Fibonacci Sums Modulo 5

Adegoke, Kunle; Frontczak, Robert; Goy, Taras

doi:10.2478/amsil-2024-0015

Full Article

1.

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: $\begin{array}{l} F_{n} = F_{n - 1} + F_{n - 2}, & F_{0} = 0, & F_{1} = 1, \\ L_{n} = L_{n - 1} + L_{n - 2}, & L_{0} = 2, & L_{1} = 1. \end{array}$ \matrix{ {{F_n} = {F_{n - 1}} + {F_{n - 2}},} \hfill & {{F_0} = 0,} \hfill & {{F_1} = 1,} \hfill \cr {{L_n} = {L_{n - 1}} + {L_{n - 2}},} \hfill & {{L_0} = 2,} \hfill & {{L_1} = 1.} \hfill \cr } For negative subscripts we have F_−n = (−1)ⁿ⁻¹F_n and L_−n = (−1)ⁿL_n.

Throughout this paper, we denote the golden ratio by $α = \frac{1 + \sqrt{5}}{2}$ \alpha = {{1 + \sqrt 5 } \over 2} and write $β = \frac{1 - \sqrt{5}}{2} = - \frac{1}{α}$ \beta = {{1 - \sqrt 5 } \over 2} = - {1 \over \alpha } . The Fibonacci and Lucas numbers possess the explicit formulas (Binet formulas) $\begin{array}{l} F_{n} = \frac{α^{n} - β^{n}}{α - β}, & L_{n} = α^{n} + β^{n}, & n \in ℤ . \end{array}$ \begin{array}{*{35}{l}}{{F}_{n}}=\frac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta }, & {{L}_{n}}={{\alpha }^{n}}+{{\beta }^{n}}, & n\in \mathbb{Z}. \\ \end{array}\

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, 5, 2, 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 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 [4].

2.

Fibonacci sums modulo 5 from the sin nx and cos nx expansions

We begin with a known lemma [9, 1.331(3) and 1.331(1)].

Lemma 2.1.

If n is a positive integer, then (2.1) $\sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1} n}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) 2^{n - 2 k - 1} {cos}^{n - 2 k} x = 2^{n - 1} {cos}^{n} x - cos nx,$ \sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}n} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){2^{n - 2k - 1}}{{\cos }^{n - 2k}}x = {2^{n - 1}}\;{{\cos }^n}\;x - \cos \;nx} , (2.2) $\sum_{k = 0}^{(n - 1) / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k - 1 \\ k \end{matrix}) 2^{n - 2 k - 1} {cos}^{n - 2 k - 1} x = \frac{sin nx}{sin x} .$ \sum\limits_{k = 0}^{\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor } {{{( - 1)}^k}\left( {\matrix{ {n - k - 1} \cr k \cr } } \right){2^{n - 2k - 1}}{{\cos }^{n - 2k - 1}}x = {{\sin nx} \over {\sin x}}.}

Lemma 2.2.

If n is an integer, then (2.3) $cos (\frac{n π}{5}) = \begin{array}{l} {(- 1)}^{n}, & if n \equiv 0 (mod 5), \\ {(- 1)}^{n - 1} α / 2, & if n \equiv 1 or 4 (mod 5), \\ {(- 1)}^{n - 1} β / 2, & if n \equiv 2 or 3 (mod 5), \end{array}$ \cos \left( {{{n\pi } \over 5}} \right) = \left\{ {\matrix{ {{{( - 1)}^n},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{n - 1}}\alpha /2,} \hfill & {if\;n \equiv 1\;\;or\;\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{n - 1}}\beta /2,} \hfill & {if\;n \equiv 2\;\;or\;\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right. (2.4) $cos (\frac{2 n π}{5}) = \begin{array}{l} 1, & if n \equiv 0 (mod 5), \\ - β / 2, & if n \equiv 1 or 4 (mod 5), \\ - α / 2, & if n \equiv 2 or 3 (mod 5) . \end{array}$ \cos \left( {{{2n\pi } \over 5}} \right) = \left\{ {\matrix{ {1,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - \beta /2,} \hfill & {if\;n \equiv 1\;\;or\;\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - \alpha /2,} \hfill & {if\;n \equiv 2\;\;or\;\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

Proof

Relations stated in (2.3) can be proved easily by elementary methods. For instance, they follow by applying the addition theorem for the cosine function $cos (a + b) = cos a cos b - sin a sin b$ \cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b combined with the special values $\begin{matrix} cos (\frac{π}{5}) = \frac{α}{2}, & cos (\frac{2 π}{5}) = - \frac{β}{2}, & cos (\frac{3 π}{5}) = \frac{β}{2}, & cos (\frac{4 π}{5}) = - \frac{α}{2} \end{matrix} .$ \matrix{ {\cos \left( {{\pi \over 5}} \right) = {\alpha \over 2},} & {\cos \left( {{{2\pi } \over 5}} \right) = - {\beta \over 2},} & {\cos \left( {{{3\pi } \over 5}} \right) = {\beta \over 2},} & {\cos \left( {{{4\pi } \over 5}} \right) = - {\alpha \over 2}} \cr } .

Relations stated in (2.4) follow directly from (2.3).

In our first main results we state Lucas (Fibonacci) identities involving binomial coefficient and additional parameter.

Theorem 2.3.

If n is a positive integer and t is any integer, then $\begin{array}{l} n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) L_{n - 2 k + t} = \begin{array}{l} L_{n + t} - {(- 1)}^{n} 2 L_{t}, & if n \equiv 0 (mod 5), \\ L_{n + t} + {(- 1)}^{n} L_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ L_{n + t} - {(- 1)}^{n} L_{t - 1}, & if n \equiv 2 or 3 (mod 5), \end{array} \\ n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) F_{n - 2 k + t} = \begin{array}{l} F_{n + t} - {(- 1)}^{n} 2 F_{t}, & if n \equiv 0 (mod 5), \\ F_{n + t} + {(- 1)}^{n} F_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ F_{n + t} - {(- 1)}^{n} F_{t - 1}, & if n \equiv 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {n\sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){L_{n - 2k + t}}} = \left\{ {\matrix{ {{L_{n + t}} - {{( - 1)}^n}2{L_t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{n + t}} + {{( - 1)}^n}{L_{t + 1}},} \hfill & {if\;n \equiv 1\;\;or\;\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{n + t}} - {{( - 1)}^n}{L_{t - 1}},} \hfill & {if\;n \equiv 2\;\;or\;\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {n\sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){F_{n - 2k + t}}} = \left\{ {\matrix{ {{F_{n + t}} - {{( - 1)}^n}2{F_t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{n + t}} + {{( - 1)}^n}{F_{t + 1}},} \hfill & {if\;n \equiv 1\;\;or\;\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{n + t}} - {{( - 1)}^n}{F_{t - 1}},} \hfill & {if\;n \equiv 2\;\;or\;\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

Set x = π/5 in (2.1) and use (2.3) and the fact that (2.5) $\begin{array}{l} 2 α^{r} = L_{r} + F_{r} \sqrt{5}, & 2 β^{r} = L_{r} - F_{r} \sqrt{5} \end{array}$ \matrix{ {2{\alpha ^r} = {L_r} + {F_r}\sqrt 5 ,} \hfill & {2{\beta ^r} = {L_r} - {F_r}\sqrt 5 } \hfill \cr } for any integer r.

We proceed with some corollaries.

Corollary 2.4.

If n is a positive integer, then $\begin{array}{l} n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) F_{2 k} = \begin{array}{l} - 2 F_{n}, & if n \equiv 0 (mod 5), \\ - F_{n - 1}, & if n \equiv 1 or 4 (mod 5), \\ F_{n + 1}, & if n \equiv 2 or 3 (mod 5), \end{array} \\ n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) F_{n - 2 k + δ} = F_{n + δ}, \end{array}$ \matrix{ {n\sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){F_{2k}}} = \left\{ {\matrix{ { - 2{F_n},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {F_{n - 1}},} \hfill & {if\;n \equiv 1\;\;or\;\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{n + 1}},} \hfill & {if\;n \equiv 2\;\;or\;\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {n\sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){F_{n - 2k + \delta }} = {F_{n + \delta }},} } \hfill \cr } where $δ = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ - 1, & if n \equiv 1 or 4 (mod 5), \\ 1, & if n \equiv 2 or 3 (mod 5) . \end{array}$ \delta = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - 1,} \hfill & {if\;n \equiv 1\;\;or\;\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {if\;n \equiv 2\;\;or\;\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

Corollary 2.5.

