Generalizations for reciprocal Fibonacci-Lucas sums of Brousseau\thanks{AMS Classification Numbers : 11B37, 11B39
\noindent We derive closed form expressions for finite and infinite \mbox{Fibonacci-Lucas sums having products of Fibonacci or Lucas numbers in the denominator of the summand. Our results generalize and extend those obtained by pioneer Brother Alfred Brousseau and later researchers.
1703.06075.tex

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\title{Generalizations for reciprocal Fibonacci-Lucas sums of Brousseau\thanks{AMS Classification Numbers : 11B37, 11B39}\vspace{10mm}} 
%\title{Generalizations for reciprocal Fibonacci-Lucas sums of Brousseau\thanks{AMS Classification Numbers : 11B37, 11B39}\vspace{10mm}\\ {\small opened:~{March 7, 2017}\\modified:~\today ,~\currenttime}} 

\author[]{Kunle Adegoke \thanks{kunle.adegoke@yandex.com, adegoke00@gmail.com}}
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\affil{Department of Physics and Engineering Physics, \mbox{Obafemi Awolowo University}, Ile-Ife, Nigeria}
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\begin{document}

\date{}

\maketitle


\begin{abstract}
\noindent We derive closed form expressions for finite and infinite \mbox{Fibonacci-Lucas} sums having products of Fibonacci or Lucas numbers in the denominator of the summand. Our results generalize and extend those obtained by pioneer Brother Alfred Brousseau and later researchers.
\end{abstract}

\tableofcontents
% \listoffigures
%\chead{\thepage}
\section{Introduction}


The Fibonacci numbers, $F_n$, and Lucas numbers, $L_n$, are defined, for $n\in\Nzero$, as usual, through the recurrence relations \mbox{$F_n=F_{n-1}+F_{n-2}$}, with $F_0=0$, $F_1=1$ and \mbox{$L_n=L_{n-1}+L_{n-2}$}, with $L_0=2$, $L_1=1$.

\bigskip

Our main aim in this paper is to derive closed form expressions for the following sums and their corresponding alternating versions, for positive integers $m$, $n$ and $q$:
\[
\sum_{k = 1}^\infty  {\frac{{L_{nk + nq} L_{nk + 2nq}  \cdots L_{nk + (m - 1)nq} }}{{F_{nk} F_{nk + nq}  \cdots F_{nk + mnq} }}} ,\quad\sum_{k = 1}^\infty  {\frac{{F_{nk + nq} F_{nk + 2nq}  \cdots F_{nk + (m - 1)nq} }}{{L_{nk} L_{nk + nq}  \cdots L_{nk + mnq} }}}\,,\quad m>1\,, 
\]
\[
\sum_{k = 1}^\infty  {\frac{1}{{F_{nk} F_{nk + nq}  \cdots F_{nk + mnq - nq} F_{nk + mnq + nq}  \cdots F_{nk + mnq + 2mnq} }}}\,, 
\]
\[
\sum_{k = 1}^\infty  {\frac{1}{{L_{nk} L_{nk + nq}  \cdots L_{nk + mnq - nq} L_{nk + mnq + nq}  \cdots L_{nk + mnq + 2mnq} }}}\,, 
\]
\[
\sum_{k = 1}^\infty  {\frac{{L_{nk + mnq} }}{{F_{nk} F_{nk + nq}  \cdots F_{nk + 2mnq} }}} ,\quad\sum_{k = 1}^\infty  {\frac{{F_{nk + mnq} }}{{L_{nk} L_{nk + nq}  \cdots L_{nk + 2mnq} }}}\,, 
\]
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq} }}{{F_{nk}^2 F_{nk + nq}^2 F_{nk + 2nq}^2  \cdots F_{nk + mnq}^2 }}} ,\quad\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq} }}{{L_{nk}^2 L_{nk + nq}^2 L_{nk + 2nq}^2  \cdots L_{nk + mnq}^2 }}}\,, 
\]
\[
\sum_{k = 1}^\infty  {\frac{{L_{nk + mnq} }}{{F_{nk}^2 F_{nk + nq}^2  \cdots F_{nk + (m - 1)nq}^2 F_{nk + mnq} F_{nk + (m + 1)q}^2  \cdots F_{nk + 2mnq}^2 }}}\,, 
\]
\[
\sum_{k = 1}^\infty  {\frac{{F_{nk + mnq} }}{{L_{nk}^2 L_{nk + nq}^2  \cdots L_{nk + (m - 1)nq}^2 L_{nk + mnq} L_{nk + (m + 1)q}^2  \cdots L_{nk + 2mnq}^2 }}}\,, 
\]
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq} L_{nk + nq}^2 L_{nk + 2nq}^2  \cdots L_{nk + (m - 1)nq}^2 }}{{F_{nk}^2 F_{nk + nq}^2 F_{nk + 2nq}^2  \cdots F_{nk + mnq}^2 }}}\,,\quad m>1\,.
\]

\bigskip

We require the following telescoping summation identities (see~\cite{adegoke})
\begin{equation}\label{equ.p2fefzx}
\sum_{k = 1}^N {\left[ {f(k) - f(k + q)} \right]}  = \sum_{k = 1}^q {f(k)}  - \sum_{k = 1}^q {f(k + N)},\quad\mbox{for $N\ge q\in\Nzero$}
\end{equation}
and
\begin{equation}\label{equ.mdgx80r}
\begin{split}
&\sum_{k = 1}^N {( - 1)^{k - 1} \left[ {f(k) + ( - 1)^{q - 1} f(k + q)} \right]}\\ 
&\quad= \sum_{k = 1}^q {( - 1)^{k - 1} f(k)}  + ( - 1)^{N - 1} \sum_{k = 1}^q {( - 1)^{k - 1} f(k + N)}\,. 
\end{split}
\end{equation}

In general, infinite sums are evaluated using
\begin{equation}\label{equ.piqoita}
\begin{split}
&\sum_{k = 1}^\infty  {\left[ {f(k) - f(k + q)} \right]},\qquad q\in\Nzero\\
&\qquad= \sum_{k = 1}^q {f(k)}  - \sum_{k = 1}^q {\mathop {\lim }_{N \to \infty } f(k + N)}
\end{split}
\end{equation}
and
\begin{equation}\label{equ.jcfrlss}
\sum_{k = 1}^\infty  {( - 1)^{k - 1} \left[ {f(k) \mp f(k + q)} \right]}  = \sum_{k = 1}^q {( - 1)^{k - 1} f(k)}\,, 
\end{equation}
where the upper sign is taken if $q$ is even and the lower if $q$ is odd.

\bigskip


If $f(N)$ approaches zero as $N$ approaches infinity, then we have, from~\eqref{equ.p2fefzx} and~\eqref{equ.mdgx80r}, the useful identities
\begin{equation}\label{equ.gy1asjs}
\sum_{k = 1}^\infty  {\left[ {f(k) - f(k + q)} \right]}  = \sum_{k = 1}^q {f(k)},\quad\mbox{$q\in\Nzero$}\,, 
\end{equation}
\begin{equation}
\sum_{k = 1}^\infty  {( - 1)^{k - 1} \left[ {f(k) \mp f(k + q)} \right]}  = \sum_{k = 1}^q {( - 1)^{k - 1} f(k)}\,,
\end{equation}
where the upper sign applies if $q$ is even and the lower if $q$ is odd.

\bigskip

The golden ratio, having the numerical value of $(\sqrt 5+1)/2$, is denoted in this paper by $\phi$. 


\bigskip

We shall require the following identities (most of which can be found in the book by Vajda~\cite{vajda}):
\begin{subequations}\label{equ.d8ulygu}
\begin{eqnarray}
L_vF_u &=& F_{u + v} + ( - 1)^vF_{u-v}\label{equ.mzt8c69} \\
F_vL_u &=& F_{u + v} - (-1)^vF_{u-v}\label{equ.yb05ue2} 
\end{eqnarray}
\end{subequations}
\begin{subequations}\label{equ.c7qhgv4}
\begin{eqnarray}
2F_{u + v} &=& L_vF_u + L_uF_v \label{equ.ejrnkwy}\\ 
(-1)^v2F_{u-v} &=& F_uL_v - L_uF_v\label{equ.moytk3x}
\end{eqnarray}
\end{subequations}
\begin{subequations}\label{equ.epvyp3u}
\begin{eqnarray}
L_vL_u &=& L_{u + v} + (-1)^vL_{u-v}\label{equ.ztsb3uk}\\ 
5F_vF_u &=& L_{u + v} - (-1)^vL_{u-v}\label{equ.q79u3q4}  
\end{eqnarray}
\end{subequations}

\begin{equation}\label{equ.yr8t5vs}
(-1)^{u-1}(F_{v+u} F_{v-u})=F_u^2(F_{v+1}F_{v-1}) - F_v^2(F_{u+1} F_{ u-1})
\end{equation}


\begin{subequations}\label{equ.hn7lvbv}
\begin{eqnarray}
( -1)^tF_u F_v=F_{t+u}F_{t+v} - F_tF_{t+u+v}\label{equ.tokbcvq}\\ 
( -1)^{t+1}5F_u F_v=L_{t+u}L_{t+v} - L_tL_{t+u+v} \label{equ.owip3z8}  
\end{eqnarray}
\end{subequations}

\begin{subequations}\label{equ.kp0k485}
\begin{eqnarray}
F_{u-v}F_{u+v} &=& F_u^2 + (-1)^{u+v-1}F_v^2\label{equ.ded0k7c}\\ 
5F_{u-v}F_{u+v} &=& L_u^2 + (-1)^{u+v-1}L_v^2\label{equ.wodrq78}  
\end{eqnarray}
\end{subequations}

\begin{subequations}\label{equ.loghvyf}
\begin{eqnarray}
F_v F_{2u + v + p} &=& F_{u + v + p} F_{u + v}  + ( - 1)^{v + 1} F_{u + p} F_u  \label{equ.cqtsjoj}\\ 
F_v L_{2u + v + p} &=& L_{u + v + p} F_{u + v}  + ( - 1)^{v + 1} L_{u + p} F_u \label{equ.dt4czzk}  
\end{eqnarray}
\end{subequations}

The identities~\eqref{equ.kp0k485} and~\eqref{equ.loghvyf} were proved by Howard in~\cite{howard}.
 
\bigskip

The following limiting values are readily established using Binet's formula:
\begin{subequations}\label{equ.y4ibpqi}
\begin{eqnarray}
\mathop {\lim }_{N \to \infty } \frac{{F_{N + m} }}{{F_{N + n} }} &=& \phi ^{m - n}  = \mathop {\lim }_{N \to \infty } \frac{{L_{N + m} }}{{L_{N + n} }} ,\\
\mathop {\lim }_{N \to \infty } \frac{{F_{N + m} }}{{L_{N + n} }} &=& \frac{{\phi ^{m - n} }}{{\sqrt 5 }} = \frac{1}{5}\mathop {\lim }_{N \to \infty } \frac{{L_{N + m} }}{{F_{N + n} }}\,.  
\end{eqnarray}
\end{subequations}

\bigskip

We shall adopt the following conventions for empty sums and empty products:
\[
\sum_{k = 1}^0 {f(k)}  = 0,\quad\prod_{k = 1}^0 {f(k)}  = 1\,.
\]

\section{Results: Generalizations}

\subsection{Telescoping summation identities}
\begin{lemma}\label{finall}
If $m$, $n$, $q$ and $N$ are positive integers and $f(k)$ is a real sequence, then
\[
\begin{split}
&\sum_{k = 1}^N {\left\{ {\left[ {f(nk) - f(nk + mnq)} \right]\prod_{j = 1}^{m - 1} {f(nk + jnq)} } \right\}}\\
&\qquad= \sum_{k = 1}^{q} {\left\{ {\prod_{j = 0}^{m - 1} {f(nk + jnq)} } \right\}}  - \sum_{k = 1}^{q} {\left\{ {\prod_{j = 0}^{m - 1} {f(nk + nN + jnq)} } \right\}}\,.
\end{split}
\]
\end{lemma}
If the sequence $f(k)$ is convergent and we denote by $f_\infty$ the limiting value of $f(nN)$ as $N$ approaches infinity, we have
\begin{equation}\label{infall}
\begin{split}
&\sum_{k = 1}^\infty  {\left\{ {\left[ {f(nk) - f(nk + mnq)} \right]\prod_{j = 1}^{m - 1} {f(nk + jnq)} } \right\}}\\
&\qquad= \sum_{k = 1}^{q} {\left\{ {\prod_{j = 0}^{m - 1} {f(nk + jnq)} } \right\}}  - f_\infty{}^mq\,.
\end{split}
\end{equation} 
\begin{proof}
We have
\[
\begin{split}
&\left[ {f(nk) - f(nk + mnq)} \right]\prod_{j = 1}^{m - 1} {f(nk + jnq)}\\
&\qquad= f(nk)\prod_{j = 1}^{m - 1} {f(nk + jnq)}  - f(nk + mnq)\prod_{j = 1}^{m - 1} {f(nk + jnq)}\\
&\qquad = \prod_{j = 0}^{m - 1} {f(nk + jnq)}  - \prod_{j = 1}^{m} {f(nk + jnq)}
\end{split}
\]
\begin{equation}\label{equ.tejlmvh}
\begin{split}
&\qquad= \prod_{j = 0}^{m - 1} {f(nk + jnq)}  - \prod_{j = 0}^{m - 1} {f(nk + jnq + nq)}\\
&\qquad\equiv \prod_{j = 0}^{m - 1} {f(nk + jnq)}  - \left. {\prod_{j = 0}^{m - 1} {f(nk + jnq)} } \right|_{k \to k + q}\,.
\end{split}
\end{equation}
The result follows by summing both sides of identity~\eqref{equ.tejlmvh}, using the identity~\eqref{equ.p2fefzx} to perform the telescopic summation on the right hand side.



