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\begin{center}
\vskip 1cm{\LARGE\bf
Finite Reciprocal Sums Involving Summands \\
\vskip .03in
That are Balanced Products of Generalized\\
\vskip .14in
Fibonacci Numbers}

\vskip 1cm
\large
R. S. Melham\\
School of Mathematical Sciences\\
University of Technology, Sydney\\
NSW 2007\\
Australia\\
\href{mailto:ray.melham@uts.edu.au}{\tt ray.melham@uts.edu.au}
\end{center}

\vskip .2 in
\begin{abstract}
In this paper we find closed forms, in terms of rational numbers, for certain finite sums. The denominator of each summand is a
finite product of terms drawn from two sequences
that are generalizations of the Fibonacci and Lucas numbers.
\end{abstract}


\section{Introduction}\label{sec1}

In \cite{mel1,mel2} we considered certain types of finite reciprocal sums involving generalized Fibonacci numbers. Indeed we gave closed forms,
in terms of rational numbers, for these sums. Our purpose here is to give closed forms for finite reciprocal sums that are of a different type than
those considered in \cite{mel1,mel2}, thereby extending the work in \cite{mel1,mel2}. As in \cite{mel1,mel2}, our results
can be used to produce finite reciprocal sums that involve the Fibonacci and Lucas numbers.

We begin by introducing the three pairs of integer sequences that feature in this paper. Let $a\geq 0$ and $b\geq 0$ be integers with
$\left(a,b\right)\neq \left(0,0\right)$. For $p$ a positive integer we define, for all integers $n$, the sequences $\left\{W_{n}\right\}$ and $\left\{\overline{W}_{n}\right\}$ by
\begin{equation*}
    W_{n}=p W_{n-1}+W_{n-2},~~W_{0}=a,~~W_{1}=b,
\end{equation*}
and 
\begin{equation*}
\overline{W}_{n}=W_{n-1}+W_{n+1}.
\end{equation*}

For $(a, b, p)=(0, 1, 1)$ we have $\left\{W_{n}\right\}$=$\left\{F_{n}\right\}$, and $\left\{\overline{W}_{n}\right\}$=$\left\{L_{n}\right\}$, which are the Fibonacci and Lucas
numbers, respectively. Retaining the parameter $p$, and taking $(a, b)=(0, 1)$, we write  $\left\{W_{n}\right\}$=$\left\{U_{n}\right\}$, and $\left\{\overline{W}_{n}\right\}$=$\left\{V_{n}\right\}$, which
are integer sequences that generalize the Fibonacci and Lucas numbers, respectively.

Let $\alpha$ and $\beta$ denote the two distinct real roots of $x^2 -p x-1=0$. Set $A=b-a\beta$ and $B=b-a\alpha$. Then the closed forms (the Binet forms)
for $\left\{W_{n}\right\}$ and $\left\{\overline{W}_{n}\right\}$ are, respectively,
\begin{equation*}
	W_{n}=\frac{A\alpha^{n}-B\beta^{n}}{\alpha-\beta},
\end{equation*}
and
\begin{equation*}
	\overline{W}_{n}=A\alpha^{n}+B\beta^{n}.
\end{equation*}
We require also the constants $e_{W}=A B=b^{2}-p a b-a^{2}$, and $\Delta=p^2+4$.


Throughout this paper we take $k\geq1$, $m\geq0$, and $n\geq2$ to be integers. Let $m_{1}< m_{2}$ and $m_{3} < m_{4}$ be non-negative integers with 
$m_{1}+ m_{2}=m_{3}+ m_{4}$. We begin by giving a closed form for the finite sum
\begin{equation*}
	S_{4}(k,m,n,m_{1},\ldots,m_{4})=\sum_{i=1}^{n-1}\frac{1}{W_{k(i+m_{1})+m}W_{k(i+m_{2})+m}\overline{W}_{k(i+m_{3})+m}\overline{W}_{k(i+m_{4})+m}}.
\end{equation*}