If n is a positive integer, then $\begin{array}{l} n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) L_{n - 2 k + 1} = \begin{array}{l} L_{n - 1} - {(- 1)}^{n} 3, & if n \equiv 2 or 3 (mod 5), \\ L_{n - 1} + {(- 1)}^{n} 2, & otherwise, \end{array} \\ n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) L_{n - 2 k + 1} = \begin{array}{l} L_{n + 1} + {(- 1)}^{n} 3, & if n \equiv 1 or 4 (mod 5), \\ L_{n + 1} - {(- 1)}^{n} 2, & otherwise, \end{array} \\ n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) L_{n - 2 k} = \begin{array}{l} L_{n} - {(- 1)}^{n} 4, & if n \equiv 0 (mod 5), \\ L_{n} + {(- 1)}^{n}, & otherwise . \end{array} \end{array}$ \matrix{ {n\sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){L_{n - 2k + 1}}} = \left\{ {\matrix{ {{L_{n - 1}} - {{( - 1)}^n}3,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{n - 1}} + {{( - 1)}^n}2,} \hfill & {otherwise,} \hfill \cr } } \right.} \hfill \cr {n\sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){L_{n - 2k + 1}}} = \left\{ {\matrix{ {{L_{n + 1}} + {{( - 1)}^n}3,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{n + 1}} - {{( - 1)}^n}2,} \hfill & {otherwise,} \hfill \cr } } \right.} \hfill \cr {n\sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){L_{n - 2k}}} = \left\{ {\matrix{ {{L_n} - {{( - 1)}^n}4,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_n} + {{( - 1)}^n},} \hfill & {otherwise.} \hfill \cr } } \right.} \hfill \cr }

Lemma 2.6.

If n is an integer, then (2.6) $\frac{sin (n π / 5)}{sin (π / 5)} = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ {(- 1)}^{n / 5⌋}, & if n \equiv 1 or 4 (mod 5), \\ {(- 1)}^{n / 5⌋} α, & if n \equiv 2 or 3 (mod 5), \end{array}$ {{\sin \left( {n\pi /5} \right)} \over {\sin \left( {\pi /5} \right)}} = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }},} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{{\left\lfloor {n/5} \right\rfloor }}}\alpha ,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right. (2.7) $\frac{sin (3 n π / 5)}{sin (3 π / 5)} = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ {(- 1)}^{n / 5⌋}, & if n \equiv 1 or 4 (mod 5), \\ {(- 1)}^{n / 5⌋} β, & if n \equiv 2 or 3 (mod 5) . \end{array}$ {{\sin \left( {3n\pi /5} \right)} \over {\sin \left( {3\pi /5} \right)}} = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }},} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }}\beta ,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

From Lemma 2.6 we can deduce the following Lucas and Fibonacci binomial identities modulo 5.

Theorem 2.7.

If n is a positive integer and t is any integer, then (2.8) $\sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k \\ k \end{matrix}) L_{n - 2 k + t} = \begin{array}{l} {(- 1)}^{(n + 1) / 5⌋} L_{t}, & if n \equiv 0 or 3 (mod 5), \\ {(- 1)}^{(n + 1) / 5⌋} L_{t + 1}, & if n \equiv 1 or 2 (mod 5), \\ 0, & if n \equiv 4 (mod 5), \end{array}$ \sum\limits_{k = 0}^{^{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}\left( {\matrix{{n - k} \cr k \cr } } \right){L_{n - 2k + t}}} = \left\{ {\matrix{{{{( - 1)}^{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor }}{L_t},} \hfill & {if\;n \equiv 0\;or\;3\;\;\;\;\left( {{\rm mod} \;5} \right),} \hfill \cr {{{( - 1)}^{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor }}{L_{t + 1}},} \hfill & {if\;n \equiv 1\;or\;2\;\;\;\;\left( {{\rm mod} \;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 4\;\;\;\;\left( {{\rm mod} \;5} \right),} \hfill \cr } } \right. (2.9) $\sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k \\ k \end{matrix}) F_{n - 2 k + t} = \begin{array}{l} {(- 1)}^{(n + 1) / 5⌋} F_{t}, & if n \equiv 0 or 3 (mod 5), \\ {(- 1)}^{n / 5⌋} F_{t + 1}, & if n \equiv 1 or 2 (mod 5), \\ 0, & if n \equiv 4 (mod 5) . \end{array}$ \sum\limits_{k = 0}^{^{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}\left( {\matrix{{n - k} \cr k \cr } } \right){F_{n - 2k + t}}} = \left\{ {\matrix{{{{( - 1)}^{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor }}{F_t},} \hfill & {if\;n \equiv 0\;or\;3\;\;\;\;\left( {{\rm mod} \;5} \right),} \hfill \cr {{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }}{F_{t + 1}},} \hfill & {if\;n \equiv 1\;or\;2\;\;\;\;\left( {{\rm mod} \;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 4\;\;\;\;\left( {{\rm mod} \;5} \right).} \hfill \cr } } \right.

Proof

Set x = π/5 in (2.2), use (2.6), (2.5) and simplify.

A variant of the Lucas and Fibonacci sums with even subscripts is stated as the next corollary.

Corollary 2.8.

If n is a positive integer, then $\begin{array}{l} \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} (\begin{matrix} n - k \\ k \end{matrix}) L_{2 k} = \begin{array}{l} {(- 1)}^{(n + 1) / 5⌋} L_{n}, & if n \equiv 0 or 3 (mod 5), \\ {(- 1)}^{(n + 1) / 5⌋ + 1} L_{n - 1}, & if n \equiv 1 or 2 (mod 5), \\ 0, & if n \equiv 4 (mod 5), \end{array} \\ \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} (\begin{matrix} n - k \\ k \end{matrix}) F_{2 k} = \begin{array}{l} {(- 1)}^{(n + 1) / 5⌋} F_{n}, & if n \equiv 0 or 3 (mod 5), \\ {(- 1)}^{(n + 1) / 5⌋ + 1} F_{n - 1}, & if n \equiv 1 or 2 (mod 5), \\ 0, & if n \equiv 4 (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){L_{2k}} = } \left\{ {\matrix{ {{{( - 1)}^{{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor }}}{L_n},} \hfill & {if\;n \equiv 0\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor + 1}}}{L_{n - 1}},} \hfill & {if\;n \equiv 1\;or\;2\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){F_{2k}} = } \left\{ {\matrix{ {{{( - 1)}^{{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor }}}{F_n},} \hfill & {if\;n \equiv 0\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor + 1}}}{F_{n - 1}},} \hfill & {if\;n \equiv 1\;or\;2\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 4\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Corollary 2.9.

If n is a positive integer, then $\begin{array}{l} \sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k \\ k \end{matrix}) F_{n - 2 k + 1} = \begin{array}{l} 0, & if n \equiv 4 (mod 5), \\ {(- 1)}^{(n + 1) / 5⌋}, & otherwise, \end{array} \\ \sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k \\ k \end{matrix}) F_{n - 2 k - δ} = 0, \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}\left( {\matrix{ {n - k} \cr k \cr } } \right){F_{n - 2k + 1}}} = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{{\left\lfloor {\left( {n + 1} \right)/5} \right\rfloor }}},} \hfill & {otherwise,} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}\left( {\matrix{ {n - k} \cr k \cr } } \right){F_{n - 2k - \delta }}} = 0,} \hfill \cr } where $δ = \begin{array}{l} 0, & if n \equiv 0 or 3 & (mod 5), \\ 1, & if n \equiv 1 or 2 & (mod 5) . \end{array}$ \delta = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;or\;3} \hfill & {\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {\;if\;n \equiv 1\;or\;2} \hfill & {\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

3.

Fibonacci sums modulo 5 from Waring formulas

This section is based on utilizing the following trigonometric identities with the use of Waring formulas.

Lemma 3.1.

(3.1) $\sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) 2^{n - 2 k - 1} {cos}^{n - 2 k} x = cos nx,$ \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){2^{n - 2k - 1}}{{\cos }^{n - 2k}}x} = \cos nx, (3.2) $\sum_{k = 0}^{(n - 1) / 2} {(- 1)}^{(n - 1) / 2 - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) 2^{n - 2 k - 1} {sin}^{n - 2 k} x = sin nx, n odd,$ \sum\limits_{k = 0}^{{\left( {n - 1} \right)/2}} {{{( - 1)}^{\left( {n - 1} \right)/2 - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){2^{n - 2k - 1}}{{\sin }^{n - 2k}}} x = \sin nx,\;\;\;n\;odd, (3.3) $\sum_{k = 0}^{n / 2} {(- 1)}^{n / 2 - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) 2^{n - 2 k - 1} {sin}^{n - 2 k} x = cos nx, n even .$ \sum\limits_{k = 0}^{{n/2}} {{{( - 1)}^{n/2 - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){2^{n - 2k - 1}}{\rm{si}}{{\rm{n}}^{n - 2k}}x = {{\cos}}\,nx,} \;\;\;n\;even.

Proof

Consider the Waring formula $\sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) {(x_{1} + x_{2})}^{n - 2 k} {(x_{1} x_{2})}^{k} = x_{1}^{n} + x_{2}^{n} .$ \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){{({x_1} + {x_2})}^{n - 2k}}{{({x_1}{x_2})}^k} = x_1^n + x_2^n} . Let i be the imaginary unit. The choice x₁ = e^ix/2, x₂ = e^−ix/2 produces (3.1), while the choice x₁ = e^ix/(2i), x₂ = −e^−ix/(2i) gives x₁ + x₂ = sin x, x₁x₂ = 1/4, and $x_{1}^{n} + x_{2}^{n} = \begin{array}{l} {(- 1)}^{(n - 1) / 2} 2^{1 - n} sin nx, & if n is odd, \\ {(- 1)}^{n / 2} 2^{1 - n} cos nx, & if n is even, \end{array}$ x_1^n + x_2^n = \left\{ {\matrix{ {{{( - 1)}^{\left( {n - 1} \right)/2}}{2^{1 - n}}\;\sin nx,} \hfill & {{\rm{if}}\;n\;{\rm{is}}\;{\rm{odd}},} \hfill \cr {{{( - 1)}^{n/2}}{2^{1 - n}}\cos nx,} \hfill & {{\rm{if}}\;n\;{\rm{is}}\;{\rm{even}},} \hfill \cr } } \right. and hence (3.2) and (3.3).

