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\begin{center}
\vskip 1cm{\Large\bf
Binomial Fibonacci Sums from Chebyshev Polynomials\\
\vskip .1in } \vskip 1cm \large
Kunle Adegoke\\
Department of Physics and Engineering Physics\\
Obafemi Awolowo University\\
220005 Ile-Ife\\
Nigeria\\
\href{mailto:adegoke00@gmail.com}{\tt adegoke00@gmail.com} \\
\vskip .2 in
Robert Frontczak\\
Reutlingen\\
Germany\\
\href{mailto:robert.frontczak@web.de}{\tt robert.frontczak@web.de}\\
\vskip .2 in
Taras Goy\\
Faculty of Mathematics and Computer Science\\
Vasyl Stefanyk Precarpathian National University \\
Shevchenka Str.\ 57\\
76018 Ivano-Frankivsk \\
Ukraine\\
\href{mailto:taras.goy@pnu.edu.ua}{\tt taras.goy@pnu.edu.ua} \\
\end{center}

\vskip .2 in

\begin{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.
\end{abstract}


\section{Preliminaries}

As usual, the Fibonacci numbers $F_n$ and the Lucas numbers $L_n$ are defined, for \text{$n\in\mathbb Z$},
through the recurrence relations $F_n = F_{n-1}+F_{n-2}$, $n\ge 2$, 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)
\begin{equation}\label{binet}
F_n = \frac{\alpha^n - \beta^n }{\alpha - \beta },\quad L_n = \alpha^n + \beta^n,\quad n\in\mathbb Z,
\end{equation}
with $\alpha=(1+\sqrt{5})/2$ being the golden ratio and $\beta=-1/\alpha$.
The sequences $(F_n)_{n\geq 0}$ and $(L_n)_{n\geq 0}$ are indexed in the On-Line Encyclopedia of Integer Sequences \cite{OEIS} as entries \seqnum{A000045} and \seqnum{A000032}, respectively. For more information we refer to Koshy \cite{Koshy} and Vajda \cite{Vajda}, who have written excellent books dealing with Fibonacci and Lucas numbers.

For an integer $n\geq0$, the Chebyshev polynomials $\big(T_n(x)\big)_{n\geq0}$ of the first kind are defined
by the second-order recurrence  relation \cite{Mason}
\begin{equation*}
	T_0(x) = 1,\quad T_1(x) = x, \quad T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x),
\end{equation*}
while the Chebyshev polynomials $\big(U_n(x)\big)_{n\geq0}$ of the second kind are defined by
\begin{equation*}
	U_0(x) = 1,\quad U_1(x) = 2x, \quad U_{n+1}(x) = 2xU_n(x) - U_{n-1}(x).
\end{equation*}
The Chebyshev polynomials possess the representations
\begin{align}
	T_n(x) &amp;= \sum_{k=0}^{\lfloor{n}/{2}\rfloor} {n\choose 2k}(x^2-1)^kx^{n-2k}, \label{ChebRep1} \\
	U_n(x) &amp;= \sum_{k=0}^{\lfloor{n}/{2}\rfloor} {n+1\choose 2k+1}(x^2-1)^kx^{n-2k}. \label{ChebRep2}
\end{align}
Also, the sequences $T_n(x)$ and $U_n(x)$ have the Binet-like  formulas
\begin{align}
	T_n (x) &amp;= \frac{1}{2} \left( (x + \sqrt {x^2 - 1} )^n + (x - \sqrt {x^2 - 1} )^n \right)\!, \label{eq.lgfu65w} \\
	U_n (x) &amp;= \frac{1}{2\sqrt {x^2  - 1}} \left( (x + \sqrt {x^2 - 1} )^{n + 1} - (x - \sqrt {x^2 - 1} )^{n + 1} \right)\!.
\end{align}

The Chebyshev polynomials of the first and second kind are connected by
the equations
\begin{align}
	T_{n + 1}(x) &amp;= xT_n(x) - (1 - x^2) U_{n - 1}(x),\nonumber\\[4pt]
\label{eq.w83lskz}
	U_{n + 1}(x) &amp;= xU_n(x) + T_{n + 1}(x),
\end{align}
from which we also get
\begin{equation*}
	T_n(x)=\frac12\big(U_n(x) - U_{n - 2}(x)\big).
\end{equation*}

The properties of Chebyshev polynomials have been studied extensively in the literature. In the recent papers \cite{Abd,Fan,Frontczak,Gao,Kilic2,Kim,Li-Wenpeng,Li,Zhang} 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 \cite{Adegoke1,Adegoke3,Adegoke2,Bai}. 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
\begin{equation*}
	\sum_{k=0}^n \frac {n}{n + k} \binom{n + k}{n - k} x_k \qquad\mbox{and}\qquad \sum_{k=0}^n \frac {k}{n + k} \binom{n + k}{n - k} x_k,
\end{equation*}
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 \cite{Gould}, Jennings \cite{Jennings}, and also Kilic and Ioanescu \cite{Kilic1}.
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 
\begin{equation*}
	\sum_{k=0}^n  \frac{\binom {n+k-1} {k} \binom {n}{k}}{\binom {2k}{k}}\, x_k,
\end{equation*}
and the second is similar.

\section{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.
\begin{theorem}\label{thm1}
For all $x\in\mathbb{C}$ and non-negative integers $n$ and $m$, we have the following identities:
\begin{align}\label{main_id1}
	\sum_{k=0}^n (-2)^k\frac n{n + k}\binom{n + k}{n - k} (1\mp x)^k &amp;= (\pm 1)^n T_n(x), \\
	\label{main_id2}
	\sum_{k=0}^n (-2)^k \frac n{n + k}\binom{n + k}{n - k} \big(1\mp T_m(x)\big)^k &amp;= (\pm 1)^n T_{n m}(x).
	\end{align}
\end{theorem}
\begin{proof} It suffices to prove \eqref{main_id1}. Let $P_n^{(a,b)}(x)$, $a, b &gt; -1$, be the Jacobi polynomial. Then $P_n^{(a,b)}(x)$ has the truncated series representation
\begin{equation*}
	P_n^{(a,b)}(x) = \frac{\Gamma(a+n+1)}{ n!\Gamma(a+b+n+1)} \sum_{k=0}^n \binom {n}{k} \frac{\Gamma(a+b+n+k+1)}{2^k \Gamma(a+k+1)} (x-1)^k.
	\end{equation*}
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 $\Gamma\big(n+\frac{1}{2} \big) = \frac{(2n)! \sqrt{\pi}}{4^n n!}$.

Identity \eqref{main_id2} follows immediately if $x\rightarrow T_{m}(x)$, because $T_n\big(T_m(x)\big)=T_{nm}(x)$.
\end{proof}

We note that the relation $T_n(2x^2-1)=T_{2n}(x)$ in conjunction with \eqref{main_id1} yields
\begin{equation}\label{main_id3}
	\sum_{k=0}^n 4^k \frac n{n + k}\binom{n + k}{n - k} (x^2-1)^k = T_{2n}(x)
\end{equation}
and
\begin{equation}\label{main_id4}
	\sum_{k=0}^n (-4)^k \frac n{n + k}\binom{n + k}{n - k} x^{2k} = (-1)^n T_{2n}(x).
\end{equation}
\begin{example}
Setting $x=0$ and $x=-1$ in \eqref{main_id1} we get
\begin{align*}
	\sum_{k=0}^n (-2)^k \frac{n}{n + k} \binom{n + k}{n - k} &amp;=
	\begin{cases}
	0, &amp; \text{if $n$ is odd;} \\
	(-1)^{n/2}, &amp; \text{otherwise;}
	\end{cases} \\
	\sum_{k=0}^n (-4)^k \frac n{n + k}\binom{n + k}{n - k} &amp;= (-1)^{n}.
\end{align*}
\end{example}
\begin{example}\label{example2}
Directly from \eqref{main_id1} we have the following interesting sums:
\begin{align*}
	\sum_{k=0}^n (-2)^k \frac {n}{n + k} \binom{n + k}{n - k} L_k &amp;= T_n(\alpha) + T_n(\beta),\\
	\sum_{k=0}^n (-2)^{k} \frac {n}{n + k} \binom{n + k}{n - k} F_k &amp;=   \frac{T_n(\alpha)-T_n(\beta)}{- \sqrt{5}}.
\end{align*}
It follows from these formulas that $T_n(\alpha)+T_n(\beta)$ and $\frac{1}{\sqrt5}\big(T_n(\alpha)-T_n(\beta)\big)$ are integers.
The sequence $\big(\frac{1}{\sqrt5}\big(T_n(\alpha)-T_n(\beta)\big)\big)_{n\geq 0}=\big\{0, 1, 2, 5, 16, 45, 130, 377, 1088, 3145, \ldots\big\}$ is sequence \seqnum{A138573} in the OEIS \cite{OEIS}. But more is true as we show in the next theorem.
\end{example}
\begin{theorem}
For an integer $s$, the sequences $\big(T_n(\alpha^s)+T_n(\beta^s)\big)_{n\geq 0}$ and
$\big(\frac{T_n(\alpha^s)-T_n(\beta^s)}{\sqrt5}\big)_{n\geq 0}$ are integers. Moreover, the sequences $\big(U_n(\alpha^s)+U_n(\beta^s)\big)_{n\geq 0}$ and $\big(\frac{U_n(\alpha^s)-U_n(\beta^s)}{\sqrt5}\big)_{n\geq 0}$
are also integer sequences for each $s$.
\end{theorem}
\begin{proof}
For a complex variable $x$, from \eqref{ChebRep1} we have
\begin{align*} 
	T_n(\alpha^sx) = \sum_{k = 0}^{\lfloor{n/2}\rfloor} \sum_{j = 0}^{k}(-1)^{j} \binom{n}{2k} \binom{k}{j} (\alpha^sx)^{n-2j},\\
	T_n(\beta^sx) = \sum_{k = 0}^{\lfloor{n/2}\rfloor} \sum_{j = 0}^{k}(-1)^{j} \binom{n}{2k}\binom{k}{j} (\beta^sx)^{n-2j}.
	\end{align*}
In particular,
	\begin{align*}
	T_n(\alpha^s) &amp;=
	\begin{cases}
	\sum\limits_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n}{2k} L^k_s \alpha^{s(n-k)}, &amp; \text{if  $s$ is odd;} \\
	\sum\limits_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n}{2k} (\sqrt5 F_s)^k \alpha^{s(n-k)}, &amp; \text{otherwise,}
	\end{cases}\\
	T_n(\beta^s) &amp;=
	\begin{cases}
	\sum\limits_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n}{2k} L^k_s \beta^{s(n-k)}, &amp; \text{if  $s$ is odd;} \\
	\sum\limits_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n}{2k}(-\sqrt5 F_s)^k \beta^{s(n-k)}, &amp; \text{otherwise.}
	\end{cases}
\end{align*}
Thus,
\begin{align*}
	T_n(\alpha^s)+T_n(\beta^s) &amp;= \sum_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n}{2k}L^k_s L_{s(n-k)},\qquad s\text{ odd},\\
	\frac{T_n(\alpha^s)-T_n(\beta^s)}{\sqrt5} &amp;= \sum_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n}{2k}L^k_s F_{s(n-k)},\qquad s\text{ odd},
\end{align*}
and
\begin{align*}
	T_n(\alpha^s) + T_n(\beta^s) &amp;= \sum_{k = 0}^{\lfloor{n/4}\rfloor} 5^k F^{2k}_s\left(\binom{n}{4k}L_{s(n-2k)}
	+ 5 \binom{n}{4k+2} F_s F_{s(n-2k-1)}\right)\!,\qquad s\text{ even},\\
	\frac{T_n(\alpha^s) - T_n(\beta^s)}{\sqrt5} &amp;= \sum_{k = 0}^{\lfloor{n/4}\rfloor} 5^k F^{2k}_s\left(\binom{n}{4k}F_{s(n-2k)}	+ \binom{n}{4k+2} F_s L_{s(n-2k-1)}\right)\!,\qquad s\text{ even}.
	\end{align*}

