}))}{2(F_{p+q}^{2} + F_{p}F_{p+q}L_{q} + (−1)^{q}F_{p}^{2})},

∑_{k=0}^{n}(−1)^{q(n−k)}\binom{n + k}{n − k}F_{p}^{n−k}F_{p+q}^{n−k}F_{q}^{2k}F_{(2p+q)k+t} = \frac{1}{2} (F_{p+q}^{2n}F_{t+qn} − F_{p}^{2n}F_{t−qn})

+ \frac{F_{q}(F_{p+q}^{2n}(F_{p+q}F_{p+t+qn} + F_{p}F_{q+p+t+qn}) − F_{p}^{2n}(F_{p+q}F_{t+p−qn} + F_{p}F_{q+p+t−qn}))}{2(F_{p+q}^{2} + F_{p}F_{p+q}L_{q} + (−1)^{q}F_{p}^{2})},

∑_{k=0}^{n}(−1)^{q(n−k)}\binom{n + k}{n − k}L_{p}^{n−k}L_{p+q}^{n−k}F_{q}^{2k}5^{k}F_{(2p+q)k+t} = \frac{1}{2} (L_{p+q}^{2n}F_{t+qn} + L_{p}^{2n}F_{t−qn})

+ \frac{F_{q}(L_{p+q}^{2n}(L_{p+q}L_{p+t+qn} + L_{p}L_{q+p+t+qn}) − L_{p}^{2n}(L_{p+q}L_{p+t−qn} + L_{p}L_{q+p+t−qn}))}{2(L_{p+q}^{2} + L_{p}L_{p+q}L_{q} + (−1)^{q}L_{p}^{2})},

∑_{k=0}^{n}(−1)^{q(n−k)}\binom{n + k}{n − k}L_{p}^{n−k}L_{p+q}^{n−k}F_{q}^{2k}5^{k}L_{(2p+q)k+t} = \frac{1}{2} (L_{p+q}^{2n}L_{t+qn} + L_{p}^{2n}L_{t−qn})

+ \frac{5F_{q} (L_{p+q}^{2n+1}(L_{p+q}F_{p+t+qn} + L_{p}F_{q+p+t+qn}) − L_{p}^{2n}(L_{p+q}F_{p+t−qn} + L_{p}F_{q+p+t−qn}))}{2(L_{p+q}^{2} + L_{p}L_{p+q}L_{q} + (−1)^{q}L_{p}^{2})}.

Example 39. If n is a non-negative integer and t is an integer, then we have

∑_{k=0}^{n} (−1)^{n−k}\binom{n + k}{n − k}L_{3k+t} = \begin{cases}L_{t+n} + L_{t+1}L_{n}, & if n is odd; \\ L_{t+n} + 5F_{t+1}F_{n}, & otherwise; \end{cases}

∑_{k=0}^{n} (−1)^{n−k}\binom{n + k}{n − k}F_{3k+t} = \begin{cases}F_{t+n} + F_{t+1}L_{n}, & if n is odd; \\ F_{t+n} + L_{t+1}F_{n}, & otherwise. \end{cases}

In particular,

∑_{k=0}^{n} (−1)^{n−k}\binom{n + k}{n − k}L_{3k} = \begin{cases}2L_{n}, & if n is odd; \\ 2L_{n+1}, & otherwise; \end{cases}

∑_{k=0}^{n} (−1)^{n−k}\binom{n + k}{n − k}F_{3k} = \begin{cases}2F_{n+1}, & if n is odd; \\ 2F_{n}, & otherwise; \end{cases}

∑_{k=0}^{n} (−1)^{n−k}\binom{n + k}{n − k}L_{3k−1} = \begin{cases}L_{n+2}, & if n is odd; \\ L_{n−1}, & otherwise; \end{cases}

∑_{k=0}^{n} (−1)^{n−k}\binom{n + k}{n − k}F_{3k−1} = \begin{cases}F_{n−1}, & if n is odd; \\ F_{n+2}, & otherwise. \end{cases}

Example 40. If n is a non-negative integer and t is an integer, then we have

∑_{k=0}^{n} (−2)^{n−k}\binom{n + k}{n − k}L_{5k+t} = 2^{2n+1}F_{t+n+1} − F_{t−n},
∑_{k=0}^{n} (−2)^{n−k}\binom{n + k}{n − k}F_{5k+t} = \frac{1}{5} (2^{2n+1}L_{t+n+1} − L_{t−n}).