Lemma 3.2.

If n is a positive integer, then (3.4) $\begin{array}{l} \sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k \\ k \end{matrix}) 2^{n - 2 k} {cos}^{n - 2 k} x = \frac{sin ((n + 1) x)}{sin x}, \\ \sum_{k = 0}^{(n - 1) / 2} {(- 1)}^{(n - 1) / 2 - k} (\begin{matrix} n - k \\ k \end{matrix}) 2^{n - 2 k} {sin}^{n - 2 k} x = \frac{sin ((n + 1) x)}{cos x}, n odd, \\ \sum_{k = 0}^{n / 2} {(- 1)}^{n / 2 - k} (\begin{matrix} n - k \\ k \end{matrix}) 2^{n - 2 k} {sin}^{n - 2 k} x = \frac{cos ((n + 1) x)}{cos x}, n even . \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}\left( {\matrix{ {n - k} \cr k \cr } } \right){2^{n - 2k}}{{\cos }^{n - 2k}}x} = {{\sin \left( {\left( {n + 1} \right)x} \right)} \over {\sin x}},} \hfill \cr {\sum\limits_{k = 0}^{{\left( {n - 1} \right)/2}} {{{( - 1)}^{\left( {n - 1} \right)/2 - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){2^{n - 2k}}{{\sin }^{n - 2k}}x} = {{\sin \left( {\left( {n + 1} \right)x} \right)} \over {\cos x}},\;\;\;n\;odd,} \hfill \cr {\sum\limits_{k = 0}^{{n/2}} {{{( - 1)}^{n/2 - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){2^{n - 2k}}{{\sin }^{n - 2k}}x} = {{\cos \left( {\left( {n + 1} \right)x} \right)} \over {\cos x}},\;\;\;n\;even.} \hfill \cr }

Proof

Similar to the proof of Lemma 3.1. We use the dual to the Waring formula $\sum_{k = 0}^{n / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k \\ k \end{matrix}) {(x_{1} + x_{2})}^{n - 2 k} {(x_{1} x_{2})}^{k} = \frac{x_{1}^{n + 1} - x_{2}^{n + 1}}{x_{1} - x_{2}} .$ \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^k}\left( {\matrix{ {n - k} \cr k \cr } } \right){{({x_1} + {x_2})}^{n - 2k}}{{({x_1}{x_2})}^k}} = {{x_1^{n + 1} - x_2^{n + 1}} \over {{x_1} - {x_2}}}.

Theorem 3.3.

If n is a positive integer and t is any integer, then $\begin{array}{l} \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) F_{n - 2 k + t} = \begin{array}{l} 2 F_{t}, & if n \equiv 0 (mod 5), \\ - F_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ F_{t - 1}, & if n \equiv 2 or 3 (mod 5), \end{array} \\ \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) L_{n - 2 k + t} = \begin{array}{l} 2 L_{t}, & if n \equiv 0 (mod 5), \\ - L_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ L_{t - 1}, & if n \equiv 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){F_{n - 2k + t}}} = \left\{ {\matrix{ {2{F_t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {F_{t + 1}},} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{t - 1}},} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){L_{n - 2k + t}}} = \left\{ {\matrix{ {2{L_t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {L_{t + 1}},} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{t - 1}},} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

We apply equation (3.1). Inserting x = π/5 and x = 3π/5, respectively, and keeping in mind the trigonometric identity cos 3x = 4 cos³ x−3 cos x we end with $\sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) α^{n - 2 k + t} = \begin{array}{l} 2 α^{t}, & if n \equiv 0 (mod 5), \\ - α^{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ α^{t - 1}, & if n \equiv 2 or 3 (mod 5), \end{array}$ \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){\alpha ^{n - 2k + t}}} = \left\{ {\matrix{ {2{\alpha ^t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {\alpha ^{t + 1}},} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{\alpha ^{t - 1}},} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right. and $\sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) β^{n - 2 k + t} = \begin{array}{l} 2 β^{t}, & if n \equiv 0 (mod 5), \\ - β^{t} (α^{3} - 3 α), & if n \equiv 1 or 4 (mod 5), \\ - β^{t} (β^{3} - 3 β), & if n \equiv 2 or 3 (mod 5) . \end{array}$ \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){\beta ^{n - 2k + t}}} = \left\{ {\matrix{ {2{\beta ^t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {\beta ^t}\left( {{\alpha ^3} - 3\alpha } \right),} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {\beta ^t}\left( {{\beta ^3} - 3\beta } \right),} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right. To complete the proof simplify the terms in brackets and combine according the Binet formulas.

From Theorem 3.3 we can immediately obtain the following finite binomial sums.

Corollary 3.4.

If n is a positive integer, then $\begin{array}{l} \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) F_{n - 2 k} = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ - 1, & if n \equiv 1 or 4 (mod 5), \\ 1, & if n \equiv 2 or 3 (mod 5), \end{array} \\ \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) L_{n - 2 k} = \begin{array}{l} 4, & if n \equiv 0 (mod 5), \\ - 1, & otherwise, \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){F_{n - 2k}}} = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - 1,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){L_{n - 2k}}} = \left\{ {\matrix{ {4,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - 1,} \hfill & {otherwise,} \hfill \cr } } \right.} \hfill \cr } and $\begin{array}{l} \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) F_{n + 1 - 2 k} = \begin{array}{l} 2, & if n \equiv 0 (mod 5), \\ - 1, & if n \equiv 1 or 4 (mod 5), \\ 0, & if n \equiv 2 or 3 (mod 5), \end{array} \\ \sum_{k = 0}^{n / 2⌋} {(- 1)}^{n - k} \frac{n}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) L_{n + 1 - 2 k} = \begin{array}{l} 2, & if n \equiv 0, 2 or 3 (mod 5), \\ - 3, & otherwise . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){F_{n + 1 - 2k}}} = \left\{ {\matrix{ {2,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - 1,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{( - 1)}^{n - k}}{n \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){L_{n + 1 - 2k}}} = \left\{ {\matrix{ {2,} \hfill & {if\;n \equiv 0,2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - 3,} \hfill & {otherwise.} \hfill \cr } } \right.} \hfill \cr }

Remark

Identities (2.8) and (2.9) in Theorem 2.7 can also be obtained straightforwardly by evaluating the trigonometric identity (3.4) at x = π/5 and x = 3π/5, respectively, while using (2.6) and (2.7).

4.

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] $\begin{array}{l} T_{n + 1} (x) = 2 x T_{n} (x) - T_{n - 1} (x), & n \geq 2, & T_{0} (x) = 1, & T_{1} (x) = x, \end{array}$ \matrix{ {{T_{n + 1}}\left( x \right) = 2x{T_n}\left( x \right) - {T_{n - 1}}\left( x \right),} \hfill & {n \ge 2,} \hfill & {{T_0}\left( x \right) = 1,} \hfill & {{T_1}\left( x \right) = x,} \hfill \cr } , while the Chebyshev polynomials {U_n(x)}_n≥0 of the second kind are defined by $\begin{array}{l} U_{n + 1} (x) = 2 x U_{n} (x) - U_{n - 1} (x), & n \geq 2, & U_{0} (x) = 1, & U_{1} (x) = 2 x . \end{array}$ \matrix{ {{U_{n + 1}}\left( x \right) = 2x{U_n}\left( x \right) - {U_{n - 1}}\left( x \right),} \hfill & {n \ge 2,} \hfill & {{U_0}\left( x \right) = 1,} \hfill & {{U_1}\left( x \right) = 2x.} \hfill \cr } , The Chebyshev polynomials possess the representations $\begin{array}{l} T_{n} (x) = \sum_{k = 0}^{n / 2⌋} (\begin{matrix} n \\ 2 k \end{matrix}) {(x^{2} - 1)}^{k} x^{n - 2 k}, \\ U_{n} (x) = \sum_{k = 0}^{n / 2⌋} (\begin{matrix} n + 1 \\ 2 k + 1 \end{matrix}) {(x^{2} - 1)}^{k} x^{n - 2 k}, \end{array}$ \matrix{ {{T_n}\left( x \right) = \sum\limits_{k = 0}^{\left\lfloor {n/2} \right\rfloor } {\left( {\matrix{ n \cr {2k} \cr } } \right){{({x^2} - 1)}^k}{x^{n - 2k}}} ,} \hfill \cr {{U_n}\left( x \right) = \sum\limits_{k = 0}^{\left\lfloor {n/2} \right\rfloor } {\left( {\matrix{ {n + 1} \cr {2k + 1} \cr } } \right){{({x^2} - 1)}^k}{x^{n - 2k}}} ,} \hfill \cr } and have the exact Binet-like formulas $\begin{array}{l} T_{n} (x) = \frac{1}{2} ({(x + \sqrt{x^{2} - 1})}^{n} + {(x - \sqrt{x^{2} - 1})}^{n}), \\ U_{n} (x) = \frac{1}{2 \sqrt{x^{2} - 1}} ({(x + \sqrt{x^{2} - 1})}^{n + 1} - {(x - \sqrt{x^{2} - 1})}^{n + 1}) . \end{array}$ \matrix{ {{T_n}\left( x \right) = {1 \over 2}\left( {{{(x + \sqrt {{x^2} - 1} )}^n} + {{(x - \sqrt {{x^2} - 1} )}^n}} \right),} \hfill \cr {{U_n}\left( x \right) = {1 \over {2\sqrt {{x^2} - 1} }}\left( {{{(x + \sqrt {{x^2} - 1} )}^{n + 1}} - {{(x - \sqrt {{x^2} - 1} )}^{n + 1}}} \right).} \hfill \cr }

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 4.1.