\end{proof}
\begin{lemma}\label{finqeven}
If $f(k)$ is a real sequence and $m$, $n$, $q$ and $N$ are positive integers such that $q$ is even, then
\[
\begin{split}
&\sum_{k = 1}^N {( - 1)^{k - 1} \left\{ {\left[ {f(nk) - f(nk + mnq)} \right]\prod_{j = 1}^{m - 1} {f(nk + jnq)} } \right\}}\\
&\quad= \sum_{k = 1}^q {( - 1)^{k - 1} \left\{ {\prod_{j = 0}^{m - 1} {f(nk + jnq)} } \right\}}\\
&\qquad + ( - 1)^{N - 1} \sum_{k = 1}^q {( - 1)^{k - 1} \left\{ {\prod_{j = 0}^{m - 1} {f(nk + nN + jnq)} } \right\}}\,. 
\end{split}
\]
\end{lemma}
\begin{proof}
Multiply through the identity~\eqref{equ.tejlmvh} by $(-1)^{k-1}$ and use identity~\eqref{equ.mdgx80r}.
\end{proof}
\begin{lemma}\label{finqodd}
If $f(k)$ is a real sequence and $m$, $n$, $q$ and $N$ are positive integers such that $q$ is odd, then
\[
\begin{split}
&\sum_{k = 1}^N {( - 1)^{k - 1} \left\{ {\left[ {f(nk) + f(nk + mnq)} \right]\prod_{j = 1}^{m - 1} {f(nk + jnq)} } \right\}}\\
&\quad= \sum_{k = 1}^q {( - 1)^{k - 1} \left\{ {\prod_{j = 0}^{m - 1} {f(nk + jnq)} } \right\}}\\
&\qquad + ( - 1)^{N - 1} \sum_{k = 1}^q {( - 1)^{k - 1} \left\{ {\prod_{j = 0}^{m - 1} {f(nk + nN + jnq)} } \right\}}\,. 
\end{split}
\]


\end{lemma}
\begin{proof}
We have the identity
\begin{equation}
\begin{split}
&\left[ {f(nk) + f(nk + mnq)} \right]\prod_{j = 1}^{m - 1} {f(nk + jnq)}\\
&\qquad= \prod_{j = 0}^{m - 1} {f(nk + jnq)}  + \prod_{j = 0}^{m - 1} {f(nk + jnq + nq)}\\
&\qquad\equiv \prod_{j = 0}^{m - 1} {f(nk + jnq)}  + \left. {\prod_{j = 0}^{m - 1} {f(nk + jnq)} } \right|_{k \to k + q}\,,
\end{split}
\end{equation}
from which the result follows after multiplying through by $(-1)^{k-1}$ and summing over $k$, making use of the identity~\eqref{equ.mdgx80r}.
\end{proof}
If the sequence $f(k)$ is convergent and $f(2Nn)$ and $f((2N-1)n)$ both have the same limiting value as $N$ approaches infinity, we have
\begin{equation}\label{infallalt}
\begin{split}
&\sum_{k = 1}^\infty {( - 1)^{k - 1} \left\{ {\left[ {f(nk) \mp f(nk + mnq)} \right]\prod_{j = 1}^{m - 1} {f(nk + jnq)} } \right\}}\\
&\quad= \sum_{k = 1}^q {( - 1)^{k - 1} \left\{ {\prod_{j = 0}^{m - 1} {f(nk + jnq)} } \right\}}\,,
\end{split}
\end{equation}
where the upper sign is taken if $q$ is even and the lower if $q$ is odd.

\subsection{Sums with $F_{nk}F_{nk+nq}\cdots F_{nk+mnq}$ or \\\mbox{$F_{nk}F_{nk+nq}\cdots F_{nk+(m-1)nq}F_{nk+(m+1)nq}\cdots F_{nk+2mnq}$} in the denominator}\label{sec.u8lgixm}
\begin{thm}\label{thm.vn8ph53}
If $m$, $n$ and $q$ are positive integers, then
\[
\sum_{k = 1}^\infty  {\left[ {( - 1)^{nk - 1} \frac{{\prod_{j = 1}^{m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right]}  = \frac{{q\sqrt{ 5^m} }}{{2F_{mnq} }} - \frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} }\,, 
\]
so that
\begin{equation}
\sum_{k = 1}^\infty  {\left[ {\frac{{\prod_{j = 1}^{m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right]}  = \frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} }  - \frac{{q\sqrt {5^m } }}{{2F_{mnq} }},\quad\mbox{$n$ even}
\end{equation}
and
\begin{equation}
\sum_{k = 1}^\infty  {\left[ {( - 1)^{k - 1} \frac{{\prod_{j = 1}^{m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right]}  = \frac{{q\sqrt {5^m } }}{{2F_{mnq} }} - \frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} } ,\quad\mbox{$n$ odd}\,.
\end{equation}
\end{thm}
In particular,
\begin{equation}\label{equ.xk94y11}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{nk - 1} }}{{F_{nk} F_{nk + nq} }}}  = \frac{{q\sqrt 5 }}{{2F_{nq} }} - \frac{1}{{2F_{nq} }}\sum_{k = 1}^q {\frac{{L_{nk} }}{{F_{nk} }}}\,.
\end{equation}
Brousseau's result (equation~(3) of~\cite{brousseau2}, also rederived in various equivalent forms by other authors, see for example reference~\cite{rabinowitz}) corresponds to setting $n=1$ in~\eqref{equ.xk94y11}, but with a different, but equivalent, form for the right hand side. Bruckman and Good's result (equation~(19) of~\cite{bruckman}) is also a special case of~\eqref{equ.xk94y11}, corresponding to setting $q=1$.
\begin{thm}\label{thm.hxvgf6w}
If $m$, $n$ and $q$ are integers such that $n$ is odd and $q$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{\prod_{j = 1}^{m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right]}  = \frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\left[ {( - 1)^k \prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} } \right]}\,. 
\]
\end{thm}
In particular,
\begin{equation}\label{equ.lldx63d}
\sum_{k = 1}^\infty  {\frac{1}{{F_{nk} F_{nk + nq} }} = \frac{1}{{2F_{nq} }}\sum_{k = 1}^q {\left[ {( - 1)^k \frac{{L_{nk} }}{{F_{nk} }}} \right]} },\quad\mbox{$n$ odd, $q$ even}\,,
\end{equation}
which generalizes the result obtained by Rabinowitz (the second of equation~(26) of~\cite{rabinowitz}), the latter corresponding to the special case $n=1$ in the identity~\eqref{equ.lldx63d}, but with a different, but equivalent, form for the right hand side.
\subsubsection*{Proof of Theorem~\ref{thm.vn8ph53} and Theorem~\ref{thm.hxvgf6w}}
Dividing through the identity~\eqref{equ.moytk3x} by $F_uF_v$ and setting $u=nk+mnq$ and $v=nk$, the following identity is established for $k$, $m$, $n$ and $q$ positive integers:
\[
\frac{{( - 1)^{nk - 1} 2F_{mnq} }}{{F_{nk} F_{nk + mnq} }} = \frac{{L_{nk + mnq} }}{{F_{nk + mnq} }} - \frac{{L_{nk} }}{{F_{nk} }}\,.
\]
Using $f(k)=L_k/F_k$ in identity~\eqref{finall} we get the finite summation identity
\begin{equation}\label{equ.oy4215f}
\begin{split}
&2F_{mnq} \sum_{k = 1}^N {\left[ {( - 1)^{nk - 1} \frac{{\prod_{j = 1}^{m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right]}\\
&\quad  = \sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + nN + jnq} }}{{F_{nk + nN + jnq} }}} } \right]}  - \sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} } \right]}\,,
\end{split}
\end{equation}
which yields Theorem~\ref{thm.vn8ph53} in the limit $N$ approaches infinity. Theorem~\ref{thm.hxvgf6w} is proved by using $f(k)=L_k/F_k$ in identity~\eqref{infallalt}.

\bigskip

\begin{thm}\label{thm.xd29ih0}
If $m$, $n$ and $q$ are positive \underline{odd} integers, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + njq} } \prod_{j = m + 1}^{2m} {F_{nk + njq} } }}} \right]}  = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{2m - 1} {F_{nk + njq} } }}} \right]}\,. 
\]

\end{thm}
Below are a few explicit examples from Theorem~\ref{thm.xd29ih0}:
\begin{equation}
\begin{split}
&\mbox{At $m=1$}:\\
&\sum_{k = 1}^\infty  {\frac{1}{{F_{nk} F_{nk + 2nq} }} = \frac{1}{{L_{nq} }}\sum_{k = 1}^q {\frac{1}{{F_{nk} F_{nk + nq} }}} },\quad\mbox{$nq$ odd}\,.
\end{split}
\end{equation}
\[
\begin{split}
&\mbox{At $(m,n,q)=(1,1,1)$ and $(m,n,q)=(1,1,3)$:}\\
&\sum_{k = 1}^\infty  {\frac{1}{{F_k F_{k + 2} }}}  = 1,\quad \sum_{k = 1}^\infty  {\frac{1}{{F_k F_{k + 6} }}}  =\frac{143}{960}\,,
\end{split}
\]
corresponding to Formulas~(4) and~(6) of Brousseau in reference~\cite{brousseau1}.
\[
\begin{split}
&\mbox{At $(m,n,q)=(3,1,3)$:}\\
&\sum_{k = 1}^\infty  {\frac{1}{{F_k F_{k + 3} F_{k + 6} F_{k + 12} F_{k + 15} F_{k + 18} }}}  = \frac{{{\rm 938359017897442612}}}{{{\rm 5579104720519492358676480}}}\,.
\end{split}
\]
\[
\begin{split}
&\mbox{At $(m,n,q)=(5,3,1)$:}\\
&\sum_{k = 1}^\infty  {\frac{1}{{F_{3k} F_{3k + 3} F_{3k + 6} F_{3k + 9} F_{3k + 12} F_{3k + 18} F_{3k + 21} F_{3k + 24} F_{3k + 27} F_{3k + 30} }}}\\
&\qquad= \frac{{\rm 1}}{{{\rm 13970032097862115517068710877593600}}}\,.
\end{split}
\]
\begin{thm}\label{thm.i3apd20}
If $q$, $m$ and $n$ are positive integers such that $q$ is odd and $nm$ is even, then
\[
\begin{split}
\sum_{k = 1}^\infty {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + njq} } \prod_{j = m + 1}^{2m} {F_{nk + njq} } }}} \right]}  &= \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{2m - 1} {F_{nk + njq} } }}} \right]}\,.
\end{split}
\]


\end{thm}
Examples from Theorem~\ref{thm.i3apd20} include
\begin{equation}
\begin{split}
&\mbox{At $m=2$}:\\
&\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} }}{{F_{nk} F_{nk + nq} F_{nk + 3nq} F_{nk + 4nq} }} = \frac{1}{{L_{2nq} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{F_{nk} F_{nk + nq} F_{nk + 2nq} F_{nk + 3nq} }}} },\quad\mbox{$q$ odd}\,.
\end{split}
\end{equation}
\[
\begin{split}
&\mbox{At $(m,n,q)=(1,2,1)$ and $(m,n,q)=(2,1,1)$:}\\
&\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} }}{{F_{2k} F_{2k + 4} }}}  = \frac{1}{9},\quad\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} }}{{F_k F_{k + 1} F_{k + 3} F_{k + 4} }}}  = \frac{1}{18}\,.
\end{split}
\]
\[
\begin{split}
&\mbox{At $(m,n,q)=(2,6,1)$:}\\
&\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} }}{{F_{6k} F_{6k + 6} F_{6k + 18} F_{6k + 24} }}}  = \frac{1}{{{\rm 44444622716928}}}\,.
\end{split}
\]


\subsubsection*{Proof of Theorem~\ref{thm.xd29ih0} and Theorem~\ref{thm.i3apd20}}
With $v=mnq$ and $u=nk+mnq$ in identity~\eqref{equ.mzt8c69}, the following identity is established:
\begin{equation}
\frac{{L_{mnq} F_{nk + mnq} }}{{F_{nk} F_{nk + 2mnq} }} = \frac{1}{{F_{nk} }} + \frac{{( - 1)^{mnq} }}{{F_{nk + 2mnq} }}\,,
\end{equation}
so that
\begin{equation}\label{equ.eo7cizi}
\frac{{L_{mnq} F_{nk + mnq} }}{{F_{nk} F_{nk + 2mnq} }} = \frac{1}{{F_{nk} }} - \frac{1}{{F_{nk + 2mnq} }},\quad\mbox{$mnq$ odd}
\end{equation}
and
\begin{equation}\label{equ.w1u72rp}
\frac{{L_{mnq} F_{nk + mnq} }}{{F_{nk} F_{nk + 2mnq} }} = \frac{1}{{F_{nk} }} + \frac{1}{{F_{nk + 2mnq} }},\quad\mbox{$mnq$ even}\,.
\end{equation}
If $m$, $n$ and $q$ are positive \underline{odd} integers, then from~\eqref{equ.eo7cizi} and using $f(k)=1/F_k$ in Lemma~\ref{finall} (with $m\to 2m$), we have the following definite summation identity
\begin{equation}
\begin{split}
&\sum_{k = 1}^N {\left[ {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq} } \prod_{j = m + 1}^{2m} {F_{nk + jnq} } }}} \right]}\\
&\qquad = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{2m - 1} {F_{nk + jnq} } }}} \right]}  - \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{2m - 1} {F_{nk + nN + jnq} } }}} \right]}\,,
\end{split}
\end{equation}
from which Theorem~\ref{thm.xd29ih0} follows as $N$ approaches infinity.