For Chebyshev polynomials of  the second kind similar formulas hold. From \eqref{ChebRep2} using Binet's formulas \eqref{binet} we have
\begin{align*}
	U_n(\alpha^s)+U_n(\beta^s) &amp;= \sum_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n+1}{2k+1}L^k_s L_{s(n-k)},\qquad s\text{ odd},\\
	\frac{U_n(\alpha^s)-U_n(\beta^s)}{\sqrt5} &amp;= \sum_{k = 0}^{\lfloor{n/2}\rfloor} \binom{n+1}{2k+1}L^k_s F_{s(n-k)},\qquad s\text{ odd},
\end{align*}
and
\begin{align*}
	U_n(\alpha^s) + U_n(\beta^s) &amp;= \sum_{k = 0}^{\lfloor{n/4}\rfloor}
	5^k F^{2k}_s \left(\binom{n+1}{4k+1}L_{s(n-2k)} + 5 \binom{n+1}{4k+3} F_s F_{s(n-2k-1)}\right)\!,\qquad s\text{ even},\\
	\frac{U_n(\alpha^s) - U_n(\beta^s)}{\sqrt5} &amp;= \sum_{k = 0}^{\lfloor{n/4}\rfloor}
	5^k F^{2k}_s\left(\binom{n+1}{4k+1}F_{s(n-2k)} + \binom{n+1}{4k+3} F_s L_{s(n-2k-1)}\right)\!,\qquad s\text{ even}.
\end{align*}
\end{proof}
\begin{example}
Here we proceed with additional interesting sums of the same kind as in Example \ref{example2}:
\begin{align*}
	\sum_{k=0}^n (-2)^k \frac {n}{n + k} \binom{n + k}{n-k} L_{2k} &amp;= (-1)^n \big(T_n(\alpha)+T_n(\beta)\big),\\
	\sum_{k=0}^n (-2)^k \frac {n}{n + k} \binom{n + k}{n-k} F_{2k} &amp;= \frac{(-1)^n}{\sqrt5}\big(T_n(\alpha)-T_n(\beta)\big),\\
	\sum_{k=0}^n (-2)^k \frac {n}{n + k}\binom{n + k}{n-k} L_{3k} &amp;= (-1)^n \big(T_n(2\alpha)+T_n(2\beta)\big),\\
	\sum_{k=0}^n (-2)^k \frac {n}{n + k}\binom{n + k}{n-k} F_{3k} &amp;= \frac{(-1)^n}{\sqrt5} \big(T_n(2\alpha)-T_n(2	\beta)\big),\\
	\sum_{k=0}^n \frac {n}{n + k} \binom{n + k}{n-k} L_k  &amp;= T_n\Big(\frac{\sqrt5\alpha}2\Big)+T_n\Big(\frac{\sqrt5\beta}2\Big),\\
	\sum_{k=0}^n \frac {n}{n + k} \binom{n + k}{n-k} F_k  &amp;= \frac{1}{\sqrt5}\Big( T_n\Big(\frac{\sqrt5\alpha}2\Big)-T_n\Big(\frac{\sqrt5\beta}2\Big)\Big),\\
	\sum_{k=0}^n 4^k \frac {n}{n + k} \binom{n + k}{n-k} L_{k} &amp;= T_{2n}(\alpha)+T_{2n}(\beta),\\
	\sum_{k=0}^n 4^k \frac {n}{n + k}\binom{n + k}{n-k} F_{k} &amp;= \frac{1}{\sqrt5}\big(T_{2n}(\alpha)-T_{2n}(\beta)\big).
	\end{align*}
\end{example}

By substituting $x=L_p/2$ and $x=\sqrt 5F_p/2$ in \eqref{eq.lgfu65w}, in turn, we obtain the results stated in Lemma \ref{lem.v25emwx}.
\begin{lemma}\label{lem.v25emwx}
If $p$ is an integer, then we have
\begin{equation}
	T_n\Big(\frac{L_p}2\Big) = \frac12 L_{pn},\qquad\text{\rm $p$ even,}\label{eq.xgyt5jf}
	\end{equation}
and
\begin{equation}
	T_n \Big( {\frac{\sqrt 5F_p}{2}} \Big) = \begin{cases}
	\frac12 
	L_{pn},&amp;\text{\rm if $p$ is odd and $n$ is even;}  \\
	\frac{\sqrt 5}{2} F_{pn},&amp;\text{\rm if $p$ is odd and $n$ is odd.} \label{eq.r1jl3d5}
	\end{cases}
\end{equation}
In particular, we have
\begin{align}
	T_n\Big(\frac32\Big) &amp;= \frac12L_{2n},\nonumber\\
	T_n \Big( {\frac{\sqrt 5 }{2}} \Big) &amp;=  \begin{cases}
	\frac12L_{n},&amp;\text{\rm if $n$ is even;}  \\
	\frac{\sqrt 5}2F_n , &amp;\text{\rm  otherwise;}  \\
	\end{cases}\label{eq.a8aqw1m}\\
	T_n \big(\sqrt 5\big) &amp;=  \begin{cases}
	\frac12 L_{3n},&amp;\text{\rm if $n$ is even;}  \\
	\frac{\sqrt 5}2F_{3n}, &amp; \text{\rm  otherwise.}
	\end{cases}\label{eq.l6oienp}
\end{align}
\end{lemma}
\begin{theorem}\label{thm_xxx}
If $n$ is a positive integer and $p$ is an integer, then the following identities hold:
\begin{align*}
	\sum_{k = 0}^n (-1)^{(p-1)(n-k)} \frac{n}{n + k} \binom{n + k}{n-k}  L_p^{2k} &amp;= \frac{L_{2pn}}{2},\\
	\sum_{k = 0}^n (-1)^{p(n-k)} \frac{n}{n + k} \binom{n + k}{n-k} 5^k F_p^{2k} &amp;= \frac{L_{2pn}}{2}.
	\end{align*}
\end{theorem}
\begin{proof}
Set $x=L_{2p}/2$ in \eqref{main_id1}, use \eqref{eq.xgyt5jf} and the fact that
\begin{equation}\label{eq.r1j2cpo}
	L_{2p} - 2 =  \begin{cases}
	5F_p^2,&amp;\text{if $p$ is even}; \\
	L_p^2, &amp;\text{otherwise};\\
	\end{cases}\qquad
	L_{2p} + 2 =  \begin{cases}
	5F_p^2,&amp; \text{if $p$ is odd};\\
	L_p^2,&amp; \text{otherwise};
	\end{cases}
	\end{equation}
to get
\begin{align*}
	\sum_{k = 0}^n \frac{n}{n + k} \binom{n + k}{n - k}  L_p^{2k} &amp;= \frac{1 }{2}L_{2pn}, \qquad\text{$p$ odd},\\
	\sum_{k = 0}^n \frac{n}{n + k} \binom{n + k}{n - k} ( -1)^k L_p^{2k} &amp;= \frac{(- 1)^n}{2}L_{2pn},\qquad\text{$p$ even},
\end{align*}
and
\begin{align*}	
	\sum_{k = 0}^n \frac{n}{n + k} \binom{n + k}{n - k} ( - 5)^k F_p^{2k} &amp;= \frac{{( - 1)^n }}{2}L_{2pn},\qquad \text{$p$ odd},\\
	\sum_{k = 0}^n \frac{n}{n + k} \binom{n + k}{n - k} 5^k F_p^{2k} &amp;= \frac{1}{2}L_{2pn},\qquad{\text{$p$ even}},
	\end{align*}
from which the stated identities follow.
\end{proof}

Theorem \ref{thm_xxx} can be generalized in the following way.
\begin{theorem}\label{thm_xxx2}
If $n$ is a positive integer and $p$ is an integer, then we have
\begin{align*}
	\sum_{k = 0}^n (-1)^{n-k} \frac{n}{n + k} \binom{n + k}{n - k}(2 \pm \sqrt{5}F_{pm})^k &amp;= \frac{\pm \sqrt{5}}{2}  F_{p m n},
	\qquad \text{\rm $p$, $m$, $n$ odd}, \\
	\sum_{k = 0}^n (-1)^{n-k} \frac{n}{n + k} \binom{n + k}{n - k}(2 \pm L_{pm})^k &amp;= \frac{( \pm 1)^n}{2} L_{pmn},
	\qquad \text{\rm $p$ odd, $m$ even}, \\
	\sum_{k = 0}^n (-1)^{n-k} \frac{n}{n + k} \binom{n + k}{n - k} (2 \pm L_{pm})^k &amp;= \frac{( \pm 1)^n }{2} L_{pmn},\qquad \text{\rm $p$ even.}
	\end{align*}
\end{theorem}
\begin{proof}
Combine \eqref{main_id2} with \eqref{eq.xgyt5jf} and \eqref{eq.r1jl3d5}.
\end{proof}