The identities in this section and many similar results can be obtained directly from Lemma 41 which is a consequence of (6), (10), and (29). Lemma 41. If n is a non-negative integer and x is a complex variable, then

∑_{k=0}^{n} (−4)^{k}\binom{n + k}{n − k}x^{2k} = (−1)^{n}U_{2n}(x).  (38)

Theorem 42. If n is a non-negative integer and p is a non-zero integer, then we have

∑_{k=0}^{n} (−1)^{n−k}\binom{n + k}{n − k} (\frac{2L_{2p}}{\sqrt{5}F_{p}} )^{2k} = (−1)^{p} \frac{L_{p}^{4n+2} − 5^{2n+1}F_{p}^{4n+2}}{4 · 5^{n}F_{2p}^{2n}} .

Proof. Set x = L_{2p}/(\sqrt{5}F_{2p}) in (38) and use (31). Theorem 43. If n is a non-negative integer and p and q are non-zero integers, then we have

∑_{k=0}^{n} (−1)^{(p−q)(n−k)}\binom{n + k}{n − k} (\frac{F_{p+q}F_{p−q}}{F_{p}F_{q}} )^{2k} = \frac{F_{p}^{4n+2} + (−1)^{p−q}F_{q}^{4n+2}}{(F_{q}F_{p})^{2n}(F_{p}^{2} + (−1)^{p−q}F_{q}^{2})}.

Proof. Set x = \frac{(−i)^{p−q+1}F_{p+q}F_{p−q}}{2F_{q}F_{p}} in (38) and use (32).

5 Related combinatorial sums

This section does not deal with Fibonacci numbers but is inspired by a recent paper by Chu and Guo [6]. In this paper the authors study combinatorial sums of the form ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n+λ \atop 2k+δ} & ) \\ ( & {n \atop k} & ) \end{matrix}, where λ and n are nonnegative integers, δ ∈ {0; 1}. We show how these sums are related to the sums studied in this paper via Theorem 1. Theorem 44. For 0 < n + m ≤ 2n + 1, we have the relation

∑_{k=0}^{n+m} \frac{(−2)^{k}}{(n + m + k)(n − m + k + 2)} \begin{matrix}( & {n+m+k \atop n+m−k} & ) \\ ( & {n−m+k+1 \atop k} & ) \end{matrix} = \frac{1}{2(n + 1)(n + m)} ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n+m \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix}.

In particular,

∑_{k=0}^{n} \frac{(−2)^{k}}{(n + k)(n + k + 1)(n + k + 2)} \begin{matrix}( & {n+k \atop n−k} & ) \\ ( & {n+k \atop k} & ) \end{matrix} = \frac{1}{2n(n + 1)^{2}} ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix},
∑_{k=0}^{2n} \frac{(−2)^{k}}{(2n + k)(k + 1)(k + 2)}\binom{2n + k}{2n − k} = \frac{1}{4n(n + 1)} ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {2n \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix}.

Proof. From Theorem 1 upon replacing n by n + m we get

∑_{k=0}^{n+m}(−2)^{k} \frac{n + m}{n + m + k} \binom{n + m + k}{n + m − k}(1 − x)^{k} = T_{n+m}(x).

Multiplying through by x^{n−m+1} and integrating from 0 to 1 results in

∑_{k=0}^{n+m}(−2)^{k} \frac{n + m}{n + m + k} \binom{n + m + k}{n + m − k} ∫_{0}^{1} x^{n−m+1}(1 − x)^{k}dx = ∫_{0}^{1} x^{n−m+1}T_{n+m}(x)dx.