Because of the conditions on the $m_{i}$ we consider $S_{4}$ to be the most intriguing sum that we present in this paper. We also give closed forms for
similar sums that have longer products in the denominator, and to this end we introduce some notation. For integers $0<m_{1}< m_{2}$ write
\begin{eqnarray*}
	&& P_{6}(W,\overline{W},k,m,i,0,m_{1},m_{2})\\ \nonumber
	&& =  W_{k i+m}W_{k(i+m_{1})+m}W_{k(i+m_{2})+m}\overline{W}_{k i+m}\overline{W}_{k(i+m_{1})+m}\overline{W}_{k(i+m_{2})+m}.
\end{eqnarray*}
Likewise, for integers $0<m_{1}< m_{2}< m_{3}$ write
\begin{eqnarray*}
	&& P_{8}(W,\overline{W},k,m,i,0,m_{1},m_{2},m_{3})\\ \nonumber
	&& =  W_{k i+m}W_{k(i+m_{1})+m}\cdots W_{k(i+m_{3})+m}\overline{W}_{k i+m}\overline{W}_{k(i+m_{1})+m}\cdots \overline{W}_{k(i+m_{3})+m}.
\end{eqnarray*}

We say that $P_{8}$ consists of a {\it balanced} product of eight terms that are drawn from the sequences $\left\{W_{n}\right\}$ and $\left\{\overline{W}_{n}\right\}$.
We also consider such products where the terms are drawn from the sequences $\left\{U_{n}\right\}$ and $\left\{V_{n}\right\}$, and where the product $U_{k i+m}V_{k i+m}$
is not included. For instance $P_{4}\left(U,V,k,m,i,m_{1},m_{2}\right)$ denotes the product of $U_{k(i+m_{1})+m}U_{k(i+m_{2})+m}$ and $V_{k(i+m_{1})+m}V_{k(i+m_{2})+m}$. Since
$U_{n}V_{n}=U_{2n}$, $P_{4}$ can be shortened to a product of two terms from the sequence $\left\{U_{n}\right\}$. However, we choose to retain the longer form for $P_{4}$ (and
for expressions analogous $P_{4}$) in order to highlight the relationship between the numerator and the denominator of the summand. Later, however, when giving examples
of our results that involve $F_{n}$ and $L_{n}$, we present these examples in simplified form.

Using the notation that we have just introduced, we now define three finite sums whose closed forms we give in this paper. Let $0<m_{1}< m_{2}$ be integers. Define
\begin{equation*}
	S_{6}^{0}(k,m,n,m_{1},m_{2})=\sum_{i=1}^{n-1}\frac{P_{2}(U,V,k,m,i,0)}{P_{6}(W,\overline{W},k,m,i,0,m_{1},m_{2})},
\end{equation*}
\begin{equation*}
	S_{6}^{1}(k,m,n,m_{1},m_{2})=\sum_{i=1}^{n-1}\frac{P_{2}(U,V,k,m,i,m_{1})}{P_{6}(W,\overline{W},k,m,i,0,m_{1},m_{2})},
\end{equation*}
and
\begin{equation*}
	S_{6}^{2}(k,m,n,m_{1},m_{2})=\sum_{i=1}^{n-1}\frac{P_{2}(U,V,k,m,i,m_{2})}{P_{6}(W,\overline{W},k,m,i,0,m_{1},m_{2})}.
\end{equation*}
In each of these three cases the numerator of the summand consists of a product of two terms, and the denominator of the summand
consists of a product of six terms.

We evaluate each of the finite sums that we consider in this paper in terms of rational numbers. In addition to the four finite sums defined above, we
consider analogous finite sums where the denominator of the summand consists of a product of eight, or ten, or twelve terms. In each case the numerator of
the summand is either unity, or is a product of terms drawn from the sequences $\left\{U_{n}\right\}$ and $\left\{V_{n}\right\}$ and is defined in terms
of the $P$ notation that we have introduced. Furthermore, we consider only finite sums where {\it the number of terms that constitute the product in
the denominator of the summand exceeds the number of terms that constitute the product in the numerator of the summand by a multiple of four}. Indeed
these are the only types of sums, with the structure described earlier in this paragraph, for which we have been able to find closed forms.