Some particular cases of Theorems \ref{thm_xxx} and \ref{thm_xxx2} stated in the next Example.
\begin{example}\label{ex.lhwdd75} We have
	\begin{align*}
		\sum_{k=0}^n \frac{n}{n + k}\binom{n + k}{n - k} \alpha^{-3k} &amp;= \sum_{k=0}^n (-1)^{k+1} \frac{n}{n + k} \binom{n + k}{n - k} \alpha^{3k}
		= \frac{\sqrt{5}F_{n}}{2}, \qquad \text{\rm $n$ odd},\\
		\sum_{k=0}^n \frac{n}{n + k}\binom{n + k}{n - k} 4^k \alpha^{-k} &amp;= \sum_{k=0}^n (-1)^{k+1} \frac{n}{n + k} \binom{n + k}{n - k} 4^k \alpha^{k}
		= \frac{\sqrt{5}F_{3n}}{2}, \qquad \text{\rm $n$ odd},\\
		\sum_{k=0}^n \frac{n}{n + k}\binom{n + k}{n - k} &amp;= \frac{L_{2n}}{2},\qquad\quad
		\sum_{k=0}^n (-1)^{n-k} \frac{n}{n + k}\binom{n + k}{n - k} 5^k = \frac{L_{2n}}{2},\\
		\sum_{k=0}^n \frac{n}{n + k}\binom{n + k}{n - k} 5^k &amp;= \frac{L_{4n}}{2},\qquad
		\quad\sum_{k=0}^n (-1)^{n-k} \frac n{n + k}\binom{n + k}{n - k} 9^k = \frac{L_{4n}}{2}.
	\end{align*}
\end{example}
\begin{lemma}
If $p$ is a non-zero integer, then
\begin{align}
	T_n \Big( {\frac{{ \sqrt 5F_p }}{L_p }} \Big) &amp;= \cosh\!\left( {n\arctanh\Big( {\frac{2}{ \sqrt 5F_p }}\Big)} \right)\!,\qquad\text{\rm $p$ odd},\label{eq.qefdypb}\\
	T_n \Big( {\frac{L_p}{ \sqrt 5F_p }} \Big) &amp;= \cosh\!\left( {n\arctanh\Big( {\frac{2}{{L_p }}} \Big)} \right)\!,\qquad\text{\rm $p$ even}. \label{eq.y8mmtsi}
	\end{align}
\end{lemma}
\begin{proof}
Setting $x=\sqrt 5F_p/L_p$ in \eqref{eq.lgfu65w} and making use of $5F_p^2 - 4(-1)^{p + 1}=L_p^2$ with $p$ odd produces
\begin{equation}\label{eq.h51c7s8}
	T_n \Big( {\frac{ \sqrt 5F_p }{L_p}} \Big) = \frac{{(\sqrt 5F_p   - 2)^n  + (\sqrt 5F_p + 2)^n }}{2L_p^n }
	\end{equation}
from which \eqref{eq.qefdypb} follows upon using the identity
\begin{equation*}
	(x - y)^n  + (x + y)^n  = 2\big( {\sqrt {x^2  - y^2 } } \big)^n \cosh\!\bigg( {n\arctanh\left( {\frac{y}{x}} \right)} \bigg).
	\end{equation*}
\end{proof}

For low values of $p$, it is easier to use \eqref{eq.h51c7s8} directly for evaluation.
Thus, at $p=1$ we recover \eqref{eq.l6oienp} while $p=3$ gives \eqref{eq.a8aqw1m}.
On account  of \eqref{eq.r1j2cpo}, Formula \eqref{eq.y8mmtsi} also implies
\begin{equation}\label{eq.wry7jin}
	T_n \Big ( {\frac{{L_{2p} }}{\sqrt 5F_{2p} }} \Big) = \frac{{(5F_p^2 )^n + (L_p^2 )^n }}{{2(\sqrt 5F_{2p}  )^n }}
\end{equation}
for every non-zero integer $p$. 

We also note that
\begin{align*}
	\sum_{k = 0}^{2n} \Big (\frac{4}{5}\Big )^k \binom{2n + k}{2n-k} \frac {4^k - (-1)^k L^{2k}_{2p} }{(2n + k)F^{2k}_{2p}} &amp;= 0,\qquad p\ne0,\\
	\sum_{k = 0}^{2n} \Big (\frac{4}{5}\Big )^k  \binom{2n + k}{2n-k} \frac {4^k + (-1)^k L^{2k}_{2p}}{(2n +k)F_{2p}^{2k}}
	&amp;= \frac{625^n F_p^{8n}+L_p^{8n}}{2n \,25^n F_{2p}^{4n}}, \qquad p\ne0.
\end{align*}
\begin{theorem}
If $p$ is a non-zero integer and $n$ is a positive integer, then we have
\begin{align*}
	\sum_{k = 0}^n \Big ( \frac{16}{5}\Big )^k \frac{n}{n + k} \binom{n + k}{n - k} F_{2p}^{2(n-k)}
	&amp;= \frac{25^n F_p^{4n} + L_p^{4n}}{2\cdot5^n},\\
	\sum_{k = 0}^n (-1)^{n-k}\Big ( \frac{4}{5}\Big )^k \frac {n}{n + k} \binom{n + k}{n - k}\Big ( \frac{F_{2p}}{L_{2p}}\Big )^{2(n-k)}
	&amp;= \frac{25^n F_p^{4n} + L_p^{4n}}{2\cdot 5^n L_{2p}^{2n}}.
	\end{align*}
In particular, we obtain
\begin{align*}
	\sum_{k = 0}^n \Big ( \frac{16}{5}\Big )^k \frac {n}{n + k} \binom{n + k}{n - k} &amp;= \frac{25^n + 1}{2\cdot 5^n},\\
	\sum_{k = 0}^n (-1)^{n-k} \Big ( \frac{36}{5}\Big )^k \frac {n}{n + k} \binom{n + k}{n - k} &amp;= \frac{25^n + 1}{2\cdot 5^n}.
	\end{align*}
\end{theorem}
\begin{proof}
Combine \eqref{eq.wry7jin} with \eqref{main_id3} and \eqref{main_id4}, respectively, while setting $x=L_{2p}/(\sqrt{5}F_{2p})$.
\end{proof}
\begin{remark}
We note the following relations:
\begin{align*}
	\sum_{k = 0}^n \Big ( \frac{5}{16}\Big )^{n-k} \frac {n}{n + k} \binom{n + k}{n - k} F_{2p}^{2(n-k)}
	&amp;=  \sum_{k=0}^n \binom {2n}{2k} \frac{L_{2p}^{2(n-k)}}{4^{2n-k}},\\
	\sum_{k = 0}^n \Big ( \!-\frac{5}{4}\Big )^{n-k} \frac {n}{n + k} \binom{n + k}{n - k} \Big (\frac{F_{2p}}{L_{2p}}\Big )^{2(n-k)}
	&amp;= \sum_{k=0}^n \binom {2n}{2k} \frac{4^{k-n}}{L^{2k}_{2p}}.
	\end{align*}
\end{remark}
\begin{theorem}\label{thm.fepmhxp}
If $p$ is an odd integer and $n$ is a positive integer, then
\begin{align*}
	\sum_{k = 0}^n  (- 4)^k \frac {n}{n + k}\binom{n + k}{n - k} \frac{L_{pk + t}}{L_p^{k}} &amp;= \cosh\!\left( {n\arctanh\left( {\frac{2}{\sqrt 5 F_p}} \right)} \right) \cdot
	\begin{cases}
	L_t,&amp;\text{\rm if $n$ is even};  \\
	- \sqrt 5 F_t ,&amp;\text{\rm otherwise};
	\end{cases}\\
	\sum_{k = 0}^n (- 4)^k \frac {n}{n + k}\binom{n + k}{n - k} \frac{F_{pk + t}}{L_p^{k}} &amp;= \cosh\! \left( {n\arctanh\left( {\frac{2}{\sqrt 5 F_p  }} \right)} \right) \cdot  \begin{cases}
	F_t,&amp;\text{\rm if $n$ is even};  \\
	- \frac{L_t}{\sqrt 5},&amp;\text{\rm otherwise}. \\
	\end{cases}
\end{align*}
\end{theorem}
\begin{proof}
Set $x=\sqrt 5F_p/L_p$ in \eqref{main_id1}, taking the upper signs, and then use \eqref{eq.qefdypb}.
\end{proof}
\begin{corollary}
If $p$ is an odd integer and $n$ is a positive integer, then
\begin{align*}
	\sum_{k = 0}^n (- 4)^k \frac {n}{n + k}\binom{n + k}{n - k} \frac{L_{pk}}{L_p^{k}} &amp;= 0,\qquad\text{\rm $n$ odd}, \\
	\sum_{k = 0}^n (- 4)^k \frac{n}{n + k}\binom{n + k}{n - k} \frac{F_{pk}}{L_p^{k}} &amp;= 0,\qquad\text{\rm $n$ even}.
	\end{align*}
\end{corollary}
\begin{corollary}
If $n$ is a positive integer, then we have the following combinatorial results:
\begin{align*}
	\sum_{k = 0}^n (- 4)^k \frac {n}{n + k}\binom{n + k}{n - k} L_{k + t} = \begin{cases}
	\frac12 L_tL_{3n},&amp;\text{\rm if $n$ is even};  \\
	- \frac52F_tF_{3n},&amp;\text{\rm otherwise};
	\end{cases}\\
	\sum_{k = 0}^n (-4)^k \frac {n}{n + k}\binom{n + k}{n - k}F_{k + t} = \begin{cases}
	\frac12 F_tL_{3n},&amp;\text{\rm if $n$ is even};  \\
	- \frac12 L_tF_{3n},&amp;\text{otherwise}.
	\end{cases}
\end{align*}
\end{corollary}

The corresponding identities to Theorem \ref{thm.fepmhxp} for $p$ even are stated next.
\begin{theorem}
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:
\begin{align*}
	\sum_{k = 0}^{\lfloor{n/2}\rfloor}  \Big(\frac{16}{5F_p^2}\Big)^k&amp;
	\Bigg( \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}\Bigg) \\
	&amp;\qquad\qquad=\frac{1}{n}\cosh\!\left( n\arctanh\Big( \frac{2}{L_p}\Big) \right)\cdot
	\begin{cases}
	L_t,&amp;\text{\rm if $n$ is even};  \\
	- \sqrt5 F_t,&amp;\text{\rm otherwise};
	\end{cases}\\
	\sum_{k = 0}^{\lfloor{n/2}\rfloor}   \Big(\frac{16}{5F^2_p}\Big)^k&amp;
	\Bigg( \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}\Bigg) \\
	&amp;\qquad\qquad=\frac{1}{\sqrt5\, n}\cosh\!\left( n\arctanh\Big( \frac{2}{L_p}\Big) \right)\cdot
	\begin{cases}
	\sqrt5 F_t,&amp;\text{\rm if $n$ is even};  \\
	- L_t,&amp;\text{\rm otherwise}.
	\end{cases}
	\end{align*}
\end{theorem}

The non-alternating versions of the identities in Theorem  \ref{thm.fepmhxp} 
are derived from \eqref{eq.qefdypb}
and \eqref{main_id1} by considering the bottom sign combinations. This is left to the interested reader.
\begin{example}\label{eq.e5fgbi7}
Noting that
\begin{equation}\label{help_ch_id1}
	\sum_{k=0}^n (-2)^k \frac n{n + k}\binom{n + k}{n - k} \big ( (1+x)^k \mp (1-x)^k \big )= \big((-1)^n \mp 1\big)T_n(x),
	\end{equation}
with $x=\sqrt{5}/2$ and \eqref{eq.a8aqw1m} we get
\begin{equation}\label{eq.hpzpjoo}
	\sum_{k=1}^n (-1)^{k-1} \frac n{n + k}\binom{n + k}{n - k} F_{3k} =\frac{1-(-1)^n}{2} F_n,
\end{equation}
as well as
\begin{equation}\label{eq.rsep1ey}
	\sum_{k=0}^n (-1)^{k} \frac n{n + k}\binom{n + k}{n - k} L_{3k} =
	\frac{1+(-1)^n}{2}L_n.
\end{equation}
\end{example}