\bigskip

If $q$ is an odd positive integer and either $m$ or $n$ is even, then from~\eqref{equ.w1u72rp} and Lemma~\ref{finqodd} (with $m\to 2m$) we have the summation identity
\begin{equation}
\begin{split}
&\sum_{k = 1}^N {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + njq} } \prod_{j = m + 1}^{2m} {F_{nk + njq} } }}} \right]}\\
&\qquad= \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{2m - 1} {F_{nk + njq} } }}} \right]}
+\frac{{( - 1)^{N - 1} }}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{2m - 1} {F_{nk + nN + njq} } }}} \right]}\,,
\end{split}
\end{equation}
from which Theorem~\ref{thm.i3apd20} follows in the limit that $N$ approaches infinity.
\begin{thm}\label{thm.cgjiqk9}
If $m$, $n$ and $q$ are positive integers such that $q$ is odd, then
\[
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + mnq} \prod_{j = 1}^{m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}}  = \frac{1}{2}\sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} }\,. 
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + nq} }}{{F_{nk} F_{nk + nq} }}}  = \frac{1}{2}\sum_{k = 1}^q {( - 1)^{k - 1} \frac{{L_{nk} }}{{F_{nk} }}},\quad\mbox{$q$ odd}\,.
\end{equation}
\begin{cor}\label{thm.ee6ees5}
If $m$, $n$ and $q$ are positive integers such that $q$ is odd, then
\[
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} L_{nk + mnq}^2 \prod_{j = 1}^{m - 1} {L_{nk + jnq} } \prod_{j = m + 1}^{2m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq} } \prod_{j = m + 1}^{2m} {F_{nk + jnq} } }}}  = \frac{1}{2}\sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{2m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} }\,.
\]

\end{cor}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} L_{nk + nq}^2 }}{{F_{nk} F_{nk + 2nq} }}}  = \frac{1}{2}\sum_{k = 1}^q {( - 1)^{k - 1} \frac{{L_{nk} L_{nk + nq} }}{{F_{nk} F_{nk + nq} }}},\quad\mbox{$q$ odd}\,.
\end{equation}
\subsubsection*{Proof of Theorem~\ref{thm.cgjiqk9} and Corollary~\ref{thm.ee6ees5}}
Dividing through the identity~\eqref{equ.ejrnkwy} by $F_uF_v$ and setting $u=nk+mnq$ and $v=nk$ we have the identity
\begin{equation}\label{equ.klrep5j}
2\frac{{F_{2nk + mnq} }}{{F_{nk} F_{nk + mnq} }} = \frac{{L_{nk + mnq} }}{{F_{nk + mnq} }} + \frac{{L_{nk} }}{{F_{nk} }}\,.
\end{equation}
If $q$ is odd, then from~\eqref{equ.klrep5j} and with $f(k)=L_k/F_k$ in Lemma~\ref{finqodd} we have
\begin{equation}\label{equ.bo7m7gb}
\begin{split}
&2\sum_{k = 1}^N {\frac{{( - 1)^{k - 1} F_{2nk + mnq} \prod_{j = 1}^{m - 1} {L_{nk + jnq} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}}\\
&\qquad\qquad= \sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq} }}{{F_{nk + jnq} }}} }  + ( - 1)^{N - 1} \sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{L_{nk + nN + jnq} }}{{F_{nk + nN + jnq} }}} }\,,
\end{split}
\end{equation}
from which Theorem~\eqref{thm.cgjiqk9} follows in the limit as $N$ approaches infinity. Corollary~\eqref{thm.ee6ees5} is obtained by specifically requiring $m$ to be even in Theorem~\eqref{thm.cgjiqk9}.
\begin{thm}\label{thm.f9a8imt}
If $m$, $n$, $q$ and $p$ are positive integers, then
\[
\begin{split}
&\sum_{k = 1}^\infty  {\left\{ {\frac{{( - 1)^{nk - 1} \prod_{j = 1}^{m - 1} {F_{nk + jnq + np} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right\}}\\
&\qquad= \frac{{\phi ^{mnp} q}}{{F_{mnq} F_{np} }} - \frac{1}{{F_{mnq} F_{np} }}\sum_{k = 1}^q {\left\{ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + np} }}{{F_{nk + jnq} }}} } \right\}}\,.
\end{split}
\]

\end{thm}
In particular we have
\begin{equation}
\sum_{k = 1}^\infty  {\left\{ {\frac{{( - 1)^{nk - 1} \prod_{j = m + 1}^{p + m - 1} {F_{nk + jn} } }}{{\prod_{j = 0}^p {F_{nk + jn} } }}} \right\}}  = \frac{{\phi ^{mnp} }}{{F_{mn} F_{pn} }} - \frac{1}{{F_{mn} F_{pn} }}\prod_{j = 0}^{m - 1} {\frac{{F_{jn + np + n} }}{{F_{jn + n} }}}
\end{equation}
and
\begin{equation}\label{equ.b2p05i3}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{nk - 1} }}{{F_{nk} F_{nk + nq} }}}  = \frac{{\phi ^{n} q}}{{F_{nq} F_{n} }} - \frac{1}{{F_{nq} F_{n} }}\sum_{k = 1}^q {\frac{{F_{nk + n} }}{{F_{nk} }}}\,.
\end{equation}
Observe that identity~\eqref{equ.b2p05i3} is equivalent to identity~\eqref{equ.xk94y11} but with a different form for the right hand side. Since $2\phi=\sqrt 5+1$, \mbox{$F_{n-1}+F_{n+1}=L_n$} and $\phi^n = \phi F_n+F_{n-1}$, both identities can be combined to yield the following interesting summation identity which is valid for all non-zero integers~$n$ and non-negative integers~$q$:
\begin{equation}
\frac{1}{{F_n }}\sum\limits_{k = 1}^q {\left[ {\frac{{F_{nk + n} }}{{F_{nk} }}} \right]}  - \frac{1}{2}\sum\limits_{k = 1}^q {\left[ {\frac{{L_{nk} }}{{F_{nk} }}} \right]}  = \frac{q}{2}\frac{{L_n }}{{F_n }}\,.
\end{equation}
\begin{thm}\label{thm.vdv26oc}
If $m$, $n$, $q$ and $p$ are positive integers such that $n$ is odd and $q$ is even, then
\[
\begin{split}
&\sum_{k = 1}^\infty  {\left\{ {\frac{{ \prod_{j = 1}^{m - 1} {F_{nk + jnq + np} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right\}}\\
&\qquad=\frac{1}{{F_{mnq} F_{np} }}\sum_{k = 1}^q {\left\{ {( - 1)^{k}\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + np} }}{{F_{nk + jnq} }}} } \right\}}\,.
\end{split}
\]

\end{thm}
In particular,
\begin{equation}\label{equ.y9fdju9}
\sum_{k = 1}^\infty  {\frac{1}{{F_{nk} F_{nk + nq} }} = \frac{1}{{F_nF_{nq} }}\sum_{k = 1}^q {\left[ {( - 1)^k \frac{{F_{nk+n} }}{{F_{nk} }}} \right]} },\quad\mbox{$n$ odd, $q$ even}\,.
\end{equation}
From identity~\eqref{equ.lldx63d} and identity~\eqref{equ.y9fdju9} we have the interesting result
\begin{equation}
\frac{1}{2}\sum_{k = 1}^q {\left[ {( - 1)^k \frac{{L_{nk} }}{{F_{nk} }}} \right]}  = \frac{1}{{F_n }}\sum_{k = 1}^q {\left[ {( - 1)^k \frac{{F_{nk + n} }}{{F_{nk} }}} \right]}\,,\quad\mbox{$q$ even}\,.
\end{equation}
\subsection*{Proof of Theorem~\ref{thm.f9a8imt} and Theorem~\ref{thm.vdv26oc}}
Dividing through identity~\eqref{equ.tokbcvq} by $F_{t+u}F_t$ and choosing $t=nk$, $u=mnq$ and $v=np$ we obtain the identity:
\begin{equation}\label{equ.oi2ipiz}
\frac{{( - 1)^{nk - 1} F_{mnq} F_{np} }}{{F_{nk} F_{nk + mnq} }} = \frac{{F_{nk + mnq + np} }}{{F_{nk + mnq} }} - \frac{{F_{nk + np} }}{{F_{nk} }}\,.
\end{equation}
With $f(k)=F_{k+np}/F_k$ in Lemma~\ref{finall} and using the identity~\eqref{equ.oi2ipiz}, we have the finite summation identity
\begin{equation}
\begin{split}
&F_{mnq} F_{np} \sum_{k = 1}^N {\left\{ {\frac{{( - 1)^{nk - 1} \prod_{j = 1}^{m - 1} {F_{nk + jnq + np} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right\}}\\
&\quad= \sum_{k = 1}^q {\left\{ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + nN + jnq + np} }}{{F_{nk + nN + jnq} }}} } \right\}}  - \sum_{k = 1}^q {\left\{ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + np} }}{{F_{nk + jnq} }}} } \right\}}\,,
\end{split}
\end{equation}
from which Theorem~\ref{thm.f9a8imt} follows in the limit as $N$ approaches infinity. 

\bigskip

Using $f(k)=F_{k+np}/F_k$ in Lemma~\ref{finqeven} gives
\begin{equation}
\begin{split}
&F_{mnq} F_{np} \sum_{k = 1}^N {\left\{ {\frac{{\prod_{j = 1}^{m - 1} {F_{nk + jnq + np} } }}{{\prod_{j = 0}^m {F_{nk + jnq} } }}} \right\}}\\ 
&\quad = ( - 1)^N \sum_{k = 1}^q {\left\{ {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{F_{nk + nN + jnq + np} }}{{F_{nk + nN + jnq} }}} } \right\}}\\ 
&\qquad - \sum_{k = 1}^q {\left\{ {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + np} }}{{F_{nk + jnq} }}} } \right\}}\,, 
\end{split}
\end{equation}
from which Theorem~\ref{thm.vdv26oc} follows.
\subsection{Sums with $L_{nk}L_{nk+nq}\cdots L_{nk+mnq}$ or\\ $L_{nk}L_{nk+nq}\cdots L_{nk+mnq-nq}L_{nk+mnq+nq}\cdots L_{nk+2mnq}$ in the denominator}
The derivations here proceed in the same fashion as in the previous section. The theorems will therefore be stated without proof. The analogous identity to~\eqref{equ.oy4215f} is 
\begin{equation}\label{equ.rcy3566}
\begin{split}
&2F_{mnq} \sum_{k = 1}^N {\left[ {( - 1)^{nk - 1} \frac{{\prod_{j = 1}^{m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right]}\\
&\quad  = \sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} } \right]}-\sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + nN + jnq} }}{{L_{nk + nN + jnq} }}} } \right]}\,.
\end{split}
\end{equation}
\begin{thm}\label{thm.v8d2lop}
If $m$, $n$ and $q$ are positive integers, then
\[
\sum_{k = 1}^\infty  {\left[ {( - 1)^{nk - 1} \frac{{\prod_{j = 1}^{m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right]}  = \frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} }-\frac{{q }}{{2F_{mnq}\sqrt{ 5^m} }}\,, 
\]
so that
\begin{equation}
\sum_{k = 1}^\infty  {\left[ {\frac{{\prod_{j = 1}^{m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right]}  = \frac{{q }}{{2F_{mnq}\sqrt{ 5^m} }}-\frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} },\quad\mbox{$n$ even}
\end{equation}
and
\begin{equation}
\sum_{k = 1}^\infty  {\left[ {( - 1)^{k - 1} \frac{{\prod_{j = 1}^{m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right]}  = \frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} }-\frac{{q }}{{2F_{mnq}\sqrt{ 5^m} }} ,\quad\mbox{$n$ odd}\,.
\end{equation}