In Section \ref{sec2} we present one result, namely the closed form for $S_{4}$, and give a proof. In Section \ref{sec3} we present the closed forms
for $S_{6}^{0}$, $S_{6}^{1}$, and $S_{6}^{2}$. In subsequent sections we present a selection of the results that we have found that involve longer products
in the denominator of the summand. Indeed, we limit the scope of this paper to finite sums that have four, or six, or eight, or ten, or twelve products
in the denominator of the summand.


There are two finite sums that feature throughout. For integers $0\leq l_{1}<l_{2}$ these finite sums are
\begin{equation*}
		\Omega_{W}(k,m,n,l_{1},l_{2})=\sum_{i=l_{1}}^{l_{2}-1}\frac{(-1)^{k i}}{W_{k(i+2)+m}W_{k(i+n)+m}},
\end{equation*}
and
\begin{equation*}
		\Omega_{\overline{W}}(k,m,n,l_{1},l_{2})=\sum_{i=l_{1}}^{l_{2}-1}\frac{(-1)^{k i}}{\overline{W}_{k(i+2)+m}\overline{W}_{k(i+n)+m}}.
\end{equation*}

To prevent the presentation from becoming too unwieldy, we suppress certain arguments from quantities when there is no danger of confusion.
For instance $S_{4}(n)$ will denote $S_{4}(k,m,n,m_{1},m_{2},m_{3},m_{4})$ when we want $n$ to vary and the other parameters to remain fixed.
Likewise $\Omega_{W}(k,m,n,l_{1},l_{2})$ will be denoted by $\Omega_{W}(l_{1},l_{2})$ when $l_{1}$ and $l_{2}$ vary and the other parameters remain fixed.

We now give two identities involving $\Omega_{W}$ and $\Omega_{\overline{W}}$ that are required for the proofs of all the theorems in this paper. We state these identities as lemmas.
\begin{lemma}\label{lem1}  With $\Omega_{W}$ as defined above, we have
\begin{equation*}
   U_{k (n-1)}\Omega_{W}(n+1)-U_{k (n-2)}\Omega_{W}(n)=\frac{(-1)^{k (n+l_{1})}U_{k(l_{2}-l_{1})}}{W_{k(n+l_{1})+m}W_{k(n+l_{2})+m}}.
\end{equation*}
\end{lemma}
\begin{lemma}\label{lem2}  With $\Omega_{\overline{W}}$ as defined above, we have
\begin{equation*}
   U_{k (n-1)}\Omega_{\overline{W}}(n+1)-U_{k (n-2)}\Omega_{\overline{W}}(n)=\frac{(-1)^{k (n+l_{1})}U_{k(l_{2}-l_{1})}}{\overline{W}_{k(n+l_{1})+m}\overline{W}_{k(n+l_{2})+m}}.
\end{equation*}
\end{lemma}

The proof of Lemma \ref{lem1} was given in \cite{mel2}, and since the proof of Lemma \ref{lem2} is similar we refrain from giving it here.