As we will see, identities \eqref{eq.hpzpjoo} and \eqref{eq.rsep1ey} are special cases of Theorem \ref{thm.iq16bty} below.
\begin{example}
We derive the inverse relation of \eqref{eq.hpzpjoo} and \eqref{eq.rsep1ey}. Starting with \eqref{help_ch_id1} we set $x=-\sqrt{5}$, and make use of $T_n(-\sqrt{5})=(-1)^nT_n(\sqrt{5})$ together with  \eqref{eq.l6oienp} to derive the following identities:
\begin{align*}
	\sum_{k=1}^n (-4)^{k} \frac n{n + k}\binom{n + k}{n - k} F_{k} &amp;= \frac{(-1)^n-1}{2}F_{3n},\\
	\sum_{k=0}^n (-4)^{k} \frac n{n + k}\binom{n + k}{n - k} L_{k} &amp;= \frac{1+(-1)^n}{2} L_{3n}.
	\end{align*}
\end{example}
\begin{example}
It is obvious that from \eqref{help_ch_id1} more appealing relations can be derived. We give just four examples:
\begin{align*}
	\sum_{k = 0}^{n} (-4)^k \frac{n}{n + k} \binom{n+k}{n-k}\big(L_{2k}\pm (-1)^k L_{k}\big)
	&amp;= \big((-1)^n\pm 1\big) \big(T_n(\alpha^3)+T_n(\beta^3)\big) \\
	&amp;=\big((-1)^n\pm 1\big) \sum_{k = 0}^{\lfloor{n/2}\rfloor}\binom{n}{2k}4^kL_{3(n-k)},\\
	\sum_{k = 0}^{n} (-4)^k \frac{n}{n + k} \binom{n+k}{n-k}\big(F_{2k}\pm(-1)^kF_{k}\big)
	&amp;= \frac{(-1)^n\pm1}{\sqrt{5}} \big(T_n(\alpha^3)-T_n(\beta^3)\big) \\
	&amp;= \frac{(-1)^n\pm1}{\sqrt{5}} \sum_{k = 0}^{\lfloor{n/2}\rfloor}\binom{n}{2k}4^k F_{3(n-k)},\\
	\sum_{k = 0}^{n} \Big(\!-\frac{4}{3}\Big)^k \frac{n}{n + k} \binom{n+k}{n-k} L_{2k} &amp;= \big((-1)^n +1\big)T_n\Big(\frac{\sqrt5}3\Big),\\
	\sum_{k = 0}^{n} \Big(\!-\frac{4}{3}\Big)^k \frac{n}{n + k} \binom{n+k}{n-k} F_{2k} &amp;= \frac{(-1)^n -1}{\sqrt{5}}\,T_n\Big(\frac{\sqrt5}3\Big).
\end{align*}
\end{example}
\begin{theorem}\label{thm.iq16bty}
If $n$ is a positive integer and $t$ is an integer, then we have
\begin{align*}
	\sum_{k = 0}^n {( - 1)^{n-k} \frac{n}{{n + k}}\binom{n + k}{n - k}L_{3k + t} } &amp;= \begin{cases}
	\frac52 F_t F_n,&amp;\text{\rm if  $n$ is odd};  \\
	\frac12 L_t L_n,&amp;\text{\rm otherwise};  \\
	\end{cases} \\
	\sum_{k = 0}^n {( - 1)^{n-k} \frac{n}{{n + k}}\binom{n + k}{n - k}F_{3k + t} } &amp;=  \begin{cases}
	\frac12 L_t F_n,&amp;\text{\rm if $n$ is odd};  \\
	\frac12 F_t L_n,&amp;\text{\rm otherwise}.
	\end{cases}
\end{align*}
\end{theorem}
\begin{proof}
Set $x=\sqrt{\alpha^3}/2$ and $x=\sqrt{\beta^3}/2$, in turn, in \eqref{main_id4} and use
\begin{equation*}
	T_{2n} \bigg( \frac{{\sqrt {\alpha ^3 } }}{2}\bigg) =  \begin{cases}
	\frac{\sqrt5}2 F_n,&amp;\text{\rm if  $n$ is odd};  \\
	\frac12 L_n,&amp;\text{\rm otherwise};  \\
	\end{cases}\qquad
	T_{2n} \bigg( {\frac{{\sqrt {\beta ^3 } }}{2}} \bigg) =  \begin{cases}
	-\frac{\sqrt5}2 F_n,&amp;\text{\rm if  $n$ is odd};  \\
	\frac12 L_n,&amp;\text{\rm otherwise}.  \\
	\end{cases}
\end{equation*}
to obtain
\begin{equation*}
	\sum_{k = 0}^n {( - 1)^{n-k} \frac{n}{{n + k}}\binom{n + k}{n - k}(\alpha ^{3k + t}  + \lambda \beta ^{3k + t} )}= \begin{cases}
\frac{\sqrt 5}{2} F_n (\alpha ^t  - \lambda \beta ^t ) ,&amp; \text{if $n$ is odd};\\
	\frac12 L_n (\alpha ^t - \lambda \beta ^t ),&amp; \text{otherwise};\\
	\end{cases}
	\end{equation*}
from which the stated results follow upon setting $\lambda=1$ and $\lambda=-1$.
\end{proof}

Note that Example \ref{eq.e5fgbi7} is the special case ($t=0$) of Theorem \ref{thm.iq16bty}.
\begin{theorem}
If $n$ is a positive integer and $p$ is a non-zero integer, then we have
\begin{equation*}
	\sum_{k = 0}^n (- 1)^{n-k} \frac{n}{{n + k}} \binom{n + k}{n - k} \Big( \frac{{2L_{2p} }}{\sqrt5F_{2p}} \Big)^{2k}
	= \frac12\bigg(\Big( \frac{\sqrt5 F_p }{L_p } \Big)^{2n} + \Big( \frac{{L_p}}{\sqrt5 F_p } \Big)^{2n}\bigg).
	\end{equation*}
\end{theorem}
\begin{proof}
Set $ x=L_{2p}/(\sqrt 5 F_{2p})$ in \eqref{main_id4} and then use \eqref{eq.wry7jin}.
\end{proof}
\begin{theorem}
If $n$ is a non-negative integer and $p$ and $q$ are non-zero integers, then we have
\begin{equation*}
	\sum_{k = 0}^n ( - 1)^{(p - q)(n-k)}  \frac{n}{{n + k}}\binom{n + k}{n - k}\left( {\frac{{F_{p + q} F_{p - q} }}{{F_p F_q }}} \right)^{2k}  = \frac{{F_p^{4n}  + F_q^{4n} }}{2(F_q F_p)^{2n}}.
\end{equation*}
\end{theorem}
\begin{proof}
Set $\displaystyle x=\frac{{(-i)^{p-q+1}F_{p + q} F_{p - q} }}{2 F_q F_p }$, where  $i=\sqrt{-1}$, in \eqref{main_id4} and use
\begin{equation*}
	T_n \left( {\frac{i^{p - q + 1}F_{p + q} F_{p - q}}{2 F_q F_p }} \right)
	= i^{n(p - q + 1)}\frac{{F_p^{2n}  + ( - 1)^{n(p - q + 1)} F_q^{2n} }}{2 (F_qF_p)^n }.
	\end{equation*}
\end{proof}
\begin{theorem}
If $n$ is a non-negative integer and $p$ is a non-zero integer, then we have
\begin{align*}
	\sum_{k = 0}^n (- 1)^{p(n-k)} \frac n{{n + k}}\binom{n + k}{n - k}\left( {\frac{F_{3p}}{F_{2p}}} \right)^{2k} &amp;= \frac{{L_p^{4n} + 1}}{2L_p^{2n} },\\
	\sum_{k = 0}^n ( - 1)^{(p + 1)(n-k)} \frac{n}{n + k}\binom{n + k}{n - k}5^{n-k} \left( \frac{{L_{3p} }}{F_{2p} } \right)^{2k} &amp;= \frac{5^{2n}F_p^{4n} + 1}{2 F_p^{2n}}.
	\end{align*}
\end{theorem}
\begin{proof}
Set $\displaystyle
x = \frac{(-i)^{p+1}{F_{3p} }}{{2F_{2p} }}$ and $\displaystyle x=\frac{(-i)^pL_{3p} }{2\sqrt 5F_{2p}}$,
in turn, in \eqref{main_id4} and use
\begin{align*}
	T_n \left( \frac{(-i)^{p+1}{F_{3p} }}{2F_{2p}} \right) &amp;= (-i)^{n(p + 1)} \frac{L_p^{2n}  + ( - 1)^{n(p + 1)} }{2L_p^n},\\
	T_n \left(\frac{(-i)^pL_{3p} }{2\sqrt 5F_{2p}} \right) &amp;= (-i)^{np} \frac{5^nF_p^{2n}   + ( - 1)^{np}}{2 \sqrt{5^n}F_p^n}.
	\end{align*}
\end{proof}
\begin{lemma}
If $p$ and $q$ are integers, then the following identities involving Chebyshev polynomials $T_n(x)$ hold:
\begin{align}
	T_{2n} \left( {\sqrt{ \frac{(- 1)^{q + 1}F_q^2 }{4F_p F_{p + q}}\alpha ^{2p + q} }} \right) &amp;= \frac{{( - 1)^{n} }}{2}\!\left((-1)^{nq}\frac{{F_{p + q}^n }}{{F_p^n }}\alpha ^{qn} + \frac{{F_p^n }}{{F_{p + q}^n }}\beta ^{qn}\right)\!,\quad \text{$p\ne 0$, $p\ne -q$}, \label{eq.in8cl0x}\\
	T_{2n} \left(\sqrt { \frac{(- 1)^{q + 1}F_q^2}{4F_p F_{p + q}}\beta ^{2p + q}} \right) &amp;= \frac{( - 1)^{n} }{2}\!\left((-1)^{nq}\frac{{F_{p + q}^n }}{{F_p^n }}\beta ^{qn} + \frac{{F_p^n }}{{F_{p + q}^n }}\alpha ^{qn}\right)\!,\quad \text{$p\ne 0$, $p\ne -q$},\nonumber\\
	T_{2n} \left( \sqrt{\frac{5( - 1)^{q + 1}F_q^2 }{4L_p L_{p + q} }\alpha ^{2p + q} } \right) &amp;= \frac{( - 1)^{n}}{2}\!\left((-1)^{nq}\frac{{L_{p + q}^n }}{{L_p^n }}\alpha ^{qn}  + \frac{{L_p^n }}{{L_{p + q}^n }}\beta ^{qn}\right)\!, \nonumber\\
	T_{2n} \left(\sqrt{ \frac{5(- 1)^{q + 1}F_q^2 }{4L_p L_{p + q} }\beta ^{2p + q} } \right) &amp;= \frac{( - 1)^{n}}{2}\!\left((-1)^{nq}\frac{{L_{p + q}^n }}{{L_p^n }}\beta ^{qn}  + \frac{{L_p^n }}{{L_{p + q}^n }}\alpha ^{qn}\right)\!.\nonumber
	\end{align}
\end{lemma}
\begin{theorem}\label{thm.awwyoac}
If $n$ is a positive integer and $p$, $q$ and $t$ are integers with $p\ne0$ and $q\neq0$, then we have the following combinatorial results:
\begin{align}
	\sum_{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} } &amp;= \frac{1}{2}\left(F_{p + q}^{2n} L_{t+qn}  + F_p^{2n} L_{t-qn}\right)\!,\label{eq.sqpsz3a}\\
	\sum_{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} &amp;= \frac{1}{2}\left( F_{p + q}^{2n} F_{t+qn}  + F_p^{2n} F_{t-qn}\right)\!,\label{eq.uqixdew}\\
	\sum_{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} } &amp;= \frac{1}{2}\left(L_{p + q}^{2n} L_{t+qn}  + L_p^{2n} L_{t-qn}\right)\!,\label{eq.tm4ir7g}\\
	\sum_{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} } &amp;= \frac{1}{2}\left(L_{p + q}^{2n} F_{t+qn}  + L_p^{2n} F_{t-qn}\right)\!.\label{eq.h4xq8wt}
	\end{align}
\end{theorem}
\begin{proof}
In \eqref{main_id4} set
$\displaystyle 	x=\sqrt{ \frac{( - 1)^{q + 1}F_q^2 }{4F_p F_{p + q} }\alpha ^{2p + q}}$ and use \eqref{eq.in8cl0x} to obtain
\begin{equation*}
	\sum_{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} \alpha^{k(2p + q) + t} }
	= \frac{{( - 1)^{nq} }}{2}F_{p + q}^{2n} \alpha^{qn + t}  + \frac{{( - 1)^t }}{2}F_p^{2n} \beta^{qn - t},
\end{equation*}
from which \eqref{eq.sqpsz3a} and \eqref{eq.uqixdew} follow. The proof of \eqref{eq.tm4ir7g} and \eqref{eq.h4xq8wt} is similar; in \eqref{main_id4}, set $\displaystyle x=\sqrt{ \frac{5( - 1)^{q + 1}F_q^2 }{4L_p L_{p + q} }\alpha ^{2p + q}}$.
\end{proof}
\begin{example}
If $n$ is a positive integer and $t$ is an integer, then we get
\begin{align*}
	\sum_{k = 0}^n (-2)^{n-k} \frac{n}{n + k}\binom{n + k}{n - k} L_{5k + t} &amp;= \frac{1}{2}\big(L_{t-n} + 4^{n} L_{n + t}\big), \\
	\sum_{k = 0}^n (-2)^{n-k} \frac{n}{n + k}\binom{n + k}{n - k} F_{5k + t} &amp;= \frac{1}{2}\big(F_{t-n} + 4^{n} F_{n + t}\big),\\
	\sum_{k = 0}^n (-2)^{n-k} \frac{n}{n + k}\binom{n + k}{n - k} 5^k L_{k + t} &amp;= \frac{1}{2}\big(4^{n} L_{t-n} + L_{n + t} \big),\\
	\sum_{k = 0}^n (-2)^{n-k}\frac{n}{{n + k}}\binom{n + k}{n - k} 5^kF_{k + t} &amp;= \frac{1}{2} \big(4^{n} F_{t-n} + F_{n + t}\big).
	\end{align*}
\end{example}