\end{thm}
In particular,
\begin{equation}\label{equ.elnysff}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{nk - 1} }}{{L_{nk} L_{nk + nq} }}}  =\frac{1}{{2F_{nq} }}\sum_{k = 1}^q {\frac{{F_{nk} }}{{L_{nk} }}}-\frac{{q }}{{2F_{nq}\sqrt{ 5} }},\quad n,q\in\Zplus\,.
\end{equation}
The case $n=3,q=1$ in~\eqref{equ.elnysff} was mentioned by Brousseau (equation~(14) of~\cite{brousseau1}).
\begin{thm}\label{thm.llpbu3h}
If $m$, $n$ and $q$ are integers such that $n$ is odd and $q$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{\prod_{j = 1}^{m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right]}  = \frac{1}{{2F_{mnq} }}\sum_{k = 1}^q {\left[ {( - 1)^{k-1} \prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} } \right]}\,. 
\]
\end{thm}
In particular,
\begin{equation}\label{equ.x6dcfpg}
\sum_{k = 1}^\infty  {\frac{1}{{L_{nk} L_{nk + nq} }} = \frac{1}{{2F_{nq} }}\sum_{k = 1}^q {\left[ {( - 1)^{k-1} \frac{{F_{nk} }}{{L_{nk} }}} \right]} },\quad\mbox{$n$ odd, $q$ even}\,.
\end{equation}

\begin{thm}\label{thm.jysuj3n}
If $m$, $n$ and $q$ are positive \underline{odd} integers, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + njq} } \prod_{j = m + 1}^{2m} {L_{nk + njq} } }}} \right]}  = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{2m - 1} {L_{nk + njq} } }}} \right]}\,. 
\]


\end{thm}

\begin{thm}\label{thm.eniunx3}
If $q$, $m$ and $n$ are positive integers such that $q$ is odd and $nm$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{nk + njq} } \prod_{j = m + 1}^{2m} {L_{nk + njq} } }}} \right]}  = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{2m - 1} {L_{nk + njq} } }}} \right]}\,. 
\]
\end{thm}

Analogous identity to identity~\eqref{equ.bo7m7gb} is
\begin{equation}\label{equ.s11wekj}
\begin{split}
&2\sum_{k = 1}^N {\frac{{( - 1)^{k - 1} F_{2nk + mnq} \prod_{j = 1}^{m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}}\\
&\qquad\qquad= \sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} }  + ( - 1)^{N - 1} \sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{F_{nk + nN + jnq} }}{{L_{nk + nN + jnq} }}} }\,,
\end{split}
\end{equation}
from which we get the following theorem in the limit as $N$ approaches infinity.
\begin{thm}\label{thm.oe85fcz}
If $m$, $n$ and $q$ are positive integers such that $q$ is odd, then
\[
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + mnq} \prod_{j = 1}^{m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}}  = \frac{1}{2}\sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} }\,. 
\]
\end{thm}
\begin{cor}\label{thm.q95bigi}
If $m$, $n$ and $q$ are positive integers such that $q$ is odd, then
\[
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{nk + mnq}^2 \prod_{j = 1}^{m - 1} {F_{nk + jnq} } \prod_{j = m + 1}^{2m - 1} {F_{nk + jnq} } }}{{\prod_{j = 0}^{m - 1} {L_{nk + jnq} } \prod_{j = m + 1}^{2m} {L_{nk + jnq} } }}}  = \frac{1}{2}\sum_{k = 1}^q {( - 1)^{k - 1} \prod_{j = 0}^{2m - 1} {\frac{{F_{nk + jnq} }}{{L_{nk + jnq} }}} }\,.
\]

\end{cor}
Corresponding to identity~\eqref{equ.oi2ipiz} of section~\ref{sec.u8lgixm} we have (from identity~\eqref{equ.owip3z8})
\begin{equation}
\frac{{( - 1)^{nk - 1} 5F_{mnq} F_{np} }}{{L_{nk} L_{nk + mnq} }} = -\frac{{L_{nk + mnq + np} }}{{L_{nk + mnq} }} + \frac{{L_{nk + np} }}{{L_{nk} }}\,,
\end{equation}
leading to the summation identities
\begin{equation}
\begin{split}
&5F_{mnq} F_{np} \sum_{k = 1}^N {\left\{ {\frac{{( - 1)^{nk - 1} \prod_{j = 1}^{m - 1} {L_{nk + jnq + np} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right\}}\\
&\quad= -\sum_{k = 1}^q {\left\{ {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + nN + jnq + np} }}{{L_{nk + nN + jnq} }}} } \right\}}  + \sum_{k = 1}^q {\left\{ {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq + np} }}{{L_{nk + jnq} }}} } \right\}}
\end{split}
\end{equation}
and
\begin{equation}
\begin{split}
&5F_{mnq} F_{np} \sum_{k = 1}^N {\left\{ {\frac{{\prod_{j = 1}^{m - 1} {L_{nk + jnq + np} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right\}}\\ 
&\quad = ( - 1)^{N-1} \sum_{k = 1}^q {\left\{ {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{L_{nk + nN + jnq + np} }}{{L_{nk + nN + jnq} }}} } \right\}}\\ 
&\qquad + \sum_{k = 1}^q {\left\{ {( - 1)^{k - 1} \prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq + np} }}{{L_{nk + jnq} }}} } \right\}}\,, 
\end{split}
\end{equation}
from which Theorem~\ref{thm.lk7ts7r} and Theorem~\ref{thm.ec1t6t2} follow.
\begin{thm}\label{thm.lk7ts7r}
If $m$, $n$, $q$ and $p$ are positive integers, then
\[
\begin{split}
&\sum_{k = 1}^\infty  {\left\{ {\frac{{( - 1)^{nk - 1} \prod_{j = 1}^{m - 1} {L_{nk + jnq + np} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right\}}\\
&\qquad= -\frac{{\phi ^{mnp} q}}{{5F_{mnq} F_{np} }} + \frac{1}{{5F_{mnq} F_{np} }}\sum_{k = 1}^q {\left\{ {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq + np} }}{{L_{nk + jnq} }}} } \right\}}\,.
\end{split}
\]

\end{thm}
In particular we have
\begin{equation}
\sum_{k = 1}^\infty  {\left\{ {\frac{{( - 1)^{nk - 1} \prod_{j = m + 1}^{p + m - 1} {L_{nk + jn} } }}{{\prod_{j = 0}^p {L_{nk + jn} } }}} \right\}}  = -\frac{{\phi ^{mnp} }}{{5F_{mn} F_{pn} }} + \frac{1}{{5F_{mn} F_{pn} }}\prod_{j = 0}^{m - 1} {\frac{{L_{jn + np + n} }}{{L_{jn + n} }}}
\end{equation}
and
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{nk - 1} }}{{L_{nk} L_{nk + nq} }}}  = -\frac{{\phi ^{n} q}}{{5F_{nq} F_{n} }} + \frac{1}{{5F_{nq} F_{n} }}\sum_{k = 1}^q {\frac{{L_{nk + n} }}{{L_{nk} }}}\,.
\end{equation}


\begin{thm}\label{thm.ec1t6t2}
If $m$, $n$, $q$ and $p$ are positive integers such that $n$ is odd and $q$ is even, then
\[
\begin{split}
&\sum_{k = 1}^\infty  {\left\{ {\frac{{ \prod_{j = 1}^{m - 1} {L_{nk + jnq + np} } }}{{\prod_{j = 0}^m {L_{nk + jnq} } }}} \right\}}\\
&\qquad=\frac{1}{{5F_{mnq} F_{np} }}\sum_{k = 1}^q {\left\{ {( - 1)^{k-1}\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq + np} }}{{L_{nk + jnq} }}} } \right\}}\,.
\end{split}
\]

\end{thm}
In particular,
\begin{equation}\label{equ.a0f8aex}
\sum_{k = 1}^\infty  {\frac{1}{{L_{nk} L_{nk + nq} }} = \frac{1}{{5F_nF_{nq} }}\sum_{k = 1}^q {\left[ {( - 1)^{k-1} \frac{{L_{nk+n} }}{{L_{nk} }}} \right]} },\quad\mbox{$n$ odd, $q$ even}\,.
\end{equation}
From identity~\eqref{equ.x6dcfpg} and identity~\eqref{equ.a0f8aex} we have
\begin{equation}
\frac{1}{2}\sum_{k = 1}^q {\left[ {( - 1)^{k-1} \frac{{F_{nk} }}{{L_{nk} }}} \right]}  = \frac{1}{{5F_n }}\sum_{k = 1}^q {\left[ {( - 1)^{k-1} \frac{{L_{nk + n} }}{{L_{nk} }}} \right]}\,,\quad\mbox{$q$ even}\,.
\end{equation}

\subsection{Sums with $F_{nk}F_{nk+nq}\cdots F_{nk+2mnq}$ in the denominator}
The results in this section are obtained from identity~\eqref{equ.yb05ue2}. We have
\[
\frac{{F_v L_u }}{{F_{u - v} F_{u + v} }}=\frac{1}{{F_{u - v} }} - \frac{{( - 1)^v }}{{F_{u + v} }}\,,
\]
from which, by setting $v=mnq$ and $u=nk+mnq$, we get
\begin{equation}
\frac{{F_{mnq} L_{nk + mnq} }}{{F_{nk} F_{nk+2mnq} }} = \frac{1}{{F_{nk} }} - \frac{{( - 1)^{mnq} }}{{F_{nk+2mnq} }}\,,
\end{equation}
so that
\begin{equation}\label{equ.srgqhu2}
\frac{{F_{mnq} L_{nk + mnq} }}{{F_{nk} F_{nk+2mnq} }} = \frac{1}{{F_{nk} }} - \frac{1}{{F_{nk+2mnq} }},\quad\quad\mbox{$mnq$ even}
\end{equation}
and
\begin{equation}\label{equ.ih5bg09}
\frac{{F_{mnq} L_{nk + mnq} }}{{F_{nk} F_{nk+2mnq} }} = \frac{1}{{F_{nk} }} + \frac{1}{{F_{nk+2mnq} }},\quad\mbox{$mnq$ odd}\,.
\end{equation}
The derivations then proceed as in the previous sections.
\begin{thm}\label{thm.k2v2cn5}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{L_{nk + mnq} }}{{\prod_{j = 0}^{2m} {F_{nk + njq} } }}} \right]}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{2m - 1} {F_{nk + njq} } }}} \right]} 
\]

\end{thm}
Examples from Theorem~\ref{thm.k2v2cn5} include:
\begin{equation}\label{equ.qchc8p4}
\begin{split}
&\mbox{At $(m,n,q)=(1,2,1)$ and $(m,n,q)=(1,1,2)$:}\\
&\sum_{k = 1}^\infty  {\frac{{L_{2k + 2} }}{{F_{2k} F_{2k + 2} F_{2k + 4} }}}  = \frac{1}{3},\quad\sum_{k = 1}^\infty  {\frac{{L_{k + 2} }}{{F_k F_{k + 2} F_{k + 4} }}}  = \frac{5}{6}\,.
\end{split}
\end{equation}
The first of the identities in~\eqref{equ.qchc8p4} was also derived in~\cite{melham01} (equation~3.7).
\[
\begin{split}
&\mbox{At $(m,n,q)=(2,7,1)$:}\\
&\sum_{k = 1}^\infty  {\frac{{L_{7k + 14} }}{{F_{7k} F_{7k + 7} F_{7k + 14} F_{7k + 21} F_{7k + 28} }}}  = \frac{1}{{{\rm 6427623373464462}}}\,.
\end{split}
\]
\begin{thm}\label{thm.n3errmo}
If $m$, $n$ and $q$ are positive integers such that $q$ is even or $mnq$ is odd, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{(-1)^{k-1}L_{nk + mnq} }}{{\prod_{j = 0}^{2m} {F_{nk + njq} } }}} \right]}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{2m - 1} {F_{nk + njq} } }}} \right]} 
\]


\end{thm}
Examples from Theorem~\ref{thm.n3errmo} include:
\[
\begin{split}
&\mbox{At $(m,n,q)=(1,1,2)$ and $(m,n,q)=(1,3,2)$:}\\
&\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} L_{k + 2} }}{{F_k F_{k + 2} F_{k + 4} }}}  = \frac{1}{6}\,,\quad \sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} L_{3k + 6} }}{{F_{3k} F_{3k + 6} F_{3k + 12} }}}  = \frac{{271}}{{156672}}\,.
\end{split}
\]
\[
\begin{split}
&\mbox{At $(m,n,q)=(2,4,2)$:}\\
&\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} L_{4k + 16} }}{{F_{4k} F_{4k + 8} F_{4k + 16} F_{4k + 24} F_{4k + 32} }}}  = \frac{{{\rm 177072540680427}}}{{{\rm 166704475185956548320480}}}\,.
\end{split}
\]