For all x ∈ ℂ and a positive integer n, we have the following identities: (4.1) $n \sum_{k = 0}^{n} {(- 1)}^{k} \frac{4^{k}}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) {sin}^{2 k} (\frac{x}{2}) = cos nx,$ n\sum\limits_{k = 0}^n {{{( - 1)}^k}{{{4^k}} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\sin }^{2k}}\left( {{x \over 2}} \right)} = \cos nx, (4.2) $n \sum_{k = 0}^{n} {(- 1)}^{n - k} \frac{4^{k}}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) {cos}^{2 k} (\frac{x}{2}) = cos nx .$ n\sum\limits_{k = 0}^n {{{( - 1)}^{n - k}}{{{4^k}} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\cos }^{2k}}\left( {{x \over 2}} \right)} = \cos nx.

Proof

Identities (4.1) and (4.2) are consequences of the identity (4.3) $n \sum_{k = 0}^{n} \frac{{(- 2)}^{k}}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) {(1 \mp x)}^{k} = {(\pm 1)}^{n} T_{n} (x)$ n\sum\limits_{k = 0}^n {{{{{( - 2)}^k}} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\left( {1 \mp x} \right)}^k}} = {\left( { \pm 1} \right)^n}{T_n}\left( x \right) derived in [3].

Lemma 4.2.

If n is a non-negative integer, then $\begin{array}{l} T_{n} (- \frac{α}{2}) = \begin{array}{l} 1, & if n \equiv 0 (mod 5), \\ - α / 2, & if n \equiv 1 or 4 (mod 5), \\ - β / 2, & if n \equiv 2 or 3 (mod 5), \end{array} \\ T_{n} (- \frac{β}{2}) = \begin{array}{l} 1, & if n \equiv 0 (mod 5), \\ - β / 2, & if n \equiv 1 or 4 (mod 5), \\ - α / 2, & if n \equiv 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {{T_n}\left( { - {\alpha \over 2}} \right) = \left\{ {\matrix{ {1,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - \alpha /2,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - \beta /2,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {{T_n}\left( { - {\beta \over 2}} \right) = \left\{ {\matrix{ {1,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - \beta /2,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - \alpha /2,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

Evaluate the identity T_n(cos x) = cos nx at x = 4π/5 and x = 2π/5, in turn.

Theorem 4.3.

If n is a positive integer and t is any integer, then (4.4) $\begin{array}{l} \sum_{k = 1}^{n / 2⌉} \frac{n}{n + 2 k - 1} (\begin{matrix} n + 2 k - 1 \\ n - 2 k + 1 \end{matrix}) 5^{k} F_{2 k + t - 1} \\ - \sum_{k = 0}^{n / 2⌋} \frac{n}{n + 2 k} (\begin{matrix} n + 2 k \\ n - 2 k \end{matrix}) 5^{k} L_{2 k + t} = \begin{array}{l} - L_{t}, & if n \equiv 0 (mod 5), \\ L_{t + 1} / 2, & if n \equiv 1 or 4 (mod 5), \\ - L_{t - 1} / 2, & if n \equiv 2 or 3 (mod 5), \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 1}^{{\left\lceil {n/2} \right\rceil }} {{n \over {n + 2k - 1}}\left( {\matrix{ {n + 2k - 1} \cr {n - 2k + 1} \cr } } \right){5^k}{F_{2k + t - 1}}} } \hfill \cr { - \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{n \over {n + 2k}}\left( {\matrix{ {n + 2k} \cr {n - 2k} \cr } } \right){5^k}{L_{2k + t}}} = \left\{ {\matrix{ { - {L_t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{t + 1}}/2,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {L_{t - 1}}/2,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr } (4.5) $\begin{array}{l} \sum_{k = 1}^{n / 2⌉} \frac{n}{n + 2 k - 1} (\begin{matrix} n + 2 k - 1 \\ n - 2 k + 1 \end{matrix}) 5^{k - 1} L_{2 k + t - 1} \\ - \sum_{k = 0}^{n / 2⌋} \frac{n}{n + 2 k} (\begin{matrix} n + 2 k \\ n - 2 k \end{matrix}) 5^{k} F_{2 k + t} = \begin{array}{l} - F_{t}, & if n \equiv 0 (mod 5), \\ F_{t + 1} / 2, & if n \equiv 1 or 4 (mod 5), \\ - F_{t - 1} / 2, & if n \equiv 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 1}^{{\left\lceil {n/2} \right\rceil }} {{n \over {n + 2k - 1}}\left( {\matrix{ {n + 2k - 1} \cr {n - 2k + 1} \cr } } \right){5^{k - 1}}{L_{2k + t - 1}}} } \hfill \cr { - \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{n \over {n + 2k}}\left( {\matrix{ {n + 2k} \cr {n - 2k} \cr } } \right){5^k}{F_{2k + t}}} = \left\{ {\matrix{ { - {F_t},} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{t + 1}}/2,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {F_{t - 1}}/2,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

Using x = −α/2 and x = −β/2, in turn, in (4.3) with the upper sign gives, in view of Lemma 4.2, $\sum_{k = 0}^{n} \frac{n}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) {(\sqrt{5})}^{k} ({(- 1)}^{k + 1} λ α^{k + t} - β^{k + t}) = \begin{array}{l} - (λ α^{t} + β^{t}), & if n \equiv 0 (mod 5), \\ (λ α^{t + 1} + β^{t + 1}) / 2, & if n \equiv 1 or 4 (mod 5), \\ - (λ α^{t - 1} + β^{t - 1}) / 2, & if n \equiv 2 or 3 (mod 5), \end{array}$ \sum\limits_{k = 0}^n {{n \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\left( {\sqrt 5 } \right)}^k}\left( {{{\left( { - 1} \right)}^{k + 1}}\lambda {\alpha ^{k + t}} - {\beta ^{k + t}}} \right)} = \left\{ {\matrix{ { - \left( {\lambda {\alpha ^t} + {\beta ^t}} \right),} \hfill & {{\rm if}\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {\left( {\lambda {\alpha ^{t + 1}} + {\beta ^{t + 1}}} \right)/2,} \hfill & {{\rm if}\;n \equiv 1\;{\rm or}\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - \left( {\lambda {\alpha ^{t - 1}} + {\beta ^{t - 1}}} \right)/2,} \hfill & {{\rm if}\;n \equiv 2\;{\rm or}\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right. from which (4.4) and (4.5) now follow upon setting λ = 1 and λ = −1, in turn, and using the Binet formulas and the summation identity $\sum_{j = 0}^{n} f_{j} = \sum_{j = 0}^{n / 2⌋} f_{2 j} + \sum_{j = 1}^{n / 2⌉} f_{2 j - 1} .$ \sum\limits_{j = 0}^n {{f_j}} = \sum\limits_{j = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{f_{2j}}} + \sum\limits_{j = 1}^{{\left\lceil {n/2} \right\rceil }} {{f_{2j - 1}}} .

We observe the following special cases of the prior result.

Corollary 4.4.

If n is a positive integer, then $\sum_{k = 1}^{n / 2⌉} \frac{n}{n + 2 k - 1} (\begin{matrix} n + 2 k - 1 \\ n - 2 k + 1 \end{matrix}) 5^{k} L_{2 k + δ - 1} = \sum_{k = 0}^{n / 2⌋} \frac{n}{n + 2 k} (\begin{matrix} n + 2 k \\ n - 2 k \end{matrix}) 5^{k + 1} F_{2 k + δ},$ \sum\limits_{k = 1}^{{\left\lceil {n/2} \right\rceil }} {{n \over {n + 2k - 1}}\left( {\matrix{ {n + 2k - 1} \cr {n - 2k + 1} \cr } } \right){5^k}{L_{2k + \delta - 1}}} = \sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{n \over {n + 2k}}\left( {\matrix{ {n + 2k} \cr {n - 2k} \cr } } \right){5^{k + 1}}{F_{2k + \delta }}} , where $δ = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ - 1, & if n \equiv 1 or 4 (mod 5), \\ 1, & if n \equiv 2 or 3 (mod 5) . \end{array}$ \delta = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - 1,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

Theorem 4.5.