\section{Binomial Fibonacci and Lucas sums from identities involving $U_n(x)$}


Using the fact that $\frac{d}{dx} T_n(x)=n U_{n-1}(x)$,
from \eqref{main_id1} we get
\begin{equation}\label{main_id5}
	\sum_{k=1}^n (-2)^{k}  \frac k{n + k}\binom{n + k}{n - k} (1 \mp x)^{k-1} = \mp U_{n-1} (x),
\end{equation}
and  from \eqref{main_id3} and \eqref{main_id4}
\begin{equation*}
	\sum_{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}
\end{equation*}
and
\begin{equation}\label{main_id7}
	\sum_{k=1}^n (-4)^{k}\frac k{n + k}\binom{n + k}{n - k} x^{2k} = (-1)^n x U_{2n-1} (x).
\end{equation}

From here, setting $x=3/2$ in \eqref{main_id5} and using $U_{n-1}(3/2)=F_{2n}$, we get
\begin{equation*}
	\sum_{k=1}^n  \frac k{n + k}\binom{n + k}{n - k} = \frac{F_{2n}}{2}
\end{equation*}
and
\begin{equation*}
	\sum_{k=1}^n (-5)^{k-1} \frac k{n + k}\binom{n + k}{n - k} = \frac{(-1)^{n-1}}{2} F_{2n}.
\end{equation*}

Also, with $x=\sqrt{5}/2$ upon combining we produce
\begin{equation*}
	\sum_{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,
\end{equation*}
where we used the fact that
\begin{displaymath}
	U_n \Big (\frac{\sqrt{5}}{2} \Big ) =
	\begin{cases}
		L_{n+1}, &amp; \text{\rm if $n$ is even;} \\
		\sqrt{5} F_{n+1}, &amp; \text{\rm otherwise.}
	\end{cases}
\end{displaymath}