\section{A closed form for $S_{4}$}\label{sec2} 

Let $k$, $m$, $n$, $m_{1}$, $m_{2}$, $m_{3}$, and $m_{4}$ satisfy the constraints given earlier in the definition of $S_{4}$. Set
\begin{equation*}
	a_{0} = a_{0}(k, m_{1}, m_{2}, m_{3}, m_{4}) = e_{W} U_{(m_{4}-m_{3})k} U_{(m_{2}-m_{1})k} V_{(m_{4}-m_{1})k} V_{(m_{4}-m_{2})k}.
\end{equation*}
Then
\begin{theorem}\label{thm1} With $S_{4}$ as defined in Section \ref{sec1},
\begin{eqnarray*}
	a_{0}\left(S_{4}(n)-S_{4}(2)\right) & = & (-1)^m U_{k(n-2)}\left((-1)^{k(m_{1}+m_{3})}U_{(m_{4}-m_{3})k}\Omega_{W}(m_{1},m_{2})
	\right.\nonumber\\ & & \left.
	{}-\Delta U_{(m_{2}-m_{1})k} \Omega_{\overline{W}}(m_{3},m_{4})\right).
\end{eqnarray*}
\end{theorem}
\begin{proof} In \cite{mel2} we demonstrated two methods of proof, and both methods apply here. The first method required quite a lot to set up, relying heavily upon
generalized Fibonacci identities. The second method was more direct, and mechanical, relying upon the closed forms of the relevant sequences. We use the second method here
since it is transparent and can be used to effectively prove all the theorems in this paper. To this end, with $\alpha=\left(p+\sqrt{\Delta}\right)/2$,
it is advantageous to write the closed forms of the sequences in question as
\begin{eqnarray*}
  U_{n} &=& \left(\alpha^n+(-1)^{n+1}\alpha^{-n}\right)/\sqrt{\Delta},\\
	V_{n} &=& \alpha^n+(-1)^n\alpha^{-n},\\
	W_{n} &=& \left(\left(b+a \alpha^{-1}\right)\alpha^n+(-1)^{n+1} \left(b-a \alpha\right)\alpha^{-n}\right)/\sqrt{\Delta},\\
	\overline{W}_{n} &=& \left(b+a \alpha^{-1}\right)\alpha^n+(-1)^n \left(b-a \alpha\right)\alpha^{-n},\\
\end{eqnarray*}
where these closed forms are valid for all integers $n$. Furthermore we set $p=\alpha-\alpha^{-1}$, and $e_{W}=b^2-p a b-a^2$.

We remind the reader that all the finite sums in this paper are defined for $n\geq 2$, and so it is for these values of $n$ that the following argument holds.
In the statement of Theorem \ref{thm1}, denote the quantity on the left side by $L\left(n\right)$ and the quantity on the right side by $R\left(n\right)$. We first prove that
\begin{equation}\label{2a}
  L(n+1)-L(n) = R(n+1)-R(n).
\end{equation}

With the previously stated restrictions on the relevant parameters, we have
\begin{equation}\label{2b}
   L(n+1)-L(n) = \frac{e_{W} U_{(m_{4}-m_{3})k} U_{(m_{2}-m_{1})k} V_{(m_{4}-m_{1})k} V_{(m_{4}-m_{2})k}}{W_{k (n+m_{1})+m}W_{k(n+m_{2})+m}\overline{W}_{k(n+m_{3})+m}\overline{W}_{k(n+m_{4})+m}}.
\end{equation}

With the use of Lemma \ref{lem1} and Lemma \ref{lem2} we can write down, after some straightforward algebra, the expression for the numerator of $R(n+1)-R(n)$. This expression is
\begin{eqnarray}\label{2c}
   (-1)^{k(n+m_{3})+m}U_{(m_{4}-m_{3})k} U_{(m_{2}-m_{1})k}\left(\overline{W}_{k(n+m_{3})+m}\overline{W}_{k(n+m_{4})+m}
	\right.\nonumber\\ \left.
	-\Delta W_{k (n+m_{1})+m}W_{k(n+m_{2})+m}\right).
\end{eqnarray}
Furthermore, $L(n+1)-L(n)$ and $R(n+1)-R(n)$ have identical denominators, so to prove (\ref{2a}) it is enough to prove that
\begin{eqnarray}\label{2d}
   e_{W} V_{(m_{4}-m_{1})k} V_{(m_{4}-m_{2})k} &=& (-1)^{k(n+m_{3})+m}\left(\overline{W}_{k(n+m_{3})+m}\overline{W}_{k(n+m_{4})+m}
	\right.\nonumber\\ & &\left.
	-{}\Delta W_{k (n+m_{1})+m}W_{k(n+m_{2})+m}\right).
\end{eqnarray}
To this end we consider the difference of the expressions on the left and right sides of (\ref{2d}), replace $m_{4}$ by $m_{1}+m_{2}-m_{3}$, and express everything in terms
of the closed forms. With the use of the computer algebra system {\it Mathematica 8} we find that a factor of the resulting expression is $(-1)^{2(k(n+m_{3})+m)}-1$, and this proves (\ref{2d}).
This, together with the fact that $L(2)=R(2)=0$, establishes Theorem \ref{thm1}.
\end{proof}