If n is a positive integer and t is any integer, then $\begin{array}{l} \sum_{k = 0}^{n} {(- 1)}^{n - k} \frac{n}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) L_{2 k + t} = \begin{array}{l} L_{t}, & if n = 0 (mod 5), \\ L_{t - 1} / 2, & if n = 1 or 4 (mod 5), \\ - L_{t + 1} / 2, & if n = 2 or 3 (mod 5), \end{array} \\ \sum_{k = 0}^{n} {(- 1)}^{n - k} \frac{n}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k + t} = \begin{array}{l} F_{t}, & if n = 0 (mod 5), \\ F_{t - 1} / 2, & if n = 1 or 4 (mod 5), \\ - F_{t + 1} / 2, & if n = 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 0}^n {{{( - 1)}^{n - k}}{n \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){L_{2k + t}}} = \left\{ {\matrix{ {{L_t},} \hfill & {if\;n = 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{t - 1}}/2,} \hfill & {if\;n = 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {L_{t + 1}}/2,} \hfill & {if\;n = 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 0}^n {{{( - 1)}^{n - k}}{n \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k + t}}} = \left\{ {\matrix{ {{F_t},} \hfill & {if\;n = 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{t - 1}}/2,} \hfill & {if\;n = 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {F_{t + 1}}/2,} \hfill & {if\;n = 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

Set x = π/5 in (4.1) and use (2.3) and the fact that sin(π/10) = −β/2 to obtain $\sum_{k = 0}^{n} {(- 1)}^{n - k} \frac{n}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) β^{2 k + t} = \begin{array}{l} β^{t}, & if n = 0 (mod 5), \\ β^{t - 1} / 2, & if n = 1 or 4 (mod 5), \\ - β^{t + 1} / 2, & if n = 2 or 3 (mod 5), \end{array}$ \sum\limits_{k = 0}^n {{{( - 1)}^{n - k}}{n \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){\beta ^{2k + t}}} = \left\{ {\matrix{ {{\beta ^t},} \hfill & {{\rm if}\;n = 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{\beta ^{t - 1}}/2,} \hfill & {{\rm if}\;n = 1\;{\rm or}\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {\beta ^{t + 1}}/2,} \hfill & {{\rm if}\;n = 2\;{\rm or}\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right. from which the results follow by (2.5).

Using Theorem 4.5, we have the following binomial Fibonacci identities modulo 5.

Corollary 4.6.

If n is a positive integer, then $\sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k + δ} = 0,$ \sum\limits_{k = 0}^n {{{{{\left( { - 1} \right)}^k}} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k + \delta }}} = 0, where $δ = \begin{array}{l} 0, & if n = 0 (mod 5), \\ 1, & if n = 1 or 4 (mod 5), \\ - 1, & if n = 2 or 3 (mod 5) . \end{array}$ \delta = \left\{ {\matrix{ {0,} \hfill & {if\;n = 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {if\;n = 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - 1,} \hfill & {if\;n = 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

Lemma 4.7.

If x is a complex variable and n is a positive integer, then (4.6) $\sum_{k = 1}^{n} {(- 1)}^{k - 1} \frac{4^{k} k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) {sin}^{2 k - 2} (\frac{x}{2}) = \frac{2 sin nx}{sin x},$ \sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k - 1}}{{{4^k}k} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\sin }^{2k - 2}}\left( {{x \over 2}} \right)} = {{2\sin nx} \over {\sin x}}, (4.7) $\sum_{k = 1}^{n} {(- 1)}^{n - k} \frac{4^{k} k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) {cos}^{2 k - 2} (\frac{x}{2}) = \frac{2 sin nx}{sin x} .$ \sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{n - k}}{{{4^k}k} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\cos }^{2k - 2}}\left( {{x \over 2}} \right)} = {{2\sin nx} \over {\sin x}}.

Proof

Identities (4.6) and (4.7) come from the following identities derived in [3]: $\begin{array}{l} \sum_{k = 1}^{n} {(- 1)}^{n - k} \frac{2^{k} k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) {(1 \mp x)}^{k - 1} = {(\mp 1)}^{n - 1} U_{n - 1} (x), \\ \sum_{k = 1}^{n} {(- 1)}^{n - k} \frac{4^{k} k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) x^{2 k - 1} = U_{2 n - 1} (x) . \end{array}$ \matrix{ {\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{n - k}}{{{2^k}k} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\left( {1 \mp x} \right)}^{k - 1}}} = {{\left( { \mp 1} \right)}^{n - 1}}{U_{n - 1}}\left( x \right)}, \hfill \cr {\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{n - k}}{{{4^k}k} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){x^{2k - 1}}} = {U_{2n - 1}}\left( x \right)}. \hfill \cr }

Theorem 4.8.

If n is a positive integer and n is any integer, then $\begin{array}{l} \sum_{k = 1}^{n} {(- 1)}^{k - 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) L_{2 k + t} = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ {(- 1)}^{n / 5⌋} L_{t + 2} / 2, & if n \equiv 1 or 4 (mod 5), \\ {(- 1)}^{n / 5⌋ + 1} L_{t + 1} / 2, & if n \equiv 2 or 3 (mod 5), \end{array} \\ \sum_{k = 1}^{n} {(- 1)}^{k - 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k + t} = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ {(- 1)}^{n / 5⌋} F_{t + 2} / 2, & if n \equiv 1 or 4 (mod 5), \\ {(- 1)}^{n / 5⌋ + 1} F_{t + 1} / 2, & if n \equiv 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k - 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){L_{2k + t}} = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {\left( { - 1} \right){^{\left\lfloor {n/5} \right\rfloor }}{L_{t + 2}}/2,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{\left( { - 1} \right)}^{\left\lfloor {n/5} \right\rfloor + 1}}{L_{t + 1}}/2,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} } \hfill \cr {\sum\limits_{k = 1}^n {{{( - 1)}^{k - 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k + t}}} = \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{\left( { - 1} \right)}^{\left\lfloor {n/5} \right\rfloor }}{F_{t + 2}}/2,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{\left( { - 1} \right)}^{\left\lfloor {n/5} \right\rfloor + 1}}{F_{t + 1}}/2,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

Set x = π/5 and x = 3π/5, respectively, in (4.6), and use (2.6) and (2.7).

Remark

Theorem 4.8 can also be proved using (4.7). Using the trigonometric identities sin 2x = 2 sin x cos x and cos 3x = 4 cos³ x − 3 cos x and working with x = 2π/5 and x = 6π/5, respectively, we end with $2 \sum_{k = 1}^{n} {(- 1)}^{k - 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) L_{2 k - 1 + t} = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ {(- 1)}^{n / 5⌋} (α^{t + 1} - β^{t - 3} + 3 β^{t - 1}), & if n \equiv 1 or 4 (mod 5), \\ {(- 1)}^{n / 5⌋} (- α^{t} + β^{t + 4} - 3 β^{t + 2}), & if n \equiv 2 or 3 (mod 5), \end{array}$ 2\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k - 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){L_{2k - 1 + t}}} = \left\{ {\matrix{ {0,} \hfill & {{\rm if}\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {\left( { - 1} \right){^{\left\lfloor {n/5} \right\rfloor }}\left( {{\alpha ^{t + 1}} - {\beta ^{t - 3}} + 3{\beta ^{t - 1}}} \right),} \hfill & {{\rm if}\;n \equiv 1\;{\rm or}\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{{( - 1)}^{{\left\lfloor {n/5} \right\rfloor }}}\left( { - {\alpha ^t} + {\beta ^{t + 4}} - 3{\beta ^{t + 2}}} \right),} \hfill & {{\rm if}\;n \equiv 2\;{\rm or}3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right. and $2 \sqrt{5} \sum_{k = 1}^{n} {(- 1)}^{k - 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k - 1 + t} = \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ {(- 1)}^{n / 5⌋} (α^{t + 1} + β^{t - 3} - 3 β^{t - 1}), & if n \equiv 1 or 4 (mod 5), \\ {(- 1)}^{n / 5⌋} (- α^{t} - β^{t + 4} + 3 β^{t + 2}), & if n \equiv 2 or 3 (mod 5) . \end{array}$ 2\sqrt 5 \sum\limits_{k = 1}^n {{{( - 1)}^{k - 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k - 1 + t}}} = \left\{ {\matrix{ {0,} \hfill & {{\rm if}\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {\left( { - 1} \right){^{\left\lfloor {n/5} \right\rfloor }}\left( {{\alpha ^{t + 1}} + {\beta ^{t - 3}} - 3{\beta ^{t - 1}}} \right),} \hfill & {{\rm if}\;n \equiv 1\;{\rm or}\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {\left( { - 1} \right){^{\left\lfloor {n/5} \right\rfloor }}\left( { - {\alpha ^t} - {\beta ^{t + 4}} + 3{\beta ^{t + 2}}} \right),} \hfill & {{\rm if}\;n \equiv 2\;{\rm or}\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right. To get Theorem 4.8 simplify the terms in brackets and replace t by t + 1.

Applying Theorem 4.8 yields the following two corollaries.