In general, working with \eqref{main_id2} we get the relations
\begin{displaymath}
	\sum_{k=1}^n (-2)^{k-1} \frac k{n + k}\binom{n + k}{n - k} \big(1 \mp T_m(x)\big)^{k-1} = (\pm 1)^{n-1} \frac{U_{nm-1}(x)}{2U_{m-1}(x)},
\end{displaymath}
where we have used that
\begin{displaymath}
U_{m-1}\big(T_n(x)\big)=\frac{U_{mn-1}(x)}{U_{n-1}(x)}.
\end{displaymath}
\begin{lemma}\label{Lem.hspwn3b}
If $n$ is a positive integer  and $p$ an integer, then we have the following identities involving Chebyshev polynomials of the second kind:
\begin{align*}
	U_n\Big(\frac{L_p}2\Big) &amp;= \frac{F_{p(n + 1)}}{F_p},\quad\text{\rm $p$ even},\,\, p\ne0,\\
	U_n\Big(\frac{iL_p}2\Big) &amp;= \frac{i^nF_{p(n + 1)}}{F_p},\quad\text{\rm $p$ odd},\\
	U_n \Big( {\frac{ \sqrt 5 F_p}{2}} \Big) &amp;=  \begin{cases}
	L_{p(n + 1)} /L_p,&amp;\text{\rm if  $p$ is odd and $n$ is even};  \\
	\sqrt5 F_{p(n + 1)}/L_p,&amp;\text{\rm if $p$ and  $n$ are odd};
	\end{cases}\\
	U_n \Big( {\frac{{i\sqrt 5 F_p  }}{2}}\Big) &amp;=  \begin{cases}
	i^nL_{p(n + 1)} /L_p,&amp;\text{\rm if $p$ and $n$ are even};  \\
	i^n\sqrt 5 F_{p(n + 1)} /L_p,&amp;\text{\rm if $p$ is even and $n$ is odd}.
	\end{cases}
\end{align*}
\end{lemma}
\begin{theorem}\label{thm.nl0qyn5}
If $n$ is a positive integer and $p$ is an integer, then we have
\begin{align*}
	\sum_{k = 1}^n (-1)^{(p-1)(n-k)} \frac {k}{n + k} \binom{n + k}{n - k} L_p^{2k - 1} &amp;= \frac{F_{2np}}{2F_p},\qquad p\ne0,\\
	\sum_{k = 1}^n (-1)^{p(n-k)} \frac {k}{n + k} \binom{n + k}{n - k} 5^{k-1} F_p^{2k - 1} &amp;= \frac{F_{2np}}{2L_p}.
	\end{align*}
\end{theorem}
\begin{proof}
Evaluate \eqref{main_id7} at $x=L_p/2$, $x=iL_p/2$, $x=\sqrt 5F_p/2$, and $x=i\sqrt 5F_p/2$, in turn, using Lemma \ref{Lem.hspwn3b}.
\end{proof}
\begin{lemma}
	If $n$ is a positive integer, then
	\begin{align}
\sqrt {\frac{5}{{2\alpha }}}\, U_{2n - 1} \bigg( {\sqrt {\frac{{\alpha ^5 }}{8}} } \bigg) &amp;= 2^n\alpha^n -(-1)^n \frac{{\beta^{n} }}{2^n},\nonumber\\
\sqrt {\frac{5}{{2\beta }}}\, U_{2n - 1} \bigg( {\sqrt {\frac{{\beta ^5 }}{8}} } \bigg) &amp;= 2^n\beta^n - (-1)^n\frac{{\alpha ^{n} }}{{2^n}}\label{eq.qcelehc}.
\end{align}
\end{lemma}
\begin{theorem}
If $n$ is a positive integer and $t$ is an integer, then we have
\begin{align*}
	\sum_{k = 1}^n ( -2)^{n-k} \frac{k}{{n + k}}\binom{n + k}{n - k}L_{5k + t} &amp;= \frac12\left( 4^{n} F_{ t +n+ 3} - F_{t-n+3}\right)\!,\\
	\sum_{k = 1}^n ( -2)^{n-k}\frac{k}{{n + k}}\binom{n + k}{n - k}F_{5k + t}  &amp;= \frac1{10}\left(4^{n} L_{t+n+3} - L_{n - t - 3}\right)\!.
\end{align*}
\end{theorem}
\begin{proof}
Set $x=\sqrt{\beta^5/8}$ in \eqref{main_id7} and use \eqref{eq.qcelehc} to obtain
\begin{displaymath}
	\sqrt 5 \sum_{k = 1}^n {\frac{{( - 1)^{k - 1} }}{{2^k }}\frac{k}{{n + k}}\binom{n + k}{n - k}\beta ^{5k + t}} = (- 2)^{n - 1} \beta ^{n + t + 3}  + \frac{{( - 1)^t }}{{2^{n + 1} }}\alpha ^{n - t - 3},
	\end{displaymath}
from which the results follow.
\end{proof}
\begin{theorem}
If $n$ is a non-negative integer and $t$ is an integer, then we have
\begin{align*}
	\sum_{k = 1}^n ( - 1)^{k - 1} {\frac{k}{{n + k}}\binom{n + k}{n - k} L_{3k + t} } &amp;=  \begin{cases}
	\frac{1}{2}L_{t + 3} L_n,&amp;\text{\rm if $n$ is odd};  \\
	- \frac{5}{2}F_{t + 3} F_n,&amp;\text{\rm otherwise};
	\end{cases}\\
	\sum_{k = 1}^n ( - 1)^{k - 1}{\frac{k}{{n + k}}\binom{n + k}{n - k} F_{3k + t} } &amp;=  \begin{cases}
	\frac{1}{2}F_{t + 3} L_n,&amp;\text{\rm if $n$ is odd};  \\
	-\frac{1}{2} L_{t + 3} F_n,&amp;\text{\rm otherwise}.
	\end{cases}
\end{align*}
\end{theorem}
\begin{proof}
Set $x=\sqrt{\alpha^3}/2$ in \eqref{main_id7} and use
\begin{displaymath}
	U_{2n - 1} \Big( {\frac{{\sqrt {\alpha ^3 } }}{2}} \Big) = \begin{cases}
	\sqrt {\alpha ^3 } L_n,&amp;\text{\rm  if $n$ is odd};  \\
	\sqrt {5\alpha ^3 } F_n, &amp;\text{\rm otherwise};
	\end{cases}
\end{displaymath}
to obtain
\begin{displaymath}
	\sum_{k = 1}^n ( - 1)^{k - 1} {\frac{k}{{n + k}}\binom{n + k}{n - k} \beta^{3k + t} } = \frac{\beta^{t + 3}}{2}\cdot\begin{cases}
	L_n,&amp;\text{if $n$ is odd};  \\
	\sqrt5 F_n, &amp;\text{otherwise};
	\end{cases}
	\end{displaymath}
from which the results follow.
\end{proof} 
\begin{theorem}
If $n$ is a non-negative integer and $t$ is an integer, then we have
\begin{align*}
	\sum_{k = 1}^n ( - 4)^{k - 1} {\frac{k}{{n + k}}\binom{n + k}{n - k} L_{k + t} } &amp;=  \begin{cases}
	-\frac12 L_{t + 1} L_{3n},&amp;\text{\rm if $n$ is odd};  \\
	\frac52F_{t + 1} F_{3n},&amp;\text{\rm otherwise};  \\
	\end{cases}\\
	\sum_{k = 1}^n ( - 4)^{k - 1}{\frac{k}{{n + k}}\binom{n + k}{n - k}F_{k + t} } &amp;=  \begin{cases}
	- \frac12 F_{t + 1} L_{3n},&amp;\text{\rm if $n$ is  odd};  \\
	\frac12 L_{t + 1} F_{3n},&amp;\text{\rm otherwise}.
	\end{cases}
\end{align*}
\end{theorem}
\begin{proof}
Set $x=\sqrt{\alpha}$ in \eqref{main_id7} and use
\begin{displaymath}
		U_{2n - 1}\big( \sqrt\alpha\big) =\begin{cases}
			\frac{\sqrt\alpha }2 L_{3n},&amp;\text{\rm if $n$ is odd};  \\[2pt]
			\frac{\sqrt{5\alpha} }2 F_{3n}, &amp;\text{\rm otherwise},
		\end{cases}
	\end{displaymath}
to obtain
	\begin{displaymath}
		\sum_{k = 1}^n ( - 4)^{k - 1}{\frac{k}{{n + k}}\binom{n + k}{n - k}\beta^{k + t} } = -\frac{\beta^{t + 1}}2\cdot\begin{cases}
			L_{3n},&amp;\text{if $n$ is odd};  \\
			\sqrt 5  F_{3n},&amp;\text{otherwise},
		\end{cases}
	\end{displaymath}
from which the results follow.
\end{proof}
\begin{theorem}
	If $n$ is a non-negative integer and $p$ is a non-zero integer, then we have
	\begin{displaymath}
		\sum_{k = 1}^n {( - 1)^{n-k} \frac{k}{{n + k}}\binom{n + k}{n - k}\left( {\frac{{2L_{2p} }}{\sqrt5 F_{2p}}} \right)} ^{2k} = ( - 1)^{p} \frac{L_{2p}(L_p^{4n} - 5^{2n} F_p^{4n} )}{4 (\sqrt5 F_{2p})^{2n}}.
	\end{displaymath}
\end{theorem}
\begin{proof}
	Set $x=L_{2p}/(\sqrt 5F_{2p})$ in \eqref{main_id7} and use
	\begin{equation}\label{kc1e6ox}
		U_{n - 1} \Big( \frac{{L_{2p} }}{\sqrt 5F_{2p}  } \Big) = \frac{{( - 1)^p(L_p^{2n}  - 5^n F_p^{2n} )}}{4(\sqrt 5F_{2p}  )^{n - 1} }.
	\end{equation}
\end{proof}
\begin{theorem}
	If $n$ is a non-negative integer and $p$ and $q$ are non-zero integers, then we have
	\begin{displaymath}
		\sum_{k = 1}^n ( - 1)^{(p - q)(n-k)} {\frac{k}{{n + k}}\binom{n + k}{n - k} \left( {\frac{{F_{p + q} F_{p - q} }}{{F_p F_q }}} \right)^{2k} } = \frac{{(F_p^{4n}  - F_q^{4n} )F_{p + q} F_{p - q} }}{2\big(F_p^2  + ( - 1)^{p - q} F_q^2 \big)(F_q F_p)^{2n} }.
	\end{displaymath}
\end{theorem}
\begin{proof}
	Set $x=\displaystyle\frac{(-i)^{p - q + 1}F_{p + q} F_{p - q}}{2 F_q F_p }$ in \eqref{main_id7} and use
	\begin{equation}\label{eq.ormac9f}
		U_{n - 1} \left( \frac{i^{p - q + 1} F_{p + q} F_{p - q}}{2F_qF_p} \right) = \frac{F_p^{2n}  + ( - 1)^{p - q} F_q^{2n} }{\big(F_p^2  + ( - 1)^{p - q} F_q^2\big)(F_qF_p)^{n - 1} }.
	\end{equation}
\end{proof}
\begin{lemma}
We have
	\begin{align}
	(-1)&amp;^{(q+1)(n+3/2)+\lfloor q/2\rfloor+1} \alpha ^{p + q/2} \,U_{2n - 1}\!\left( {\sqrt{\frac{( - 1)^{q + 1} F_q^2}{4F_p F_{p + q}}\alpha ^{2p + q} }} \right) \nonumber \\
	&amp; \quad= \frac{{F_{p + q}^{2n + 1} \alpha ^{qn + p} - F_p^{2n} F_{p + q} }(-\beta)^{qn - p}+F_{p + q}^{2n} F_p \alpha ^{qn + p + q} - F_p^{2n + 1} (-\beta) ^{qn - p - q} }{\big(F_p F_{p + q}\big)^{n-1/2}\big(F_{p + q}^2  + F_p F_{p + q} L_q  + ( - 1)^qF_p^2\big) },\label{eq.r6f98dl} \\
	(-1)&amp;^{(q+1)(n+5/2)+\lfloor q/2\rfloor+p} \beta ^{p + q/2}\, U_{2n - 1}\! \left( \sqrt{\frac{( - 1)^{q + 1}F_q^2 }{{4F_p F_{p + q} }}\beta ^{2p + q}} \right), \nonumber  \\
	&amp; \quad= \frac{F_{p + q}^{2n + 1} \beta ^{qn + p} - F_p^{2n} F_{p + q} (-\alpha)^{qn - p} + F_{p + q}^{2n} F_p \beta ^{qn + p + q} - F_p^{2n + 1} (-\alpha) ^{qn - p - q} }{\big(F_pF_{p + q}\big)^{n-1/2}\big(F_{p + q}^2 + F_p F_{p + q} L_q + ( - 1)^qF_p^2\big) }, \nonumber \\
	(-1)&amp;^{(q+1)(n+3/2)+\lfloor q/2\rfloor+1} \alpha ^{p + q/2}\, U_{2n - 1}\! \left( \sqrt{\frac{( - 1)^{q + 1}5 F_q^2 }{{4L_p L_{p + q} }}\alpha ^{2p + q}} \right) \nonumber  \\
	&amp;\quad = \frac{L_{p + q}^{2n + 1} \alpha ^{qn + p} - L_p^{2n} L_{p + q} (-\beta)^{qn - p} + L_{p + q}^{2n}L_p \alpha ^{qn + p + q} - L_p^{2n + 1} (-\beta) ^{qn - p - q} }{\big(L_pL_{p + q}\big)^{n-1/2} \big(L_{p + q}^2  + L_p L_{p + q} L_q  + ( - 1)^qL_p^2\big) }, \nonumber \\
	(-1)&amp;^{(q+1)(n+3/2)+p}  \beta ^{p + q/2} \,U_{2n - 1} \!\left( \sqrt{\frac{{( - 1)^{q + 1}5F_q^2 }}{4L_p L_{p + q} }\beta ^{2p + q}}\right) \nonumber  \\
	&amp;\quad = \frac{L_p^{2n+1}\beta^{qn+p}- L_{p}^{2n} L_{p+q}(-\alpha)^{qn - p} + L_{p+q}^{2n} L_{p}\beta ^{qn + p+q} -L_p^{2n + 1} (-\alpha)^{qn - p - q}}{(L_pL_{p + q})^{n-1/2}\big(L_{p + q}^2 + L_p L_{p + q} L_q + ( - 1)^qL_p^2\big)} \nonumber .
	\end{align}
\end{lemma}
\begin{theorem}\label{thm.kyu2oro}
	If $n$ is a positive integer, $p$, $q$, and $t$ are integers, then we have the following sum evaluations
	\begin{equation}
		\begin{split}
			\sum_{k = 1}^n &amp; ( - 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} \\
			&amp; = \frac{ F_{p + q}^{2n}\big(F_{p + q} L_{ p + t + qn} + F_p L_{ q + p + t + qn}\big) - F_p^{2n}\big( F_{p + q} L_{p + t - qn}
			+ F_p L_{ q + p + t -  qn}\big)}{2\big({F_{p + q}^2  + F_p F_{p + q} L_q  + ( - 1)^q F_p^2 }\big)},
			\label{eq.tpynnr7}
		\end{split}
	\end{equation}
	\begin{equation}
		\begin{split}
			\sum_{k = 1}^n &amp; ( - 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}\\
			&amp; = \frac{F_{p +q}^{2n}\big(F_{p + q} F_{ p + t +qn}   + F_p F_{q + p + t+qn }\big) - F_p^{2n}\big( F_{p + q} F_{p+t -qn}
			+ F_pF_{p + q + t - qn}\big) }{2\big(F_{p + q}^2  + F_p F_{p + q} L_q  + ( - 1)^q F_p^2 \big)}, \label{eq.gyr3ptz}
		\end{split}
	\end{equation}
	\begin{equation}
		\begin{split}
			\sum_{k = 1}^n &amp;( - 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^kF_{(2p + q)k + t} \\
			&amp; = \frac{ L_{p + q}^{2n}\big(L_{p + q} L_{ p + t + qn}   + L_p L_{q + p + t+qn}\big) -  L_p^{2n}\big( L_{p + q} L_{p+t -qn}
			+ L_p L_{p + q + t - qn}\big) }{2\big(L_{p + q}^2  + L_p L_{p + q} L_q  + ( - 1)^q L_p^2 \big)},
		\end{split}\label{eq.z2nmrct}
	\end{equation}
	and
	\begin{equation}
		\begin{split}
	\sum_{k = 1}^n &amp; ( - 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} \\
			&amp; = \frac{L_{p + q}^{2n}\big(L_{p + q} F_{qn + p + t} +  L_p F_{ q + p + t + qn}\big)  - L_p^{2n}\big( L_{p + q} F_{p+t-qn}
			+ L_p F_{p + q + t - qn }\big)}{{2\big(L_{p + q}^2  + L_p L_{p + q} L_q  + ( - 1)^q L_p^2 \big)}}.\label{eq.rankav6}
		\end{split}
	\end{equation}
\end{theorem}
\begin{proof}
In \eqref{main_id7} set
	$\displaystyle
	x=\sqrt{ \frac{( - 1)^{q + 1}F_q^2 }{{4F_p F_{p + q} }}\alpha ^{2p + q}}
	$
 	and use \eqref{eq.r6f98dl} to obtain
	\begin{align*}
			&amp;2\sum_{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}\, \alpha^{(2p + q)k + t} }
			= \frac{{F_q }}{2\big(F_{p + q}^2  + F_p F_{p + q} L_q  + ( - 1)^q F_p^2\big)} \\
			&amp;\qquad\qquad\quad\times\!\Bigl( {( - 1)^{nq} F_{p + q}^{2n + 1}\, 2\alpha^{qn + p + t} - (- 1)^{p + t} F_p^{2n} F_{p + q}\, 2\beta^{qn - p - t} } \\
			&amp;\qquad\qquad\qquad\quad  + ( - 1)^{nq} F_{p + q}^{2n} F_p\, 2\alpha^{qn + q + p + t} - (- 1)^{p + q + t} F_p^{2n + 1}\, 2\beta^{qn - p - q - t} \Bigr),
		\end{align*}
	from which \eqref{eq.tpynnr7} and \eqref{eq.gyr3ptz} follow. The proof of \eqref{eq.z2nmrct} and \eqref{eq.rankav6} is similar; in \eqref{main_id7}, set
	$\displaystyle
	x =\sqrt{ \frac{{( - 1)^{q + 1}5F_q^2 }}{{4L_p L_{p + q} }}\alpha ^{2p + q} }
	$.
\end{proof}