\subsection{Sums with $L_{nk}L_{nk+nq}\cdots L_{nk+2mnq}$ in the denominator}
The results here follow from the identity~\eqref{equ.q79u3q4}.
\begin{thm}\label{thm.h6ute1y}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{F_{nk + mnq} }}{{\prod_{j = 0}^{2m} {L_{nk + jnq} } }}} \right]}  =\frac1{5F_{mnq}} \sum_{k = 1}^q {\left[ {\prod_{j = 0}^{2m - 1} {\frac{1}{{L_{nk + jnq} }}} } \right]}\,. 
\]
\end{thm}

\begin{thm}\label{thm.gkxbzjk}
If $m$, $n$ and $q$ are positive integers such that $q$ is even or $mnq$ is odd, then
\[
\sum_{k = 1}^\infty  {(-1)^{k-1}\left[ {\frac{{F_{nk + mnq} }}{{\prod_{j = 0}^{2m} {L_{nk + jnq} } }}} \right]}  = \frac1{5F_{mnq}}\sum_{k = 1}^q {\left[(-1)^{k-1} {\prod_{j = 0}^{2m - 1} {\frac{1}{{L_{nk + jnq} }}} } \right]}\,. 
\]
\end{thm}

\subsection{Sums with $F_{nk}F_{nk+2nq}F_{nk+4nq}F_{nk+6nq}\cdots F_{nk+2mnq}$ in the denominator}\label{sec.pgpoxnq}
\begin{thm}\label{thm.t7s7spm}
If $m$, $n$ and $q$ are positive \underline{odd} integers, then
\[
\sum_{k=1}^\infty \frac{(\pm1)^{k-1}F_{nk+mnq}}{\prod_{j=0}^m F_{nk+2jnq}}=\frac1{L_{mnq}}\sum_{k=1}^{2q}\frac{(\pm1)^{k-1}}{\prod_{j=0}^{m-1}F_{nk+2jnq}}\,.
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{F_{nk + nq} }}{{F_{nk} F_{nk + 2nq} }}}  = \frac{1}{{L_{nq} }}\sum_{k = 1}^{2q} {\frac{1}{{F_{nk} }}},\quad \mbox{$nq$ odd}\,.
\end{equation}
\begin{proof}
From identity~\eqref{equ.eo7cizi} and with $f(k)=1/F_k$ (and $q\to 2q$) in Lemma~\ref{finall} we have the finite summation identity
\begin{equation}\label{equ.u640lbl}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{F_{nk + mnq} }}{{\prod_{j = 0}^m {F_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ odd)}\\
&\qquad  - \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
From identity~\eqref{equ.eo7cizi} with $m$, $n$ and $q$ positive odd intergers and with $f(k)=1/F_k$ (and $q\to 2q$) in Lemma~\ref{finqeven} we have the alternating finite summation identity
\begin{equation}\label{equ.bdnv59h}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{( - 1)^{k - 1} F_{nk + mnq} }}{{\prod_{j = 0}^m {F_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ odd)}\\
&\qquad  + ( - 1)^{N - 1} \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + nN + 2jnq} } }}}
\end{split}
\end{equation}
Theorem~\ref{thm.t7s7spm} follows from identities~\eqref{equ.u640lbl} and~\eqref{equ.bdnv59h} in the limit as $N$ approaches infinity.
\end{proof}

\begin{thm}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k=1}^\infty \frac{(\pm1)^{k-1}L_{nk+mnq}}{\prod_{j=0}^m F_{nk+2jnq}}=\frac1{F_{mnq}}\sum_{k=1}^{2q}\frac{(\pm1)^{k-1}}{\prod_{j=0}^{m-1}F_{nk+2jnq}}\,.
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{L_{nk + nq} }}{{F_{nk} F_{nk + 2nq} }}}  = \frac{1}{{F_{nq} }}\sum_{k = 1}^{2q} {\frac{1}{{F_{nk} }}},\quad \mbox{$nq$ even}\,.
\end{equation}
\begin{proof}
As in Theorem~\ref{thm.t7s7spm}, with the identity~\eqref{equ.srgqhu2}, with $mnq$ even.

\bigskip

The corresponding finite summation identities are
\begin{equation}\label{equ.medhcfp}
\begin{split}
F_{mnq} \sum_{k = 1}^N {\frac{{L_{nk + mnq} }}{{\prod_{j = 0}^m {F_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ even)}\\
&\qquad  - \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + nN + 2jnq} } }}}
\end{split}
\end{equation}
and
\begin{equation}\label{equ.pmefb69}
\begin{split}
F_{mnq} \sum_{k = 1}^N {\frac{{( - 1)^{k - 1} L_{nk + mnq} }}{{\prod_{j = 0}^m {F_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ even)}\\
&\qquad  + ( - 1)^{N - 1} \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
\end{proof}

\subsection{Sums with $L_{nk}L_{nk+2nq}L_{nk+4nq}L_{nk+6nq}\cdots L_{nk+2mnq}$ in the denominator}
In this section we state the Lucas versions of the results given in section~\ref{sec.pgpoxnq}.

\bigskip

Here the basic identities (from identities~\eqref{equ.epvyp3u}) are:
\begin{equation}\label{equ.rgq8asw}
\frac{{L_{mnq} L_{nk + mnq} }}{{L_{nk} L_{nk + 2mnq} }} = \frac{1}{{L_{nk} }} - \frac{1}{{L_{nk + 2mnq} }},\quad\mbox{$mnq$ odd}
\end{equation}
\begin{equation}\label{equ.p4ho3jr}
\frac{{5F_{mnq} F_{nk + mnq} }}{{L_{nk} L_{nk + 2mnq} }} = \frac{1}{{L_{nk} }} - \frac{1}{{L_{nk + 2mnq} }},\quad\mbox{$mnq$ even}\,.
\end{equation}
\begin{thm}
If $m$, $n$ and $q$ are positive odd integers, then
\[
\sum_{k=1}^\infty \frac{(\pm1)^{k-1}L_{nk+mnq}}{\prod_{j=0}^m L_{nk+2jnq}}=\frac1{L_{mnq}}\sum_{k=1}^{2q}\frac{(\pm1)^{k-1}}{\prod_{j=0}^{m-1}L_{nk+2jnq}}\,.
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{L_{nk + nq} }}{{L_{nk} L_{nk + 2nq} }}}  = \frac{1}{{L_{nq} }}\sum_{k = 1}^{2q} {\frac{1}{{L_{nk} }}},\quad \mbox{$nq$ odd}\,.
\end{equation}

\begin{thm}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k=1}^\infty \frac{(\pm1)^{k-1}F_{nk+mnq}}{\prod_{j=0}^m L_{nk+2jnq}}=\frac1{5F_{mnq}}\sum_{k=1}^{2q}\frac{(\pm1)^{k-1}}{\prod_{j=0}^{m-1}L_{nk+2jnq}}\,.
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{F_{nk + nq} }}{{L_{nk} L_{nk + 2nq} }}}  = \frac{1}{{5F_{nq} }}\sum_{k = 1}^{2q} {\frac{1}{{F_{nk} }}},\quad \mbox{$nq$ even}\,.
\end{equation}
We have the following finite summation identities:
\begin{equation}\label{equ.r78m8j9}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{L_{nk + mnq} }}{{\prod_{j = 0}^m {L_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ odd)}\\
&\qquad  - \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
\begin{equation}\label{equ.ans99dd}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{( - 1)^{k - 1} L_{nk + mnq} }}{{\prod_{j = 0}^m {L_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {L_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ odd)}\\
&\qquad  + ( - 1)^{N - 1} \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {L_{nk + nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
\begin{equation}\label{equ.en25r4r}
\begin{split}
5F_{mnq} \sum_{k = 1}^N {\frac{{F_{nk + mnq} }}{{\prod_{j = 0}^m {L_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ even)}\\
&\qquad  - \sum_{k = 1}^{2q} {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
\begin{equation}\label{equ.qa1qok6}
\begin{split}
5F_{mnq} \sum_{k = 1}^N {\frac{{( - 1)^{k - 1} F_{nk + mnq} }}{{\prod_{j = 0}^m {L_{nk + 2jnq} } }}}  &= \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {L_{nk + 2jnq} } }}}\qquad\qquad\qquad\mbox{($mnq$ even)}\\
&\qquad  + ( - 1)^{N - 1} \sum_{k = 1}^{2q} {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {L_{nk + nN + 2jnq} } }}}\,.
\end{split}
\end{equation}

\subsection{Sums with $F_{2nk}F_{2nk+2nq}F_{2nk+4nq}F_{2nk+6nq}\cdots F_{2nk+2mnq}$ in the denominator}\label{sec.ffn2hfi}
\begin{thm}\label{thm.yj680l9}
If $m$, $n$ and $q$ are positive \underline{odd} integers, then
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\,. 
\]

\end{thm}
In particular,
\begin{equation}\label{equ.akqtcqn}
\sum_{k = 1}^\infty  {\frac{{F_{2nk + nq} }}{{F_{2nk} F_{2nk + 2nq} }}}  = \frac{1}{{L_{nq} }}\sum_{k = 1}^q {\frac{1}{{F_{2nk} }}}\,,
\end{equation}
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{F_{2nk + 3nq} }}{{F_{2nk} F_{2nk + 2nq} F_{2nk + 4nq} F_{2nk + 6nq} }}}  = \frac{1}{{L_{3nq} }}\sum_{k = 1}^q {\frac{1}{{F_{2nk} F_{2nk + 2nq} F_{2nk + 4nq} }}}\,.
\end{equation}
Identity~(18) of reference~\cite{frontczak16} with $n=0$ in their notation corresponds to setting $q=1$ in identity~\eqref{equ.akqtcqn}.
\begin{proof}
From identity~\eqref{equ.mzt8c69} with $v=mnq$ and $u=2nk+mnq$ comes the identity
\begin{equation}\label{equ.cx6zf5s}
\frac{{L_{mnq} F_{2nk + mnq} }}{{F_{2nk} F_{2nk + 2mnq} }} = \frac{1}{{F_{2nk} }} - \frac{1}{{F_{2nk + 2mnq} }},\quad\mbox{$mnq$ odd}\,.
\end{equation}
From identity~\eqref{equ.cx6zf5s} and Lemma~\ref{finall} with $f(k)=1/F_{2k}$ we have the finite summation identity:
\begin{equation}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{F_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\\
&\qquad  - \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2nN + 2jnq} } }}}\,,
\end{split}
\end{equation}
from which Theorem~\ref{thm.yj680l9} follows in the limit as $N$ approaches infinity.
\end{proof}
\begin{thm}\label{thm.jarxg8t}
If $m$, $n$ and $q$ are positive integers such that $q$ is odd and $mn$ is even , then
\[
\sum_{k = 1}^\infty  {\frac{(-1)^{k-1}F_{2nk + mnq} }{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\,. 
\]
\end{thm}
In particular,
\begin{equation}\label{equ.dn22fd6}
\sum_{k = 1}^\infty  {\frac{{(-1)^{k-1}F_{2nk + nq} }}{{F_{2nk} F_{2nk + 2nq} }}}  = \frac{1}{{L_{nq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{F_{2nk} }}}\,.
\end{equation}
\begin{proof}
From identity~\eqref{equ.mzt8c69} with $v=mnq$ and $u=2nk+mnq$ comes the identity
\begin{equation}\label{equ.dfe2pci}
\frac{{L_{mnq} F_{2nk + mnq} }}{{F_{2nk} F_{2nk + 2mnq} }} = \frac{1}{{F_{2nk} }} + \frac{1}{{F_{2nk + 2mnq} }},\quad\mbox{$mnq$ even}\,.
\end{equation}
From identity~\eqref{equ.dfe2pci} and Lemma~\ref{finqodd} with $f(k)=1/F_{2k}$ we have the finite summation identity:
\begin{equation}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{(-1)^{k-1}F_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\\
&\qquad  +(-1)^{N-1} \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2nN + 2jnq} } }}}\,,
\end{split}
\end{equation}
from which Theorem~\ref{thm.jarxg8t} follows in the limit as $N$ approaches infinity.
\end{proof}
\begin{thm}\label{thm.un8i8oz}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k = 1}^\infty  {\frac{{L_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\,. 
\]