Corollary 4.9.

If n is a positive integer, then $\sum_{k = 1}^{n} {(- 1)}^{k} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k - δ} = 0,$ \sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^k}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k - \delta }} = 0} , where $δ = \begin{array}{l} 2, & if n \equiv 1 or 4 & (mod 5), \\ 1, & if n \equiv 2 or 3 & (mod 5) . \end{array}$ \delta = \left\{ {\matrix{ {2,} \hfill & {if\;n \equiv 1\;or\;4} \hfill & {\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {if\;n \equiv 2\;or\;3} \hfill & {\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

Corollary 4.10.

If n is a positive integer and t is any integer, then we have:

If n ≡ 0 (mod 5), then $\begin{array}{l} \sum_{k = 0}^{(n - 1) / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k - 1 \\ k \end{matrix}) L_{n - 2 k + t} = 2 \sum_{k = 1}^{n} {(- 1)}^{k - 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) L_{2 k + t}, \\ \sum_{k = 0}^{(n - 1) / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k - 1 \\ k \end{matrix}) F_{n - 2 k + t} = 2 \sum_{k = 1}^{n} {(- 1)}^{k - 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k + t}, \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor }} {{{\left( { - 1} \right)}^k}\left( {\matrix{ {n - k - 1} \cr k \cr } } \right){L_{n - 2k + t}}} = 2\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k - 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){L_{2k + t}}} ,} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor }} {{{\left( { - 1} \right)}^k}\left( {\matrix{ {n - k - 1} \cr k \cr } } \right){F_{n - 2k + t}}} = 2\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k - 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k + t}}} ,} \hfill \cr }
if n ≡ 1 or 4 (mod 5), then $\begin{array}{l} \sum_{k = 0}^{(n - 1) / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k - 1 \\ k \end{matrix}) L_{n - 2 k + t} = 2 \sum_{k = 1}^{n} {(- 1)}^{k + 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) L_{2 k - 1 + t}, \\ \sum_{k = 0}^{(n - 1) / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k - 1 \\ k \end{matrix}) F_{n - 2 k + t} = 2 \sum_{k = 1}^{n} {(- 1)}^{k + 1} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k - 1 + t}, \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor }} {{{\left( { - 1} \right)}^k}\left( {\matrix{ {n - k - 1} \cr k \cr } } \right){L_{n - 2k + t}}} = 2\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k + 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){L_{2k - 1 + t}}} ,} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor }} {{{\left( { - 1} \right)}^k}\left( {\matrix{ {n - k - 1} \cr k \cr } } \right){F_{n - 2k + t}}} = 2\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k + 1}}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k - 1 + t}}} ,} \hfill \cr }
if n ≡ 2 or 3 (mod 5), then $\begin{array}{l} \sum_{k = 0}^{(n - 1) / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k - 1 \\ k \end{matrix}) L_{n - 2 k + t} = 2 \sum_{k = 1}^{n} {(- 1)}^{k} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) L_{2 k + 1 + t}, \\ \sum_{k = 0}^{(n - 1) / 2⌋} {(- 1)}^{k} (\begin{matrix} n - k - 1 \\ k \end{matrix}) F_{n - 2 k + t} = 2 \sum_{k = 1}^{n} {(- 1)}^{k} \frac{k}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k + 1 + t} . \end{array}$ \matrix{ {\sum\limits_{k = 0}^{{\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor }} {{{\left( { - 1} \right)}^k}\left( {\matrix{ {n - k - 1} \cr k \cr } } \right){L_{n - 2k + t}}} = 2\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^k}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){L_{2k + 1 + t}}} ,} \hfill \cr {\sum\limits_{k = 0}^{{\left\lfloor {\left( {n - 1} \right)/2} \right\rfloor }} {{{\left( { - 1} \right)}^k}\left( {\matrix{ {n - k - 1} \cr k \cr } } \right){F_{n - 2k + t}}} = 2\sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^k}{k \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k + 1 + t}}} .} \hfill \cr }

Proof

Compare Theorem 4.8 with Theorem 2.7.

Lemma 4.11.

If n is a non-negative integer, then (4.8) $\sum_{k = 0}^{n} {(- 1)}^{n - k} 4^{k} (\begin{matrix} n + k \\ n - k \end{matrix}) {cos}^{2 k} x = \frac{sin ((2 n + 1) x)}{sin x} .$ \sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^{n - k}}{4^k}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){{\cos }^{2k}}x} = {{\sin \left( {\left( {2n + 1} \right)x} \right)} \over {\sin x}}.

Proof

Evaluate the identity [3] $\sum_{k = 0}^{n} {(- 1)}^{n - k} 4^{k} (\begin{matrix} n + k \\ n - k \end{matrix}) x^{2 k} = U_{2 n} (x)$ \sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^{n - k}}{4^k}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){x^{2k}}} = {U_{2n}}\left( x \right) at x = cos x.

Lemma 4.12.

If n is an integer, then $\frac{sin ((2 n + 1) π / 5)}{sin (π / 5)} = \begin{array}{l} 1, & if n \equiv 0 & (mod 5), \\ α, & if n \equiv 1 & (mod 5), \\ 0, & if n \equiv 2 & (mod 5), \\ - α, & if n \equiv 3 & (mod 5), \\ - 1, & if n \equiv 4 & (mod 5) . \end{array}$ {{\sin \left( {\left( {2n + 1} \right)\pi /5} \right)} \over {\sin \left( {\pi /5} \right)}} = \left\{ {\matrix{ {1,} \hfill & {if\;n \equiv 0} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr {\alpha ,} \hfill & {if\;n \equiv 1} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 2} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr { - \alpha ,} \hfill & {if\;n \equiv 3} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr { - 1,} \hfill & {if\;n \equiv 4} \hfill & {\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.

From Lemmas 4.11 and 4.12 we can deduce the following Fibonacci and Lucas binomial identities modulo 5.

Theorem 4.13.

If n is a non-negative integer and t is any integer, then $\begin{array}{l} \sum_{k = 0}^{n} {(- 1)}^{n - k} (\begin{matrix} n + k \\ n - k \end{matrix}) L_{2 k + t} = \begin{array}{l} L_{t}, & if n \equiv 0 & (mod 5), \\ L_{t + 1}, & if n \equiv 1 & (mod 5), \\ 0, & if n \equiv 2 & (mod 5), \\ - L_{t + 1}, & if n \equiv 3 & (mod 5), \\ - L_{t}, & if n \equiv 4 & (mod 5), \end{array} \\ \sum_{k = 0}^{n} {(- 1)}^{n - k} (\begin{matrix} n + k \\ n - k \end{matrix}) F_{2 k + t} = \begin{array}{l} F_{t}, & if n \equiv 0 & (mod 5), \\ F_{t + 1}, & if n \equiv 1 & (mod 5), \\ 0, & if n \equiv 2 & (mod 5), \\ - F_{t + 1}, & if n \equiv 3 & (mod 5), \\ - F_{t}, & if n \equiv 4 & (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^{n - k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){L_{2k + t}}} = \left\{ {\matrix{ {{L_t},} \hfill & {if\;n \equiv 0\;} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{t + 1}},} \hfill & {if\;n \equiv 1} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 2} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr { - {L_{t + 1}},} \hfill & {if\;n \equiv 3\;} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr { - {L_t},} \hfill & {if\;n \equiv 4\;} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 0}^n {{{\left( { - 1} \right)}^{n - k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){F_{2k + t}}} = \left\{ {\matrix{ {{F_t},} \hfill & {if\;n \equiv 0} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{t + 1}},} \hfill & {if\;n \equiv 1} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr {0,} \hfill & {if\;n \equiv 2\;} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr { - {F_{t + 1}},} \hfill & {if\;n \equiv 3\;} \hfill & {\left( {{\rm mod }\;5} \right),} \hfill \cr { - {F_t},} \hfill & {if\;n \equiv 4} \hfill & {\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

Set x = π/5 in (4.8) and use Lemma 4.12.

Lemma 4.14 ([10, (41.2.16.1)]).

If n is a positive integer and x is any variable, then (4.9) $\sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{cos x - cos (π k / n)} = \frac{1}{2} (\frac{1}{1 - cos x} + \frac{{(- 1)}^{n}}{1 + cos x}) - \frac{n}{sin x sin nx} .$ \sum\limits_{k = 1}^n {{{{{( - 1)}^k}} \over {\cos x - \cos \left( {\pi k/n} \right)}}} = {1 \over 2}\left( {{1 \over {1 - \cos x}} + {{{{( - 1)}^n}} \over {1 + \cos x}}} \right) - {n \over {\sin x\sin nx}}.

Further interesting identities involving Fibonacci and Lucas numbers are stated in the next theorem.

Theorem 4.15.