\section{Some binomial-coefficient weighted Fibonacci and \break Lucas sums}


In this section we present some Fibonacci and Lucas sums having binomial coefficients as weights.
\begin{theorem}
	If $n$ is a positive integer and $p$ is an integer, then we have the following identities
	\begin{align*}
		\sum_{k = 0}^n (- 1)^{(p-1)(n-k)}\binom{n + k}{n - k}  L_p^{2k} &amp;= \frac{F_{(2n + 1)p}}{F_p}, \\
		\sum_{k = 0}^n ( - 1)^{p(n-k)} \binom{n + k}{n - k}  5^k F_p^{2k} &amp;= \frac{L_{(2n + 1)p}}{L_p}.
	\end{align*}
\end{theorem}
\begin{proof}
Addition of each identity stated in Theorem \ref{thm_xxx} and its counterpart in Theorem \ref{thm.nl0qyn5} while making use of
$L_rF_s + L_sF_r = 2F_{r + s}$ and $L_rL_s + 5F_rF_s = 2L_{r + s}$ gives the proof.
\end{proof}

The identities stated below in Theorem \ref{thm.kihhjta} follow immediately upon addition of each identity in Theorem~\ref{thm.awwyoac} to the respective corresponding identity in Theorem \ref{thm.kyu2oro}.
\begin{theorem}\label{thm.kihhjta}
If $n$ is a non-negative integer and $p$, $q$ and $t$ are integers, then we have
\begin{align*}
\sum_{k = 0}^n &amp;( - 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}\left(F_{p + q}^{2n} L_{t+qn}  + F_p^{2n} L_{t-qn}\right) \\
	&amp;\quad +\frac{F_q\big( 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}) \big)}{{2\big(F_{p + q}^2  + F_p F_{p + q} L_q  + ( - 1)^q F_p^2\big) }}, \\
\sum_{k = 0}^n &amp;( - 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}\left(F_{p + q}^{2n} F_{t+qn}  - F_p^{2n} F_{t-qn}\right) \\
	&amp;\quad +\frac{F_q\big( 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}) \big)}{2\big(F_{p + q}^2  + F_p F_{p + q} L_q + ( - 1)^q F_p^2\big)}, \\
\sum_{k = 0}^n &amp; ( - 1)^{q(n-k)} \binom{n + k}{n - k}L_p^{n - k} L_{p + q}^{n - k} F_q^{2k} 5^kF_{(2p + q)k + t} = \frac{1}{2}\left(L_{p + q}^{2n} F_{t+qn}  + L_p^{2n}F_{t-qn}\right) \\
	&amp;\quad +\frac{F_q\big( 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_pL_{q + p + t- qn}) \big)} {2\big(L_{p + q}^2  + L_p L_{p + q} L_q  + ( - 1)^q L_p^2\big) }, \\
\sum_{k = 0}^n &amp;( - 1)^{q(n-k)} \binom{n + k}{n - k}L_p^{n - k} L_{p + q}^{n - k} F_q^{2k} 5^kL_{(2p + q)k + t} = \frac{1}{2}\left(L_{p + q}^{2n} L_{t+qn}  + L_p^{2n} L_{t-qn}\right) \\
	&amp; \quad+\frac{5F_q\left(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})  \right)}{2(L_{p + q}^2  + L_p L_{p + q} L_q + ( - 1)^q L_p^2)}.
\end{align*}
\end{theorem}
\begin{example}
If $n$ is a non-negative integer and $t$ is an integer, then we have
\begin{align*}
\sum_{k = 0}^n {( - 1)^{n-k} \binom{n + k}{n - k}L_{3k + t} } &amp;= \begin{cases}
			L_{t + n}+  L_{t + 1} L_n,&amp;\text{\rm if $n$ is odd};  \\
			L_{t + n}  + 5F_{t + 1} F_n,&amp;\text{\rm otherwise};  \\
		\end{cases} \\
\sum_{k = 0}^n {( - 1)^{n-k} \binom{n + k}{n - k}F_{3k + t} } &amp;= \begin{cases}
			F_{t + n} + F_{t + 1} L_n,&amp;\text{\rm  if $n$ is odd};  \\
			F_{t + n}  + L_{t + 1} F_n,&amp;\text{\rm otherwise}.
		\end{cases}
\end{align*}
In particular, 
\begin{align*}
\sum_{k = 0}^n {( - 1)^{n-k} \binom{n + k}{n - k} L_{3k} } &amp;= \begin{cases}
			2L_n ,&amp;\text{\rm if $n$ is odd};  \\
			2L_{n + 1},&amp;\text{\rm otherwise};  \\
		\end{cases} \\
\sum_{k = 0}^n {( - 1)^{n-k} \binom{n + k}{n - k}F_{3k} } &amp;= \begin{cases}
			2F_{n + 1},&amp;\text{\rm if $n$  is odd};  \\
			2F_n,&amp;\text{\rm otherwise};
		\end{cases}\\
\sum_{k = 0}^n {( - 1)^{n-k} \binom{n + k}{n - k}L_{3k - 1} } &amp;= \begin{cases}
			L_{n + 2} ,&amp;\text{\rm if $n$ is  odd};  \\
			L_{n - 1},&amp;\text{\rm otherwise};  \\
		\end{cases}\\
\sum_{k = 0}^n {( - 1)^{n-k} \binom{n + k}{n - k}F_{3k - 1} } &amp;= \begin{cases}
			F_{n - 1},&amp;\text{\rm if $n$ is odd};  \\
			F_{n + 2},&amp;\text{\rm otherwise}.
		\end{cases}
\end{align*}
\end{example}
\begin{example}
If $n$ is a non-negative integer and $t$ is an integer, then we have
	\begin{align*}
		\sum_{k = 0}^n {( -2)^{n-k} \binom{n + k}{n - k} L_{5k + t} } &amp;= 2^{2n + 1} F_{ t +n +  1} - F_{t-n},\\
		\sum_{k = 0}^n {( - 2)^{n-k} \binom{n + k}{n - k} F_{5k + t} } &amp;= \frac15\left( 2^{2n + 1} L_{ t +n+ 1}- L_{t- n}\right)\!.
	\end{align*}
\end{example}

The identities in this section and many similar results can be obtained directly from Lemma \ref{lem.x6c6bl9} which is a consequence of \eqref{eq.w83lskz}, \eqref{main_id4}, and \eqref{main_id7}.
\begin{lemma}\label{lem.x6c6bl9}
If $n$ is a non-negative integer and $x$ is a complex variable, then
\begin{equation}\label{eq.f8j08d9}
	\sum_{k=0}^n{(-4)^{k}\binom{n + k}{n - k}x^{2k}} = (-1)^nU_{2n}(x).
\end{equation}
\end{lemma}
\begin{theorem}
If $n$ is a non-negative integer and $p$ is a non-zero integer, then we have
\begin{equation*}
	\sum_{k = 0}^n {( - 1)^{n - k} \binom{n + k}{n - k}\left( {\frac{2L_{2p}}{\sqrt5F_{p}}} \right)} ^{2k}
	= (- 1)^{p}\,\frac{L_p^{4n + 2} - 5^{2n + 1} F_p^{4n + 2}}{4\cdot5^n F_{2p}^{2n}  }.
	\end{equation*}
\end{theorem}
\begin{proof}
Set $ x=L_{2p}/(\sqrt 5 F_{2p})$ in \eqref{eq.f8j08d9} and use \eqref{kc1e6ox}.
\end{proof}
\begin{theorem}
If $n$ is a non-negative integer and $p$ and $q$ are non-zero integers, then we have
\begin{equation*}
	\sum_{k = 0}^n {( - 1)^{(p - q)(n-k)} \binom{n + k}{n - k} \left( {\frac{{F_{p + q} F_{p - q} }}{{F_p F_q }}} \right)^{2k} }
	= \frac{{F_p^{4n + 2} + ( - 1)^{p - q} F_q^{4n + 2}}}{(F_qF_p)^{2n}(F_p^2 + ( - 1)^{p - q} F_q^2 )}.
\end{equation*}
\end{theorem}
\begin{proof}
Set  $\displaystyle x=\frac{(-i)^{p - q + 1}F_{p + q} F_{p - q}}{2 F_q F_p}$ in \eqref{eq.f8j08d9} and use \eqref{eq.ormac9f}.
\end{proof}