The left-hand side evaluated using the beta function B(a, b) given by

B(a, b) = ∫_{0}^{1} x^{a−1}(1 − x)^{b−1}dx,

and equals

∑_{k=0}^{n+m}(−2)^{k} \frac{n + m}{n + m + k} \binom{n + m + k}{n + m − k}B(n − m + 2, k + 1)

with

B(n − m + 2, k + 1) = \frac{1}{n − m + k + 2}\binom{n − m + k + 1}{k}^{−1}.

The right-hand side is evaluated similarly using the representation from (2). We have

∫_{0}^{1} x^{n−m+1}T_{n+m}(x)dx = ∑_{k=0}^{⌊(n+m)/2⌋}(−1)^{k}\binom{n + m}{2k} ∫_{0}^{1} x^{2n−2k+1}(1 − x^{2})^{k}dx

= \frac{1}{2} ∑_{k=0}^{⌊(n+m)/2⌋}(−1)^{k}\binom{n + m}{2k} ∫_{0}^{1} z^{n−k}(1 − z)^{k}dz (z = x^{2})

= \frac{1}{2} ∑_{k=0}^{⌊(n+m)/2⌋}(−1)^{k}\binom{n + m}{2k}B(n − k + 1, k + 1)

= \frac{1}{2(n + 1)} ∑_{k=0}^{⌊(n+m)/2⌋}(−1)^{k} \begin{matrix}( & {n+m \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix}.

The final expression follows as \binom{n}{k} = 0 if k > n.

Keeping in mind that

\frac{n}{n + k} \binom{n + k}{n − k} = \begin{matrix}\binom{n+k−1}{k} & \binom{n}{k} \\ ( & {2k \atop k} & ) \end{matrix},

Theorem 44 is actually an identity containing four binomial coefficients

∑_{k=0}^{n+m} \frac{(−2)^{k}}{n − m + k + 2} \begin{matrix}( & {n+m+k−1 \atop k} & ) \\ ( & {n−m+k+1 \atop k} & ) \end{matrix} \begin{matrix}( & {n+m \atop k} & ) \\ ( & {2k \atop k} & ) \end{matrix} = \frac{1}{2(n + 1)} ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n+m \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix}.

The special case m = 0 gives the combinatorial relation valid for all n ≥ 1

∑_{k=0}^{n} \frac{(−2)^{k}}{(n + k)(n + k + 1)(n + k + 2)} \begin{matrix}( & {n \atop k} & ) \\ ( & {2k \atop k} & ) \end{matrix} = \frac{1}{2n(n + 1)} − \frac{1}{4(n + 1)^{2}} ∑_{k=0}^{n} \frac{1}{\binom{n}{k}},

as Chu and Guo have shown [6, Prop. 5] that

∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix} = n + 1 − \frac{n}{2} ∑_{k=0}^{n} \frac{1}{\binom{n}{k}}.

We continue with some more examples based on Chu and Guo’s results. Example 45. When m = 1, 2, 3, 4, then we can use the following evaluation from [6]

∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n+1 \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix} = 2 − ∑_{k=0}^{n} \frac{1}{\binom{n}{k}},
∑_{k=0}^{n}(−1)^{k+1} \begin{matrix}( & {n+2 \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix} = \frac{2}{n},
∑_{k=0}^{n}(−1)^{k+1} \begin{matrix}( & {n+3 \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix} = \frac{2(n − 3)}{(n − 1)n},
∑_{k=0}^{n}(−1)^{k+1} \begin{matrix}( & {n+4 \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix} = \frac{2(n^{2} − 7n + 16)}{(n − 2)(n − 1)n},

to get the following summation formulas:

∑_{k=0}^{n+1} \frac{(−2)^{k}}{(n + 1 + k)^{2}} \begin{matrix}( & {n+1+k \atop n+1−k} & ) \\ ( & {n+k \atop k} & ) \end{matrix} = \frac{1}{2(n + 1)^{2}} (2 − ∑_{k=0}^{n} \frac{1}{\binom{n}{k}}),  n ≥ 0,
∑_{k=0}^{n+2} \frac{(−2)^{k}}{(n + k)(n + 2 + k)} \begin{matrix}( & {n+2+k \atop n+2−k} & ) \\ ( & {n−1+k \atop k} & ) \end{matrix} = \frac{−1}{n(n + 1)(n + 2)},  n ≥ 1,
∑_{k=0}^{n+3} \frac{(−2)^{k}}{(n − 1 + k)(n + 3 + k)} \begin{matrix}( & {n+3+k \atop n+3−k} & ) \\ ( & {n−2+k \atop k} & ) \end{matrix} = \frac{3 − n}{(n − 1)n(n + 1)(n + 3)},  n ≥ 2,
∑_{k=0}^{n+4} \frac{(−2)^{k}}{(n − 2 + k)(n + 4 + k)} \begin{matrix}( & {n+4+k \atop n+4−k} & ) \\ ( & {n−3+k \atop k} & ) \end{matrix} = \frac{−n^{2} + 7n − 16}{(n − 2)(n − 1)n(n + 1)(n + 4)},  n ≥ 3.

Working with the second part of Theorem 1, applying the same arguments but using the elementary integral

∫_{0}^{1} x^{n−m+1}(1 + x)^{k}dx = ∑_{j=0}^{k} \frac{\binom{k}{j}}{n − m + 2 + j}

yields the next theorem involving a double sum.

Theorem 46. For m − 1 ≤ n, we have the relation

∑_{j=0}^{n+m} ∑_{k=0}^{n+m−j} \frac{(−2)^{k+j}}{(n + m + k + j)(n − m + j + 2)}\binom{n + m + k + j}{n + m − k − j}\binom{k + j}{j}

= \frac{(−1)^{n+m}}{2(n + 1)(n + m)} ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n+m \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix}.

In particular,

∑_{j=0}^{n} ∑_{k=0}^{n−j} \frac{(−2)^{k+j}}{(n + k + j)(n + j + 2)}\binom{n + k + j}{n − k − j}\binom{k + j}{j} = \frac{(−1)^{n}}{2n(n + 1)} ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {n \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix},

∑_{j=0}^{2n} ∑_{k=0}^{2n−j} \frac{(−2)^{k+j}}{(2n + k + j)(j + 2)}\binom{2n + k + j}{2n − k − j}\binom{k + j}{j} = \frac{1}{4n(n + 1)} ∑_{k=0}^{n}(−1)^{k} \begin{matrix}( & {2n \atop 2k} & ) \\ ( & {n \atop k} & ) \end{matrix}.

Similar double sums involving ratios of binomial coefficients were studied recently by Stenlund and Wan [20].

The analogous identities involving \binom{n+m}{2k+1} should follow from (28) in conjunction with (3).

6 Conclusion

In this paper, we have derived a huge number of new combinatorial identities involving the binomial coefficients

\frac{n}{n + k} \binom{n + k}{n − k} and \frac{k}{n + k} \binom{n + k}{n − k},

and Fibonacci and Lucas numbers. The basic idea behind the proofs is to relate the sums to certain relations involving the Chebyshev polynomials T_{n}(x) and U_{n}(x), respectively, evaluated at specific arguments. Theorems 25, 36, and 38 offer additional appealing identities not discussed here. In addition, we have shown a connection between the sums studied in this paper and the combinatorial sums studied recently by Chu and Guo [6].

Extensions of the results presented in the first part of the paper to Fibonacci and Lucas polynomials should be possible without many efforts.

7 Acknowledgments

The authors thank the referee for their very helpful comments that improved the exposition of the paper.

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2020 Mathematics Subject Classification: Primary 11B39; Secondary 11B37. Keywords: Fibonacci number, Lucas number, Chebyshev polynomial, binomial coefficient.

(Concerned with sequences \underline{A000032}, \underline{A000045}, and \underline{A138573}.)

Received August 5 2023; revised version received December 5 2023. Published in Journal of Integer Sequences, December 11 2023.

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