\end{thm}
In particular,
\begin{equation}\label{equ.ci55vvi}
\sum_{k = 1}^\infty  {\frac{{L_{2nk + nq} }}{{F_{2nk} F_{2nk + 2nq} }}}  = \frac{1}{{F_{nq} }}\sum_{k = 1}^q {\frac{1}{{F_{2nk} }}}\,,
\end{equation}
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{L_{2nk + 3nq} }}{{F_{2nk} F_{2nk + 2nq} F_{2nk + 4nq} F_{2nk + 6nq} }}}  = \frac{1}{{F_{3nq} }}\sum_{k = 1}^q {\frac{1}{{F_{2nk} F_{2nk + 2nq} F_{2nk + 4nq} }}}\,.
\end{equation}
\begin{proof}
From identity~\eqref{equ.yb05ue2} with $v=mnq$ and $u=2nk+mnq$ comes the identity
\begin{equation}\label{equ.cagwzby}
\frac{{F_{mnq} L_{2nk + mnq} }}{{F_{2nk} F_{2nk + 2mnq} }} = \frac{1}{{F_{2nk} }} - \frac{1}{{F_{2nk + 2mnq} }},\quad\mbox{$mnq$ even}\,.
\end{equation}
From identity~\eqref{equ.cagwzby} and Lemma~\ref{finall} with $f(k)=1/F_{2k}$ we have the finite summation identity:
\begin{equation}
\begin{split}
F_{mnq} \sum_{k = 1}^N {\frac{{L_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\\
&\qquad  - \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2nN + 2jnq} } }}}\,,
\end{split}
\end{equation}
from which Theorem~\ref{thm.un8i8oz} follows in the limit as $N$ approaches infinity.
\end{proof}
\begin{thm}\label{thm.m05lv0w}
If $m$, $n$ and $q$ are positive integers such that $q$ is even \underline{or} $mnq$ is odd , then
\[
\sum_{k = 1}^\infty  {\frac{(-1)^{k-1}L_{2nk + mnq} }{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\,. 
\]
\end{thm}
In particular,
\begin{equation}\label{equ.rdxyneo}
\sum_{k = 1}^\infty  {\frac{{(-1)^{k-1}L_{2nk + nq} }}{{F_{2nk} F_{2nk + 2nq} }}}  = \frac{1}{{F_{nq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{F_{2nk} }}}\,,\quad\mbox{$q$ even \underline{or} $nq$ odd}\,.
\end{equation}
The alternating summation identity here, valid for $q$ even or $mnq$ odd, is
\begin{equation}
\begin{split}
F_{mnq} \sum_{k = 1}^N {\frac{{(-1)^{k-1}L_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2jnq} } }}}\\
&\qquad  +(-1)^{N-1} \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {F_{2nk + 2nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
\subsection{Sums with $L_{2nk}L_{2nk+2nq}L_{2nk+4nq}L_{2nk+6nq}\cdots L_{2nk+2mnq}$ in the denominator}\label{sec.esv6wcx}
The results in this section are derived from identities~\eqref{equ.epvyp3u}. The proofs are identical to those in section~\ref{sec.ffn2hfi} and are therefore omitted.
\begin{thm}
If $m$, $n$ and $q$ are positive \underline{odd} integers, then
\[
\sum_{k = 1}^\infty  {\frac{{L_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\,. 
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{L_{2nk + nq} }}{{L_{2nk} L_{2nk + 2nq} }}}  = \frac{1}{{L_{nq} }}\sum_{k = 1}^q {\frac{1}{{L_{2nk} }}}\,,\quad\mbox{$nq$ odd}\,,
\end{equation}
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{L_{2nk + 3nq} }}{{L_{2nk} L_{2nk + 2nq} L_{2nk + 4nq} L_{2nk + 6nq} }}}  = \frac{1}{{L_{3nq} }}\sum_{k = 1}^q {\frac{1}{{L_{2nk} L_{2nk + 2nq} L_{2nk + 4nq} }}}\,.
\end{equation}
The finite summation identity is
\begin{equation}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{L_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\qquad\qquad\mbox{$mnq$ odd}\\
&\qquad  - \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
\begin{thm}
If $m$, $n$ and $q$ are positive integers such that $q$ is odd and $mn$ is even , then
\[
\sum_{k = 1}^\infty  {\frac{(-1)^{k-1}L_{2nk + mnq} }{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  = \frac{1}{{L_{mnq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\,. 
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{(-1)^{k-1}L_{2nk + nq} }}{{L_{2nk} L_{2nk + 2nq} }}}  = \frac{1}{{L_{nq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{L_{2nk} }}}\,,\qquad \mbox{$q$ odd, $n$ even}\,.
\end{equation}
The alternating finite summation identity is
\begin{equation}
\begin{split}
L_{mnq} \sum_{k = 1}^N {\frac{{(-1)^{k-1}L_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\qquad\qquad\mbox{$q$ odd, $mn$ even}\\
&\qquad  +(-1)^{N-1} \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2nN + 2jnq} } }}}\,.
\end{split}
\end{equation}
\begin{thm}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  = \frac{1}{{5F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\,. 
\]

\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{F_{2nk + nq} }}{{L_{2nk} L_{2nk + 2nq} }}}  = \frac{1}{{5F_{nq} }}\sum_{k = 1}^q {\frac{1}{{L_{2nk} }}}\,,\qquad\mbox{$nq$ even}\,,
\end{equation}
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{F_{2nk + 3nq} }}{{L_{2nk} L_{2nk + 2nq} L_{2nk + 4nq} L_{2nk + 6nq} }}}  = \frac{1}{{5F_{3nq} }}\sum_{k = 1}^q {\frac{1}{{L_{2nk} L_{2nk + 2nq} L_{2nk + 4nq} }}}\,.
\end{equation}
The finite summation identity is
\begin{equation}
\begin{split}
5F_{mnq} \sum_{k = 1}^N {\frac{{F_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\qquad\qquad\mbox{$mnq$ even}\\
&\qquad  - \sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2nN + 2jnq} } }}}\,,
\end{split}
\end{equation}
\begin{thm}
If $m$, $n$ and $q$ are positive integers such that $q$ is even \underline{or} $mnq$ is odd , then
\[
\sum_{k = 1}^\infty  {\frac{(-1)^{k-1}F_{2nk + mnq} }{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  = \frac{1}{{5F_{mnq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\,. 
\]
\end{thm}
In particular,
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{(-1)^{k-1}F_{2nk + nq} }}{{L_{2nk} F_{2nk + 2nq} }}}  = \frac{1}{{5F_{nq} }}\sum_{k = 1}^q {\frac{(-1)^{k-1}}{{L_{2nk} }}}\,,\quad\mbox{$q$ even \underline{or} $nq$ odd}\,.
\end{equation}
The alternating summation identity here, valid for $q$ even or $mnq$ odd, is
\begin{equation}
\begin{split}
5F_{mnq} \sum_{k = 1}^N {\frac{{(-1)^{k-1}F_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{2nk + 2jnq} } }}}  &= \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2jnq} } }}}\\
&\qquad  +(-1)^{N-1} \sum_{k = 1}^q {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{2nk + 2nN + 2jnq} } }}}\,.
\end{split}
\end{equation}

\subsection{Sums with $F_{nk}^2F_{nk+nq}^2\cdots F_{nk+mnq-nq}^2F_{nk+mnq}F_{nk+mnq+nq}^2\cdots F_{nk+2mnq}^2$ or $F_{nk}^2F_{nk+nq}^2\cdots F_{nk+mnq}^2$ in the denominator}
\begin{thm}\label{thm.kvn1tje}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{F_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq}^2 } }}}\,. 
\]
\end{thm}
Explicitly,
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq} }}{{F_{nk}^2 F_{nk + nq}^2  \cdots F_{nk + mnq}^2 }}}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{F_{nk}^2 F_{nk + nq}^2  \cdots F_{nk + (m - 1)nq}^2 }}}\,. 
\]
Examples include:
\begin{equation}
\begin{split}
&\mbox{At $m=1$}\\
&\sum_{k = 1}^\infty  {\frac{{F_{2nk + nq} }}{{F_{nk}^2 F_{nk + nq}^2 }}}  = \frac{1}{{F_{nq} }}\sum_{k = 1}^q {\frac{1}{{F_{nk}^2 }}},\quad \mbox{$nq$ even}\,.
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\mbox{At $(m,n,q)=(1,1,2)$ and $(m,n,q)=(1,2,1)$}:\\
&\sum_{k = 1}^\infty  {\frac{{F_{2k + 2} }}{{F_k^2 F_{k + 2}^2 }}}  = 2,\quad\sum_{k = 1}^\infty  {\frac{{F_{4k + 2} }}{{F_{2k}^2 F_{2k + 2}^2 }}}  = 1\,.
\end{split}
\end{equation}

\begin{equation}
\begin{split}
&\mbox{At $(m,n,q)=(3,2,2)$}:\\
&\sum_{k = 1}^\infty  {\frac{{F_{4k + 12} }}{{F_{2k}^2 F_{2k + 4}^2 F_{2k + 8}^2 F_{2k + 12}^2 }}}  = \frac{1288981}{35850395750400}\,.
\end{split}
\end{equation}



\begin{cor}\label{thm.rh1ujkm}
If $m$, $n$ and $q$ are positive integers, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{L_{nk + mnq} }}{{F_{nk + mnq} \prod_{j = 0}^{m - 1} {F_{nk + jnq}^2 } \prod_{j = m + 1}^{2m} {F_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{F_{2mnq} }}\sum_{k = 1}^q {\left[ {\prod_{j = 0}^{2m - 1} {\frac{1}{{F_{nk + jnq}^2 }}} } \right]}\,. 
\]

\end{cor}
Explicitly, Corollary~\ref{thm.rh1ujkm} is
\[
\begin{split}
&\sum_{k = 1}^\infty  {\frac{{L_{nk + mnq} }}{{F_{nk}^2 F_{nk + nq}^2  \cdots F_{nk + (m - 1)nq}^2 F_{nk + mnq} F_{nk + (m + 1)nq}^2  \cdots F_{nk + 2mnq}^2 }}}\\
&\qquad\qquad\qquad= \frac{1}{{F_{2mnq} }}\sum_{k = 1}^q {\frac{1}{{F_{nk}^2 F_{nk + nq}^2  \cdots F_{nk + (2m - 1)nq}^2 }}}\,.
\end{split}
\]
Below are a couple of examples:
\begin{equation}
\begin{split}
&\mbox{At $(m,n,q)=(1,1,1)$ and $(m,n,q)=(2,1,1)$}:\\
&\sum_{k = 1}^\infty  {\frac{{L_{k + 1} }}{{F_k^2 F_{k + 1} F_{k + 2}^2 }}}  = 1,\quad\sum_{k = 1}^\infty  {\frac{{L_{k + 2} }}{{F_k^2 F_{k + 1}^2 F_{k + 2} F_{k + 3}^2 F_{k + 4}^2 }}}  = \frac{1}{{108}}
\,.
\end{split}
\end{equation}
\begin{equation}
\begin{split}
&\mbox{At $(m,n,q)=(3,2,2)$}:\\
&\sum_{k = 1}^\infty  {\frac{{L_{2k + 12} }}{{F_{2k}^2 F_{2k + 4}^2 F_{2k + 8}^2 F_{2k + 12} F_{2k + 16}^2 F_{2k + 20}^2 F_{2k + 24}^2 }}}\\
&= \frac{636693716175181614930457}{1701394375843622618689225675379000792710492054565683200}\,.
\end{split}
\end{equation}

\subsubsection*{Proof of Theorem~\ref{thm.kvn1tje}}
By making appropriate choices for the indices $u$ and $v$ in the identity~\eqref{equ.ded0k7c}, it is straightforward to establish the following identity:
\begin{equation}\label{equ.g4pbkha}
\frac{{F_{mnq} F_{2nk + mnq} }}{{F_{nk}^2 F_{nk + mnq}^2 }} = \frac{1}{{F_{nk}^2 }} - \frac{{( - 1)^{mnq} }}{{F_{nk + mnq}^2 }}\,,
\end{equation}
so that
\begin{equation}\label{equ.vcnyyet}
\frac{{F_{mnq} F_{2nk + mnq} }}{{F_{nk}^2 F_{nk + mnq}^2 }} = \frac{1}{{F_{nk}^2 }} - \frac{1 }{F_{nk + mnq}^2 },\quad\mbox{$mnq$ even}
\end{equation}
and
\begin{equation}\label{equ.werykpm}
\frac{{F_{mnq} F_{2nk + mnq} }}{{F_{nk}^2 F_{nk + mnq}^2 }} = \frac{1}{{F_{nk}^2 }} + \frac{1}{{F_{nk + mnq}^2 }},\quad\mbox{$mnq$ odd}\,.
\end{equation}
From~\eqref{equ.vcnyyet}, with $f(k)=1/F_k^2$ in Lemma~\ref{finall}, we have the finite summation identity
\begin{equation}
\begin{split}
\sum_{k = 1}^N {\frac{{F_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{nk + jnq}^2 } }}}  &= \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq}^2 } }}}\\
&\qquad - \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + nN + jnq}^2 } }}},\quad\mbox{$mnq$ even}\,.
\end{split}
\end{equation}
As $N$ approaches infinity, we have Theorem~\ref{thm.kvn1tje}, while specifically requiring $m$ to be even gives Corollary~\ref{thm.rh1ujkm}.

\begin{thm}\label{thm.hblrah9}
If $m$, $n$ and $q$ are integers such that $q$ is even \underline{or} $mnq$ is odd, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{( - 1)^{k - 1} F_{2nk + mnq} }}{{\prod_{j = 0}^m {F_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq}^2 } }}} \right]}\,. 
\]
\end{thm}
\begin{proof}
If $q$ is even, then the statement of the theorem follows from~\eqref{equ.vcnyyet} and identity~\eqref{infallalt} with the upper sign or from~\eqref{equ.werykpm} and identity~\eqref{infallalt} with the lower sign if $mnq$ is odd.
\end{proof}
In particular, we have
\begin{equation}\label{equ.l14z6cr}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + nq} }}{{F_{nk}^2 F_{nk + nq}^2 }}}  = \frac{1}{{F_{nq} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{F_{nk}^2 }}},\quad \mbox{$q$ even or $nq$ odd}\,,
\end{equation}
which generalizes Brousseau's result (Formula~(6) of~\cite{brousseau1}, also derived by Melham in reference~\cite{melham16} as a special case of a more general result). Brousseau's formula~(6) corresponds to $n=1$ in~\eqref{equ.l14z6cr}.  Brousseau's formula~(15) in~\cite{brousseau1} is also contained in the identity~\eqref{equ.l14z6cr} above at \mbox{$n=3,\,q=1$}.