If n is a positive integer and t is any integer, then $\begin{array}{l} \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} (L_{t - 1} + 2 L_{t} cos (π k / n))}{4 {cos}^{2} (π k / n) - 2 cos (π k / n) - 1} \\ = \frac{1}{2} (L_{t + 2} + (- 1)^{n} F_{t - 1}) - 2 {(- 1)}^{n / 5⌋} n . \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ F_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ F_{t}, & if n \equiv 2 or 3 (mod 5), \end{array} \\ \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} (F_{t - 1} + 2 F_{t} cos (π k / n))}{4 {cos}^{2} (π k / n) - 2 cos (π k / n) - 1} \\ = \frac{1}{2} (F_{t + 2} + \frac{{(- 1)}^{n}}{5} L_{t - 1}) - \frac{2 {(- 1)}^{n / 5⌋}}{5} n \cdot \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ L_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ L_{t}, & if n \equiv 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 1}^n {{{{{( - 1)}^{k - 1}}\left( {{L_{t - 1}} + 2{L_t}\cos \left( {\pi k/n} \right)} \right)} \over {4\,{{\cos }^2}\left( {\pi k/n} \right) - 2\cos \left( {\pi k/n} \right) - 1}}} } \hfill \cr {\; = {1 \over 2}({L_{t + 2}} + \left( { - 1{)^n}{F_{t - 1}}} \right) - 2{{\left( { - 1} \right)}^{\left\lfloor {n/5} \right\rfloor }}n.\left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_{t + 1}},} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{F_t},} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 1}^n {{{{{( - 1)}^{k - 1}}\left( {{F_{t - 1}} + 2{F_t}\cos \left( {\pi k/n} \right)} \right)} \over {4\,{{\cos }^2}\left( {\pi k/n} \right) - 2\cos \left( {\pi k/n} \right) - 1}}} } \hfill \cr {\; = {1 \over 2}\left( {{F_{t + 2}} + {{{{( - 1)}^n}} \over 5}{L_{t - 1}}} \right) - {{2{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }}} \over 5}n \cdot \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_{t + 1}},} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{L_t},} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Proof

Set x = π/5 and x = 3π/5, in turn, in (4.9) to obtain $2 \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{α - 2 cos (π k / n)} = \frac{1}{2 - α} + \frac{{(- 1)}^{n}}{2 + α} - \frac{4 n α}{\sqrt{5}} \frac{sin (π / 5)}{sin (n π / 5)}$ 2\sum\limits_{k = 1}^n {{{{{( - 1)}^k}} \over {\alpha - 2\cos \left( {\pi k/n} \right)}}} = {1 \over {2 - \alpha }} + {{{{( - 1)}^n}} \over {2 + \alpha }} - {{4n\alpha } \over {\sqrt 5 }}{{\sin \left( {\pi /5} \right)} \over {\sin \left( {n\pi /5} \right)}} and $2 \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{β - 2 cos (π k / n)} = \frac{1}{2 - β} + \frac{{(- 1)}^{n}}{2 + β} + \frac{4 n β}{\sqrt{5}} \frac{sin (3 π / 5)}{sin (3 n π / 5)},$ 2\sum\limits_{k = 1}^n {{{{{( - 1)}^k}} \over {\beta - 2\cos \left( {\pi k/n} \right)}}} \; = {1 \over {2 - \beta }} + {{{{( - 1)}^n}} \over {2 + \beta }} + {{4n\beta } \over {\sqrt 5 }}{{\sin \left( {3\pi /5} \right)} \over {\sin \left( {3n\pi /5} \right)}}, from which the identities follow.

By setting t = 0 and t = 1 in Theorem 4.15, we obtain the following.

Corollary 4.16.

If n is a positive integer, then $\begin{array}{l} \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} (4 cos (π k / n) - 1)}{4 {cos}^{2} (π k / n) - 2 cos (π k / n) - 1} = \frac{3 + {(- 1)}^{n}}{2} - 2 {(- 1)}^{n / 5⌋} n \cdot \begin{array}{l} 0, & if n \equiv 0, 2 or 3 (mod 5), \\ 1, & o t h e r w i s e, \end{array} \\ \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{4 {cos}^{2} (π k / n) - 2 cos (π k / n) - 1} = \frac{5 - {(- 1)}^{n}}{10} - \frac{2 {(- 1)}^{n / 5⌋}}{5} n \cdot \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ 1, & if n \equiv 1 or 4 (mod 5), \\ 2, & if n \equiv 2 or 3 (mod 5), \end{array} \\ \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} {cos}^{2} (π k / 2 n)}{4 {cos}^{2} (π k / n) - 2 cos (π k / n) - 1} = \frac{1}{2} - \frac{{(- 1)}^{n / 5⌋}}{2} n \cdot \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ 1, & o t h e r w i s e, \end{array} \\ \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1} cos (π k / n)}{4 {cos}^{2} (π k / n) - 2 cos (π k / n) - 1} = \frac{5 + {(- 1)}^{n}}{10} - \frac{{(- 1)}^{n / 5⌋}}{5} n \cdot \begin{array}{l} 0, & if n \equiv 0 (mod 5), \\ 3, & if n \equiv 1 or 4 (mod 5), \\ 1, & if n \equiv 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {\sum\limits_{k = 1}^n {{{{{( - 1)}^{k - 1}}\left( {4\cos \left( {\pi k/n} \right) - 1} \right)} \over {4{{\cos }^2}\left( {\pi k/n} \right) - 2\cos \left( {\pi k/n} \right) - 1}}} = {{3 + {{( - 1)}^n}} \over 2} - 2{{\left( { - 1} \right)}^{\left\lfloor {n/5} \right\rfloor }}n \cdot \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0,2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {otherwise,} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 1}^n {{{{{( - 1)}^{k - 1}}} \over {4{{\cos }^2}\left( {\pi k/n} \right) - 2\cos \left( {\pi k/n} \right) - 1}}} = {{5 - {{( - 1)}^n}} \over {10}} - {{2{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }}} \over 5}n \cdot \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {2,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 1}^n {{{{{( - 1)}^{k - 1}}{{\cos }^2}\left( {\pi k/2n} \right)} \over {4{{\cos }^2}\left( {\pi k/n} \right) - 2\cos \left( {\pi k/n} \right) - 1}}} = {1 \over 2} - {{{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }}} \over 2}n \cdot \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {otherwise,} \hfill \cr } } \right.} \hfill \cr {\sum\limits_{k = 1}^n {{{{{( - 1)}^{k - 1}}\cos \left( {\pi k/n} \right)} \over {4{{\cos }^2}\left( {\pi k/n} \right) - 2\cos \left( {\pi k/n} \right) - 1}}} = {{5 + {{( - 1)}^n}} \over {10}} - {{{{( - 1)}^{\left\lfloor {n/5} \right\rfloor }}} \over 5}n \cdot \left\{ {\matrix{ {0,} \hfill & {if\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {3,} \hfill & {if\;n \equiv 1\;or\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {1,} \hfill & {if\;n \equiv 2\;or\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

5.

Some additional observations

We close this paper with some additional observations leading to possibly new series representations of the constant α involving Bernoulli polynomials. Recall that Bernoulli polynomials B_n(t), n ≥ 0, may be defined by the $B_{n} (t) = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) B_{n - k} t^{k},$ {B_n}\left( t \right) = \sum\limits_{k = 0}^n {\left( {\matrix{ n \cr k \cr } } \right){B_{n - k}}{t^k}} , where B_n is the nth Bernoulli number, defined by the power series $\frac{z}{e^{z} - 1} = \sum_{n = 0}^{\infty} B_{n} \frac{z^{n}}{n!}, z| < 2 π .$ {z \over {{e^z} - 1}} = \sum\limits_{n = 0}^\infty {{B_n}{{{z^n}} \over {n!}}} ,\;\;\;\;\left| z \right| < 2\pi . We have B_n(1) = B_n(0) = B_n for all n ≥ 2 and B_2n+1 = 0 for all n ≥ 1.

Theorem 5.1.

Let m be a non-negative integer. Then (5.1) $\begin{array}{l} \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2}) = {(- 1)}^{m - 1}, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + \frac{1}{2}) = 0, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + 1) = {(- 1)}^{m}, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + \frac{3}{2}) = {(- 1)}^{m} α, \end{array}$ \matrix{ {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2}} \right)} = {{\left( { - 1} \right)}^{m - 1}},} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + {1 \over 2}} \right)} = 0,} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + 1} \right)} = {{\left( { - 1} \right)}^m},} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + {3 \over 2}} \right)} = {{\left( { - 1} \right)}^m}\alpha ,} \hfill \cr } and (5.2) $\sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + 2) = {(- 1)}^{m} α .$ \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + 2} \right)} = {\left( { - 1} \right)^m}\alpha .