\section{Related combinatorial sums}


This section does not deal with Fibonacci numbers but is inspired by a recent paper by Chu and Guo \cite{ChuGuo}. In this paper the authors study combinatorial sums of the form
$\sum\limits_{k=0}^n (-1)^k \frac{\binom {n+\lambda}{2k+\delta}}{\binom {n}{k}}$, where $\lambda$ and $n$ are nonnegative integers,  $\delta\in\{0;1\}$. We show how these sums are related to the sums studied in this paper via Theorem \ref{thm1}.
\begin{theorem}\label{comb_thm}
	For $0&lt;n+m\leq 2n+1$, we have the relation
	\begin{equation*}
		\sum_{k=0}^{n+m}  \frac{(-2)^k}{(n+m+k)(n-m+k+2)}\frac{\binom {n+m+k}{n+m-k}}{\binom {n-m+k+1}{k}}
		= \frac{1}{2(n+1)(n+m) }\sum_{k=0}^n (-1)^k \frac{\binom {n+m}{2k}}{\binom {n}{k}}.
	\end{equation*}
In particular, 
\begin{align*}
	\sum_{k=0}^{n} \frac{(-2)^k}{(n+k)(n+k+1)(n+k+2)}\frac{\binom {n+k}{n-k}}{\binom {n+k}{k}}
	&amp;= \frac{1}{2n(n+1)^2 }\sum_{k=0}^n (-1)^k \frac{\binom {n}{2k}}{\binom {n}{k}},\\
\sum_{k=0}^{2n}  \frac{(-2)^k}{(2n+k)(k+1)(k+2)} \binom {2n+k}{2n-k}
&amp;=\frac{1}{4n(n+1)} \sum_{k=0}^n (-1)^k \frac{\binom {2n}{2k}}{\binom {n}{k}}.
\end{align*}
\end{theorem}
\begin{proof}
From Theorem \ref{thm1} upon replacing $n$ by $n+m$ we get
\begin{equation*}
	\sum_{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).
	\end{equation*}
Multiplying through by $x^{n-m+1}$ and integrating from 0 to 1 results in
\begin{equation*}
	\sum_{k=0}^{n+m} (-2)^k \frac{n+m}{n + m + k} \binom{n + m + k}{n + m - k} \int_0^1 x^{n-m+1} (1-x)^k dx = \int_0^1 x^{n-m+1} T_{n+m}(x)dx.
\end{equation*}

The left-hand side evaluated using the beta function $B(a,b)$ given by
\begin{equation*}
	B(a,b) = \int_0^1 x^{a-1} (1-x)^{b-1} dx,
	\end{equation*}
	and equals
	\begin{equation*}
		\sum_{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)
	\end{equation*}
	with
	\begin{equation*}
		B(n-m+2,k+1) = \frac{1}{n-m+k+2}\binom {n-m+k+1}{k}^{-1}.
	\end{equation*}

The right-hand side is evaluated similarly using the representation from \eqref{ChebRep1}. We have
\begin{align*}
	\int_0^1 x^{n-m+1} T_{n+m}(x)dx &amp; =  \sum_{k=0}^{\lfloor{(n+m)/2}\rfloor} (-1)^k \binom {n+m}{2k} \int_0^1 x^{2n-2k+1} (1-x^2)^k dx \\
	&amp; = \frac{1}{2} \sum_{k=0}^{\lfloor{(n+m)/2}\rfloor} (-1)^k \binom {n+m}{2k} \int_0^1 z^{n-k} (1-z)^k dz \qquad (z=x^2) \\
	&amp; = \frac{1}{2} \sum_{k=0}^{\lfloor{(n+m)/2}\rfloor} (-1)^k \binom {n+m}{2k} B(n-k+1,k+1) \\
	&amp; = \frac{1}{2(n+1)} \sum_{k=0}^{\lfloor{(n+m)/2}\rfloor} (-1)^k \frac{\binom {n+m}{2k}}{\binom {n}{k}}.
\end{align*}
The final expression follows as $\binom {n}{k}=0$ if $k&gt;n$.
\end{proof}

Keeping in mind that
$$
\frac{n}{n + k} \binom{n + k}{n - k} = \frac{\binom {n+k-1}{k} \binom {n}{k}}{\binom {2k}{k}},
$$
Theorem \ref{comb_thm} is actually an identity containing four binomial coefficients
\begin{equation*}
	\sum_{k=0}^{n+m}  \frac{(-2)^k}{n-m+k+2}\frac{\binom {n+m+k-1}{k}}{\binom {n-m+k+1}{k}}\frac{\binom {n+m}{k}}{\binom {2k}{k}}
	=\frac{1}{2(n+1)}\sum_{k=0}^n (-1)^k \frac{\binom {n+m}{2k}}{\binom {n}{k}}.
\end{equation*}

The special case $m=0$ gives the combinatorial relation valid for all $n\geq 1$
\begin{equation*}
	\sum_{k=0}^{n}  \frac{(-2)^k}{(n+k)(n+k+1)(n+k+2)} \frac{\binom {n}{k}}{\binom {2k}{k}}
	= \frac{1}{2n(n+1)} - \frac{1}{4(n+1)^2} \sum_{k=0}^n \frac{1}{\binom {n}{k}},
\end{equation*}
as Chu and Guo have shown \cite[Prop.~5]{ChuGuo} that
\begin{equation*}
	\sum_{k=0}^n (-1)^k \frac{\binom {n}{2k}}{\binom {n}{k}} = n + 1 - \frac{n}{2} \sum_{k=0}^n \frac{1}{\binom {n}{k}}.
\end{equation*}

We continue with some more examples based on Chu and Guo's results.
\begin{example}
	When $m=1,2,3,4$, then we can use the following evaluation from \cite{ChuGuo}
	\begin{align*}
	\sum_{k=0}^n (-1)^k \frac{\binom {n+1}{2k}}{\binom {n}{k}} &amp;= 2 - \sum_{k=0}^n \frac{1}{\binom {n}{k}}, \\
	\sum_{k=0}^n (-1)^{k+1} \frac{\binom {n+2}{2k}}{\binom {n}{k}} &amp;= \frac{2}{n}, \\
	\sum_{k=0}^n (-1)^{k+1} \frac{\binom {n+3}{2k}}{\binom {n}{k}} &amp;= \frac{2(n-3)}{(n-1)n}, \\
	\sum_{k=0}^n (-1)^{k+1} \frac{\binom {n+4}{2k}}{\binom {n}{k}} &amp;= \frac{2(n^2-7n+16)}{(n-2)(n-1)n},
	\end{align*}
to get the following summation formulas:
	\begin{align*}
	\sum_{k=0}^{n+1} \frac{(-2)^k }{(n+1+k)^2}\frac{\binom {n+1+k}{n+1-k}}{\binom {n+k}{k}}
		&amp;= \frac{1}{2(n+1)^2}\Big ( 2 - \sum_{k=0}^n \frac{1}{\binom {n}{k}} \Big ),\qquad n\geq 0,\\
	\sum_{k=0}^{n+2}  \frac{(-2)^{k} }{(n+k)(n+2+k)}\frac{\binom {n+2+k}{n+2-k}}{\binom {n-1+k}{k}}
		&amp;= \frac{-1}{n(n+1)(n+2)},\qquad n\geq 1,\\
	\sum_{k=0}^{n+3}  \frac{(-2)^{k} }{(n-1+k)(n+3+k)}\frac{\binom {n+3+k}{n+3-k}}{\binom {n-2+k}{k}}
		&amp;= \frac{3-n}{(n-1)n(n+1)(n+3)},\qquad n\geq 2,\\
	\sum_{k=0}^{n+4}  \frac{(-2)^{k} }{(n-2+k)(n+4+k)}\frac{\binom {n+4+k}{n+4-k}}{\binom {n-3+k}{k}}
		&amp;= \frac{-n^2+7n-16}{(n-2)(n-1)n(n+1)(n+4)}, \qquad n\geq 3.
	\end{align*}
\end{example}

Working with the second part of Theorem \ref{thm1}, applying the same arguments but using the elementary integral
\begin{equation*}
	\int_0^1 x^{n-m+1} (1+x)^k dx = \sum_{j=0}^k  \frac{\binom {k}{j}}{n-m+2+j}
\end{equation*}
yields the next theorem involving a double sum.
\begin{theorem}\label{comb_thm2}
	For $m-1\leq n$, we have the relation
	\begin{align*}
	\sum_{j=0}^{n+m}\sum_{k=0}^{n+m-j} &amp; \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} \\
		&amp; \qquad = \frac{(-1)^{n+m}}{2(n+1)(n+m) } \sum_{k=0}^n (-1)^k \frac{\binom {n+m}{2k}}{\binom {n}{k}}.
	\end{align*}
	In particular, 
	\begin{align*}
		\sum_{j=0}^{n}\sum_{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}
		&amp;= \frac{(-1)^{n}}{2n(n+1)} \sum_{k=0}^n (-1)^k \frac{\binom {n}{2k}}{\binom {n}{k}},\\
		\sum_{j=0}^{2n}\sum_{k=0}^{2n-j}  \frac{(-2)^{k+j}}{(2n+k+j)(j+2)} \binom {2n+k+j}{2n-k-j} \binom {k+j}{j}
		&amp;= \frac{1}{4n(n+1)}\sum_{k=0}^n (-1)^k \frac{\binom {2n}{2k}}{\binom {n}{k}}.
	\end{align*}
\end{theorem}

Similar double sums involving ratios of binomial coefficients were studied recently by Stenlund and Wan \cite{Sten}.

The analogous identities involving $\binom{n+m}{2k+1}$ should follow from \eqref{main_id5} in conjunction with \eqref{ChebRep2}.


\section{Conclusion}

In this paper, we have derived a huge number of new combinatorial identities involving the binomial coefficients
$$
\frac{n}{n+k} \binom{n+k}{n-k}\quad \text{and} \quad\frac{k}{n+k}\binom{n+k}{n-k},$$
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 \ref{thm.awwyoac}, \ref{thm.kyu2oro}, and \ref{thm.kihhjta} 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 \cite{ChuGuo}.

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


\section{Acknowledgments}

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

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\end{thebibliography}
\bigskip
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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B37.


\noindent \emph{Keywords:} Fibonacci  number, Lucas number, Chebyshev polynomial, binomial coefficient.

\bigskip
\hrule
\bigskip

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


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received August 5 2023;
revised version received December 5 2023.
Published in {\it Journal of Integer Sequences}, December 11 2023.

\bigskip
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\noindent
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