\bigskip

More examples from Theorem~\ref{thm.hblrah9}:
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + 3nq} }}{{F_{nk}^2 F_{nk + nq}^2 F_{nk + 2nq}^2 F_{nk + 3nq}^2 }}}  = \frac{1}{{F_{3nq} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{F_{nk}^2 F_{nk + nq}^2 F_{nk + 2nq}^2 }}},\quad \mbox{$q$ even or $nq$ odd}\,.
\end{equation}


\begin{cor}\label{thm.p79aeki}
If $q$ is a positive \underline{even} integer, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{(-1)^{k-1}L_{nk + mnq} }}{{F_{nk + mnq} \prod_{j = 0}^{m - 1} {F_{nk + jnq}^2 } \prod_{j = m + 1}^{2m} {F_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{F_{2mnq} }}\sum_{k = 1}^q {(-1)^{k-1}\left[ {\prod_{j = 0}^{2m - 1} {\frac{1}{{F_{nk + jnq}^2 }}} } \right]}\,. 
\]

\end{cor}
In particular:
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} L_{k + q} }}{{F_k^2 F_{k + q} F_{k + 2q}^2 }}}  = \frac{1}{{F_{2q} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{F_k^2 F_{k + q}^2 }}},\quad\mbox{$q$ even}\,.
\end{equation}
We note that Theorem~\ref{thm.kvn1tje}, Theorem~\ref{thm.hblrah9}, Corollary~\ref{thm.rh1ujkm} and Corollary~\ref{thm.p79aeki} correspond to setting $p=0$ in Theorem \ref{thm.byyca8f} and Theorem \ref{thm.xbwcpz1} of section~\ref{sec.cp47u43}.
\begin{thm}\label{thm.xlspejf}
If $m$, $n$ and $q$ are positive integers, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{( - 1)^{nk - 1} F_{2nk + mnq} \prod_{j = 1}^{m - 1} {L_{nk + jnq}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq}^2 } }}} \right]}  = \frac{{5^m q}}{{4F_{mnq} }} - \frac{1}{{4F_{mnq} }}\sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq}^2 }}{{F_{nk + jnq}^2 }}} } \right]}\,. 
\]

\end{thm}
In particular
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{nk - 1} F_{2nk + nq} }}{{F_{nk}^2 F_{nk + nq}^2 }}}  = \frac{{5q}}{{4F_{nq} }} - \frac{1}{{4F_{nq} }}\sum_{k = 1}^q {\frac{{L_{nk}^2 }}{{F_{nk}^2 }}}\,.
\end{equation}

\begin{thm}\label{thm.wxp5pic}
If $n$ and $q$ are positive integers such that $n$ is odd and $q$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{F_{2nk + mnq} \prod_{j = 1}^{m - 1} {L_{nk + jnq}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{4F_{mnq} }}\sum_{k = 1}^q {\left[ {( - 1)^k \prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq}^2 }}{{F_{nk + jnq}^2 }}} } \right]}\,. 
\]

\end{thm}

\subsubsection*{Proof of Theorem~\ref{thm.xlspejf} and Theorem~\ref{thm.wxp5pic}}
Multiplying identity~\eqref{equ.ejrnkwy} and identity~\eqref{equ.moytk3x} and choosing $u$ and $v$ judiciously, it is easy to establish that the following identity holds for positive integers $m$, $n$, $q$ and $k$:
\begin{equation}\label{equ.oeu2ljq}
( - 1)^{nk - 1} \frac{{4F_{mnq} F_{2nk + mnq} }}{{F_{nk}^2 F_{nk + mnq}^2 }} = \frac{{L_{nk + mnq}^2 }}{{F_{nk + mnq}^2 }} - \frac{{L_{nk}^2 }}{{F_{nk}^2 }}\,.
\end{equation}
From identity~\eqref{equ.oeu2ljq} and $f(k)=L_k^2/F_k^2$ in Lemma~\ref{finall} we have the finite summation formula:
\begin{equation}
\begin{split}
&4F_{mnq} \sum_{k = 1}^N {\frac{{( - 1)^{nk - 1} F_{2nk + mnq} \prod_{j = 1}^{m - 1} {L_{nk + jnq}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq}^2 } }}}\\
&\qquad  = \sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + nN + jnq}^2 }}{{F_{nk + nN + jnq}^2 }}} }  - \sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{L_{nk + jnq}^2 }}{{F_{nk + jnq}^2 }}} }\,,
\end{split}
\end{equation}
from which Theorem~\ref{thm.xlspejf} follows in the limit $N$ approaches infinity. Theorem~\ref{thm.wxp5pic} follows from identity~\eqref{infallalt}. 


\subsection{Sums with $L_{nk}^2L_{nk+nq}^2\cdots L_{nk+mnq-nq}^2L_{nk+mnq}L_{nk+mnq+nq}^2\cdots L_{nk+2mnq}^2$ or $L_{nk}^2L_{nk+nq}^2\cdots L_{nk+mnq}^2$ in the denominator}
The theorems in this section are the Lucas versions of those of the previous section. We omit their proofs. The basic identity is
\begin{equation}\label{equ.o1vymkb}
\frac{{5F_{mnq} F_{2nk + mnq} }}{{L_{nk}^2 L_{nk + mnq}^2 }} = \frac{1}{{L_{nk}^2 }} - \frac{{( - 1)^{mnq} }}{{L_{nk + mnq}^2 }}\,,
\end{equation}
which follows from the identity~\eqref{equ.wodrq78}.
\begin{thm}\label{thm.wfq1axk}
If $m$, $n$ and $q$ are positive integers such that $mnq$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{F_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{5F_{mnq} }}\sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{1}{{L_{nk + jnq}^2 }}} } \right]}\,. 
\]
\end{thm}
Explicitly,
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq} }}{{L_{nk}^2 L_{nk + nq}^2  \cdots L_{nk + mnq}^2 }}}  = \frac{1}{{5F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{L_{nk}^2 L_{nk + nq}^2  \cdots L_{nk + (m - 1)nq}^2 }}}\,. 
\]
\begin{cor}\label{thm.v07bx6w}
If $m$, $n$ and $q$ are positive integers, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{F_{nk + mnq} }}{{L_{nk + mnq} \prod_{j = 0}^{m - 1} {L_{nk + jnq}^2 } \prod_{j = m + 1}^{2m} {L_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{5F_{2mnq} }}\sum_{k = 1}^q {\left[ {\prod_{j = 0}^{2m - 1} {\frac{1}{{L_{nk + jnq}^2 }}} } \right]}\,. 
\]

\end{cor}
\begin{thm}\label{thm.t9u45bs}
If $m$, $n$ and $q$ are positive integers such that $q$ is even \underline{or} $mnq$ is odd, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{( - 1)^{k - 1} F_{2nk + mnq} }}{{\prod_{j = 0}^m {L_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{5F_{mnq} }}\sum_{k = 1}^q {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {L_{nk + jnq}^2 } }}} \right]}\,. 
\]

\end{thm}
In particular, we have
\begin{equation}\label{equ.gk5b94r}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + nq} }}{{L_{nk}^2 L_{nk + nq}^2 }}}  = \frac{1}{{5F_{nq} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{L_{nk}^2 }}},\quad \mbox{$q$ even or $nq$ odd}\,,
\end{equation}
\begin{cor}\label{thm.uho7fyr}
If $q$ is a positive \underline{even} integer, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{(-1)^{k-1}F_{nk + mnq} }}{{L_{nk + mnq} \prod_{j = 0}^{m - 1} {L_{nk + jnq}^2 } \prod_{j = m + 1}^{2m} {L_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{5F_{2mnq} }}\sum_{k = 1}^q {\left[ {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{2m - 1} {L_{nk + jnq}^2 } }}} \right]}\,. 
\]

\end{cor}
In particular:
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{k + q} }}{{L_k^2 L_{k + q} L_{k + 2q}^2 }}}  = \frac{1}{{5F_{2q} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{L_k^2 L_{k + q}^2 }}},\quad\mbox{$q$ even}\,.
\end{equation}

\begin{thm}\label{thm.fw565ya}
If $m$, $n$ and $q$ are positive integers, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{( - 1)^{nk - 1} F_{2nk + mnq} \prod_{j = 1}^{m - 1} {F_{nk + jnq}^2 } }}{{\prod_{j = 0}^m {L_{nk + jnq}^2 } }}} \right]}  =  \frac{1}{{4F_{mnq} }}\sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq}^2 }}{{L_{nk + jnq}^2 }}} } \right]}-\frac{{1}}{{4F_{mnq} }}\frac q{5^m}\,. 
\]

\end{thm}
In particular
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{nk - 1} F_{2nk + nq} }}{{L_{nk}^2 L_{nk + nq}^2 }}}  = \frac{1}{{4F_{nq} }}\sum_{k = 1}^q {\frac{{F_{nk}^2 }}{{L_{nk}^2 }}}-\frac{q}{{20F_{nq} }}\,.
\end{equation}

\begin{thm}\label{thm.qwgs5xu}
If $n$ and $q$ are positive integers such that $n$ is odd and $q$ is even, then
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{F_{2nk + mnq} \prod_{j = 1}^{m - 1} {F_{nk + jnq}^2 } }}{{\prod_{j = 0}^m {L_{nk + jnq}^2 } }}} \right]}  = \frac{1}{{4F_{mnq} }}\sum_{k = 1}^q {\left[ {( - 1)^{k-1} \prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq}^2 }}{{L_{nk + jnq}^2 }}} } \right]}\,. 
\]

\end{thm}
In particular
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{ F_{2nk + nq} }}{{L_{nk}^2 L_{nk + nq}^2 }}}  = \frac{1}{{4F_{nq} }}\sum_{k = 1}^q {( - 1)^{k - 1}\frac{{F_{nk}^2 }}{{L_{nk}^2 }}},\quad\mbox{$n$ odd, $q$ even}\,.
\end{equation}