Proof

Combine (2.6) with the representation [13, Eq. (2.5)] (5.3) $\frac{sin xt}{sin t} = \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} B_{2 k + 1} (\frac{1 + x}{2}) t^{2 k}, t| < π .$ {{\sin xt} \over {\sin t}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{B_{2k + 1}}\left( {{{1 + x} \over 2}} \right){t^{2k}}} ,\;\;\;\;\left| t \right| < \pi .\;

When m = 0 then from (5.1) and (5.2) we get the special series: $\begin{array}{l} \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{3}{2}) = α, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (2) = α . \end{array}$ \matrix{ {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{3 \over 2}} \right)} = \alpha ,} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( 2 \right)} = \alpha .} \hfill \cr }

From Raabe’s formula $B_{n} (ax) = a^{n - 1} \sum_{k = 0}^{a - 1} B_{n} (x + \frac{k}{a})$ {B_n}\left( {ax} \right) = {a^{n - 1}}\sum\limits_{k = 0}^{a - 1} {{B_n}} \left( {x + {k \over a}} \right) we get $B_{2 k + 1} (2) = 2^{2 k} (B_{2 k + 1} (1) + B_{2 k + 1} (\frac{3}{2}))$ {B_{2k + 1}}\left( 2 \right) = {2^{2k}}\left( {{B_{2k + 1}}\left( 1 \right) + {B_{2k + 1}}\left( {{3 \over 2}} \right)} \right) and $\begin{matrix} \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{3}{2}) = α, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{4 k + 1}}{(2 k + 1)!} \frac{π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{3}{2}) = α - 3 = \sqrt{5} β . \end{matrix}$ \matrix{ {\sum\limits_{k = 0}^\infty {{{( - 1)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{3 \over 2}} \right)} = \alpha ,} \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{4k + 1}}} \over {\left( {2k + 1} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{3 \over 2}} \right)} = \alpha - 3 = \sqrt 5 \beta .} \cr } But making use of B_n(t + 1) − B_n(t) = ntⁿ⁻¹ we see that $B_{2 k + 1} (\frac{3}{2}) = \frac{2 k + 1}{2^{2 k}}$ {B_{2k + 1}}\left( {{3 \over 2}} \right) = {{2k + 1} \over {{2^{2k}}}} and thus the series turn into (5.4) $\sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{(2 k)!} \frac{π^{2 k}}{25^{k}} = \frac{α}{2} - 1 = - \frac{β^{2}}{2}$ \sum\limits_{k = 1}^\infty {{{{{( - 1)}^k}} \over {\left( {2k} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}} = {\alpha \over 2} - 1 = - {{{\beta ^2}} \over 2} and (5.5) $\sum_{k = 1}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k)!} \frac{π^{2 k}}{25^{k}} = \sqrt{5} β .$ \sum\limits_{k = 1}^\infty {{{( - 1)}^k}{{{2^{2k + 1}}} \over {\left( {2k} \right)!}}{{{\pi ^{2k}}} \over {{{25}^k}}}} = \sqrt 5 \beta . The series (5.4) and (5.5) are essentially cosh(iπ/5) = cos(π/5) = α/2 and cosh(2iπ/5) = cos(2π/5) = −β/2 which we encountered at the beginning of the paper.

Combining (2.7) with (5.3) we have the following theorem. The details of we leave to the reader.

Theorem 5.2.

Let m be a non-negative integer. Then (5.6) $\begin{array}{l} \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{9^{k} π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2}) = {(- 1)}^{m - 1}, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k}}{(2 k + 1)!} \frac{9^{k} π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + \frac{1}{2}) = 0, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{9^{k} π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + 1) = {(- 1)}^{m}, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{9^{k} π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + \frac{3}{2}) = {(- 1)}^{m} β, \end{array}$ \matrix{ {\sum\limits_{k = 0}^\infty {{{( - 1)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{9^k}{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2}} \right)} = {{( - 1)}^{m - 1}},} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k}}} \over {\left( {2k + 1} \right)!}}{{{9^k}{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + {1 \over 2}} \right)} = 0,} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{9^k}{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + 1} \right)} = {{\left( { - 1} \right)}^m},} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{9^k}{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + {3 \over 2}} \right)} = {{\left( { - 1} \right)}^m}\beta ,} \hfill \cr } and (5.7) $\sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{9^{k} π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{5 m}{2} + 2) = {(- 1)}^{m} β .$ \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{9^k}{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{{5m} \over 2} + 2} \right)} = {\left( { - 1} \right)^m}\beta .

Finally, we obtain the following special series as a consequence of (5.6) and (5.7): $\begin{array}{l} \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{9^{k} π^{2 k}}{25^{k}} B_{2 k + 1} (\frac{3}{2}) = β, \\ \sum_{k = 0}^{\infty} {(- 1)}^{k} \frac{2^{2 k + 1}}{(2 k + 1)!} \frac{9^{k} π^{2 k}}{25^{k}} B_{2 k + 1} (2) = β . \end{array}$ \matrix{ {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{9^k}{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( {{3 \over 2}} \right)} = \beta ,} \hfill \cr {\sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{{{2^{2k + 1}}} \over {\left( {2k + 1} \right)!}}{{{9^k}{\pi ^{2k}}} \over {{{25}^k}}}{B_{2k + 1}}\left( 2 \right)} = \beta .} \hfill \cr }

6.

Concluding comments

In this paper, we presented new closed forms for some types of finite Fibonacci and Lucas sums involving different kinds of binomial coefficients and depending on the modulo 5 nature of the upper summation limit. To prove our results, we applied some trigonometric identities utilizing Waring formulas and Chebyshev polynomials of the first and second kinds.

Using similar techniques, we can generalize our findings to more common number sequences. Let us give, for example, a generalization of Theorems 2.3, 3.3 and 4.5 to the case of the gibonacci (generalized Fibonacci) sequence defined by the recurrence G_n = G_n−₁ +G_n−₂, n ≥ 2, with G₀ = a and G₁ = b, where a and b are arbitrary [12, 18]. Note that F_n corresponds to the case of G_n when a = 1 and b = 0, while L_n to the case when a = 1 and b = 2. The following identities modulo 5 hold for positive integer n and any integer t: $\begin{array}{l} n \sum_{k = 1}^{n / 2⌋} \frac{{(- 1)}^{k - 1}}{k} (\begin{matrix} n - k - 1 \\ k - 1 \end{matrix}) G_{n - 2 k + t} = \begin{array}{l} G_{n + t} - {(- 1)}^{n} 2 G_{t}, & if n \equiv 0 (mod 5), \\ G_{n + t} + {(- 1)}^{n} G_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ G_{n + t} - {(- 1)}^{n} G_{t - 1}, & if n \equiv 2 or 3 (mod 5), \end{array} \\ n \sum_{k = 0}^{n / 2⌋} \frac{{(- 1)}^{n - k}}{n - k} (\begin{matrix} n - k \\ k \end{matrix}) G_{n - 2 k + t} = \begin{array}{l} 2 G_{t}, & if n \equiv 0 (mod 5), \\ - G_{t + 1}, & if n \equiv 1 or 4 (mod 5), \\ G_{t - 1}, & if n \equiv 2 or 3 (mod 5), \end{array} \\ n \sum_{k = 0}^{n} \frac{{(- 1)}^{n - k}}{n + k} (\begin{matrix} n + k \\ n - k \end{matrix}) G_{2 k + t} = \begin{array}{l} G_{t}, & if n = 0 (mod 5), \\ G_{t - 1} / 2, & if n = 1 or 4 (mod 5), \\ - G_{t + 1} / 2, & if n = 2 or 3 (mod 5) . \end{array} \end{array}$ \matrix{ {n\sum\limits_{k = 1}^{{\left\lfloor {n/2} \right\rfloor }} {{{{{( - 1)}^{k - 1}}} \over k}\left( {\matrix{ {n - k - 1} \cr {k - 1} \cr } } \right){G_{n - 2k + t}}} = \left\{ {\matrix{ {{G_{n + t}} - {{( - 1)}^n}2{G_t},} \hfill & {{\rm{if}}\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{G_{n + t}} + {{( - 1)}^n}{G_{t + 1}},} \hfill & {{\rm{if}}\;n \equiv 1\;{\rm{or}}\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{G_{n + t}} - {{( - 1)}^n}{G_{t - 1}},} \hfill & {{\rm{if}}\;n \equiv 2\;{\rm{or}}\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {n\sum\limits_{k = 0}^{{\left\lfloor {n/2} \right\rfloor }} {{{{{( - 1)}^{n - k}}} \over {n - k}}\left( {\matrix{ {n - k} \cr k \cr } } \right){G_{n - 2k + t}}} = \left\{ {\matrix{ {2{G_t},} \hfill & {{\rm{if}}\;n \equiv 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {G_{t + 1}},} \hfill & {{\rm{if}}\;n \equiv 1\;{\rm{or}}\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{G_{t - 1}},} \hfill & {{\rm{if}}\;n \equiv 2\;{\rm{or}}\;3\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr } } \right.} \hfill \cr {n\sum\limits_{k = 0}^n {{{{{( - 1)}^{n - k}}} \over {n + k}}\left( {\matrix{ {n + k} \cr {n - k} \cr } } \right){G_{2k + t}}} = \left\{ {\matrix{ {{G_t},} \hfill & {{\rm{if}}\;n = 0\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr {{G_{t - 1}}/2,} \hfill & {{\rm{if}}\;n = 1\;{\rm{or}}\;4\;\;\;\;\left( {{\rm mod }\;5} \right),} \hfill \cr { - {G_{t + 1}}/2,} \hfill & {{\rm{if}}\;n = 2\;{\rm{or}}\;3\;\;\;\;\left( {{\rm mod }\;5} \right).} \hfill \cr } } \right.} \hfill \cr }

Fibonacci Sums Modulo 5

Full Article

Paradigm

My account