\subsection{Sums with $F_{nk}F_{nk+np}F_{nk+nq}F_{nk+nq+np}F_{nk+2nq}F_{nk+2nq+np}$$\cdots$ $F_{nk+mnq}F_{nk+mnq+np}$ in the denominator}\label{sec.cp47u43}
\begin{thm}\label{thm.byyca8f}
If $m$, $n$, $q$ are positive integers such that $mnq$ is even; and $p$ is a non-negative integer, then
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + np} } }}}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq} F_{nk + jnq + np} } }}}\,. 
\]
\end{thm}
\begin{thm}\label{thm.xbwcpz1}
If $m$, $n$, $q$ are positive integers such that $mnq$ is odd or $q$ is even; and $p$ is a non-negative integer, then
\[
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + np} } }}}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq} F_{nk + jnq + np} } }}}\,. 
\]
\end{thm}
\subsubsection*{Proof of Theorem \ref{thm.byyca8f} and Theorem \ref{thm.xbwcpz1}}
By dividing through the identity~\eqref{equ.cqtsjoj} by $F_uF_{u+p}F_{u+v}F_{u+v+p}$ and setting $u=nk$ and $v=mnq$, the following identity is established
\begin{equation}\label{equ.chzhm4b}
\frac{{F_{mnq} F_{2nk + mnq + np} }}{{F_{nk} F_{nk + np} F_{nk + mnq} F_{nk + mnq + np} }} = \frac{1}{{F_{nk} F_{nk + np} }} + \frac{{( - 1)^{mnq + 1} }}{{F_{nk + mnq} F_{nk + mnq + np} }}\,,
\end{equation}
so that
\begin{equation}\label{equ.wz80v2j}
\begin{split}
&\frac{{F_{mnq} F_{2nk + mnq + np} }}{{F_{nk} F_{nk + np} F_{nk + mnq} F_{nk + mnq + np} }}\\ 
&\qquad= \frac{1}{{F_{nk} F_{nk + np} }} - \frac{1}{{F_{nk + mnq} F_{nk + mnq + np} }},\quad\mbox{$mnq$ even}
\end{split}
\end{equation}
and
\begin{equation}\label{equ.hitwupq}
\begin{split}
&\frac{{F_{mnq} F_{2nk + mnq + np} }}{{F_{nk} F_{nk + np} F_{nk + mnq} F_{nk + mnq + np} }}\\
&\qquad = \frac{1}{{F_{nk} F_{nk + np} }} + \frac{1}{{F_{nk + mnq} F_{nk + mnq + np} }},\quad\mbox{$mnq$ odd}\,.
\end{split}
\end{equation}
From~\eqref{equ.wz80v2j} and $f(k)=1/(F_kF_{k+np})$ in Lemma~\ref{infall} we obtain the finite summation identity
\begin{equation}
\begin{split}
&F_{mnq} \sum_{k = 1}^N {\left[ {\frac{{F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + np} } }}} \right]}\\ 
&\qquad = \sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{m - 1} {F_{nk + jnq} F_{nk + jnq + np} } }}} \right]}\\
&\qquad\quad  - \sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{m-1} {F_{nk + nN + jnq} F_{nk + nN + jnq + np} } }}} \right]}\,, 
\end{split}
\end{equation}
from which Theorem~\ref{thm.byyca8f} follows in the limit as $N$ approaches infinity. If $q$ is even, then the statement of Theorem~\ref{thm.xbwcpz1} follows from~\eqref{equ.wz80v2j} and identity~\eqref{infallalt} with the upper sign or from~\eqref{equ.hitwupq} and identity~\eqref{infallalt} with the lower sign if $mnq$ is odd.




\subsection{Sums with $L_{nk}L_{nk+np}L_{nk+nq}L_{nk+nq+np}L_{nk+2nq}L_{nk+2nq+np}$$\cdots$ $L_{nk+mnq}L_{nk+mnq+np}$ in the denominator}
The results in this section are the Lucas versions of the results in the preceding section. The basic identities are
\begin{equation}
\frac{{5F_{mnq} F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {L_{nk + jnq} L_{nk + jnq + np} } }} = \frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + jnq} L_{nk + jnq + np} } }} + \frac{{( - 1)^{mnq + 1} }}{{\prod_{j = 1}^m {L_{nk + jnq} L_{nk + jnq + np} } }}\,,
\end{equation}
and, if $mnq$ is even
\begin{equation}
\begin{split}
&5F_{mnq} \sum_{k = 1}^N {\left[ {\frac{{F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {L_{nk + jnq} L_{nk + jnq + np} } }}} \right]}\\ 
&\qquad\quad = \sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + jnq} L_{nk + jnq + np} } }}} \right]}\\
&\qquad\qquad  - \sum_{k = 1}^q {\left[ {\frac{1}{{\prod_{j = 0}^{m-1} {L_{nk + nN + jnq} L_{nk + nN + jnq + np} } }}} \right]}\,, 
\end{split}
\end{equation}
while if $mnq$ is odd or $q$ is even:
\begin{equation}
\begin{split}
&5F_{mnq} \sum_{k = 1}^N {\left[ {\frac{{(-1)^{k-1}F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {L_{nk + jnq} L_{nk + jnq + np} } }}} \right]}\\ 
&\qquad\quad = \sum_{k = 1}^q {\left[ {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m - 1} {L_{nk + jnq} L_{nk + jnq + np} } }}} \right]}\\
&\qquad\qquad  +(-1)^{N-1} \sum_{k = 1}^q {\left[ {\frac{(-1)^{k-1}}{{\prod_{j = 0}^{m-1} {L_{nk + nN + jnq} L_{nk + nN + jnq + np} } }}} \right]}\,. 
\end{split}
\end{equation}
\begin{thm}\label{thm.ckvmm33}
If $m$, $n$, $q$ are positive integers such that $mnq$ is even; and $p$ is a non-negative integer, then
\[
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {L_{nk + jnq} L_{nk + jnq + np} } }}}  = \frac{1}{{5F_{mnq} }}\sum_{k = 1}^q {\frac{1}{{\prod_{j = 0}^{m - 1} {L_{nk + jnq} L_{nk + jnq + np} } }}}\,. 
\]

\end{thm}\label{thm.v7eumk9}
\begin{thm}
If $m$, $n$, $q$ are positive integers such that $mnq$ is odd or $q$ is even; and $p$ is a non-negative integer, then
\[
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + mnq + np} }}{{\prod_{j = 0}^m {L_{nk + jnq} L_{nk + jnq + np} } }}}  = \frac{1}{{5F_{mnq} }}\sum_{k = 1}^q {\frac{{( - 1)^{k - 1} }}{{\prod_{j = 0}^{m - 1} {L_{nk + jnq} L_{nk + jnq + np} } }}}\,. 
\]

\end{thm}



\subsection{Evaluation of other sums}

\begin{thm}\label{thm.jq353ph}
If $m$, $n$ and $q$ are positive integers, then 
\[
\sum_{k = 1}^\infty  {\frac{{( - 1)^{nk - 1} F_{2nk + mnq + 2} \prod_{j = 1}^{m - 1} {F_{nk + jnq + 1}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + 2} } }}}  = \frac{q}{{F_{mnq} }} - \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + 1}^2 }}{{F_{nk + jnq} F_{nk + jnq + 2} }}} }\,, 
\]
so that
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{F_{2nk + mnq + 2} \prod_{j = 1}^{m - 1} {F_{nk + jnq + 1}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + 2} } }}}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + 1}^2 }}{{F_{nk + jnq} F_{nk + jnq + 2} }}} }-\frac{q}{{F_{mnq} }},\quad\mbox{$n$ even}
\end{equation}
and
\begin{equation}
\sum_{k = 1}^\infty  {\frac{{( - 1)^{k - 1} F_{2nk + mnq + 2} \prod_{j = 1}^{m - 1} {F_{nk + jnq + 1}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + 2} } }}}  = \frac{q}{{F_{mnq} }} - \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + 1}^2 }}{{F_{nk + jnq} F_{nk + jnq + 2} }}} },\quad\mbox{$n$ odd}\,.
\end{equation}


\end{thm}

\begin{thm}\label{thm.tpgmuma}
If $m$, $n$ and $q$ are positive integers such that $n$ is odd and $q$ is even, then 
\[
\sum_{k = 1}^\infty  {\left[ {\frac{{F_{2nk + mnq + 2} \prod_{j = 1}^{m - 1} {F_{nk + jnq + 1}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + 2} } }}} \right]}  = \frac{1}{{F_{mnq} }}\sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{( - 1)^k F_{nk + jnq + 1}^2 }}{{F_{nk + jnq} F_{nk + jnq + 2} }}} } \right]}\,. 
\]


\end{thm}
\subsubsection*{Proof of Theorem~\eqref{thm.jq353ph} and Theorem~\eqref{thm.tpgmuma}}
Dividing through the identity~\eqref{equ.yr8t5vs} by $F_{v+1}F_{v-1}F_{u+1}F_{u-1}$ and setting $u=nk+1$ and $v=nk+mnq+1$ we obtain the identity
\begin{equation}\label{equ.hro9n7d}
\frac{{( - 1)^{nk - 1} F_{mnq} F_{2nk + mnq + 2} }}{{F_{nk} F_{nk + 2} F_{nk + mnq} F_{nk + mnq + 2} }} = \frac{{F_{nk + mnq + 1}^2 }}{{F_{nk + mnq + 2} F_{nk + mnq} }} - \frac{{F_{nk + 1}^2 }}{{F_{nk + 2} F_{nk} }}\,.
\end{equation}
With $f(k)=F_{k+1}^2/(F_kF_{k+2})$ in Lemma~\ref{finall} and use of the identity~\eqref{equ.hro9n7d} we get the finite summation identity
\begin{equation}
\begin{split}
&F_{mnq} \sum_{k = 1}^N {\left[ {\frac{{( - 1)^{nk - 1} F_{2nk + mnq + 2} \prod_{j = 1}^{m - 1} {F_{nk + jnq + 1}^2 } }}{{\prod_{j = 0}^m {F_{nk + jnq} F_{nk + jnq + 2} } }}} \right]}\\ 
&\qquad\qquad = \sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + nN + jnq + 1}^2 }}{{F_{nk + nN + jnq} F_{nk + nN + jnq + 2} }}} } \right]}\\
&\qquad\qquad\qquad  - \sum_{k = 1}^q {\left[ {\prod_{j = 0}^{m - 1} {\frac{{F_{nk + jnq + 1}^2 }}{{F_{nk + jnq} F_{nk + jnq + 2} }}} } \right]}\,, 
\end{split}
\end{equation}
from which Theorem~\eqref{thm.jq353ph} follows in the limit as $N$ approaches infinity. Theorem~\eqref{thm.tpgmuma} follows from the identity~\eqref{equ.hro9n7d} and taking \mbox{$f(k)=F_{k+1}^2/(F_kF_{k+2})$} in identity~\eqref{infallalt} with the upper sign.
\begin{thm}
\[
\sum_{k = 1}^\infty  {\frac{{F_{2k + 3} }}{{F_k^4 F_{k + 1}^3 F_{k + 2}^3 F_{k+3}^4 }}}  = \frac{1}{{128}},\quad\sum_{k = 1}^\infty  {\frac{{F_{2k + 3} }}{{L_k^4 L_{k + 1}^3 L_{k + 2}^3 L_{k+3}^4 }}}  = \frac{1}{{829440}}\,.
\]

\end{thm}
\begin{thm}
\[
\sum_{k = 1}^\infty  {\frac{{F_{3k + 1} F_{3k + 2} F_{6k + 3} }}{{F_{3k}^4 F_{3k + 3}^4 }}}  = \frac{1}{{128}},\quad\sum_{k = 1}^\infty  {\frac{{L_{3k + 1} L_{3k + 2} F_{6k + 3} }}{{L_{3k}^4 L_{3k + 3}^4 }}}  = \frac{1}{{10240}}\,.
\]

\end{thm}

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P.~S.~BRUCKMAN and I.~J.~GOOD (1976),
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\end{document}


1703.06075.extracted_bib
{adegoke K.~ADEGOKE (2017), \newblock Some remarkable infinite product identities involving Fibonacci and Lucas numbers, \newblock{\em arXiv:1702.08321\\\url{https://arxiv.org/abs/1702.08321.
1703.06075.extracted_bib
{brousseau1 Bro.~A.~BROUSSEAU (1969), \newblock Fibonacci-Lucas Infinite Series - Research Topic, \newblock{\em The Fibonacci Quarterly 7 (2):211--217.
1703.06075.extracted_bib
{brousseau2 Bro.~A.~BROUSSEAU (1969), \newblock Summation of infinite Fibonacci series, \newblock{\em The Fibonacci Quarterly 7 (2):143--168.
1703.06075.extracted_bib
{rabinowitz S.~RABINOWITZ (1999), \newblock Algorithmic summation of reciprocals of products of fibonacci numbers, \newblock{\em The Fibonacci Quarterly 7 (2):122--127.
1703.06075.extracted_bib
{bruckman P.~S.~BRUCKMAN and I.~J.~GOOD (1976), \newblock A generalization of a series of de Morgan, with applications of Fibonacci type, \newblock{\em The Fibonacci Quarterly 14 (3):193--196.
1703.06075.extracted_bib
{vajda S.~VAJDA (2008), \newblock Fibonacci and Lucas numbers, and the Golden Section: Theory and Applications, \newblock{\em Dover Press
1703.06075.extracted_bib
{howard F.~T.~HOWARD (2003), \newblock The Sum of the Squares of Two Generalized Fibonacci Numbers, \newblock{\em The Fibonacci Quarterly 41 (1):80--84.
1703.06075.extracted_bib
{melham16 R.~MELHAM (2016), \newblock Finite reciprocal sums in which the denominator of the summand contains squares of generalized fibonacci numbers, \newblock {\em Integers, 16:1--11.
1703.06075.extracted_bib
{frontczak16 R.~FRONTCZAK (2016), \newblock New results on reciprocal series related to fibonacci and lucas numbers with subscripts in arithmetic progression, \newblock {\em International Journal of Contemporary Mathematical Sciences, 11:509--516.
1703.06075.extracted_bib
{melham01 R.~S.~MELHAM (2001), \newblock Summation of reciprocals which involve products of terms from generalized Fibonacci sequences-Part II, \newblock{\em The Fibonacci Quarterly 39 (3):264--288.