In the proof above the key identity is (\ref{2d}). Likewise, the proof of each theorem in this paper hinges around the proof of a key identity that is
analogous to (\ref{2d}), and each such identity follows immediately by substitution of the appropriate closed forms. The  method is mechanical and is
not dependent upon and special identities. However, the use of a computer algebra system (in our case {\it Mathematica 8}) is essential. The proof above
serves as a template for the proof of each theorem in this paper, and so we state the theorems in the sections that follow without proof.

We pause to give two examples. Let $k=1$, $m=0$, and $\left(m_{1},m_{2},m_{3},m_{4}\right)=(0,3,1,2)$. Then for $W_{n}=F_{n}$ the result in Theorem \ref{thm1} becomes
\begin{equation*}
	36\sum_{i=1}^{n-1}\frac{1}{F_{i}F_{i+3}L_{i+1}L_{i+2}}=1+F_{n-2}\left(\frac{6}{F_{n}}-\frac{3}{F_{n+1}}+\frac{2}{F_{n+2}}-\frac{15}{L_{n+1}}\right).
\end{equation*}

Next let $k=2$, $m=0$, and $\left(m_{1},m_{2},m_{3},m_{4}\right)=(2,3,1,4)$. Then for $W_{n}=F_{n}$ the result in Theorem \ref{thm1} becomes
\begin{eqnarray*}
	&& 40790736\sum_{i=1}^{n-1}\frac{1}{F_{2(i+2)}F_{2(i+3)}L_{2(i+1)}L_{2(i+4)}} \\
	&& = 282+F_{2(n-2)}\left(\frac{92496}{F_{2(n+2)}}-\frac{67445}{L_{2(n+1)}}-\frac{25830}{L_{2(n+2)}}-\frac{9870}{L_{2(n+3)}}\right).
\end{eqnarray*}

\section{Closed forms for $S_{6}^{0}$, $S_{6}^{1}$, and $S_{6}^{2}$}\label{sec3}

As stated in the introduction, here, and in the sequel we take $k\geq1$, $m\geq0$, and $n\geq2$ to be integers. In this section we take $0<m_{1}< m_{2}$
to be integers.

For $0 \leq i \leq 2$ define $a_{i}=a_{i}(k,m,m_{1},m_{2})$ as
\begin{eqnarray*}
		a_{0} &=& 2(-1)^{m+1} e_{W}^3 U_{2m_{1}k}U_{2m_{2}k}U_{2(m_{2}-m_{1})k},\\
		a_{1} &=& U_{2(m_{2}-m_{1})k}W_{0}\overline{W}_{0},\\
		a_{2} &=& -U_{2m_{1}k}W_{m_{2}k}\overline{W}_{m_{2}k}.
\end{eqnarray*}
We then have
\begin{theorem}\label{thm2}  With $S_{6}^{0}$ as defined in Section \ref{sec1},
\begin{eqnarray*}
	a_{0}\left(S_{6}^{0}(n)-S_{6}^{0}(2)\right) & = & U_{k(n-2)}\left(a_{1}\Omega_{W}(0,m_{1})+a_{2}\Omega_{W}(m_{1},m_{2})
	\right.\nonumber\\ & & \left.
	{}-\Delta a_{1}\Omega_{\overline{W}}(0,m_{1})-\Delta a_{2}\Omega_{\overline{W}}(m_{1},m_{2})\right).
\end{eqnarray*}
\end{theorem}

Next, for $1 \leq i \leq 2$ define $b_{i}=b_{i}(k,m_{1},m_{2})$ by
\begin{eqnarray*}
		b_{1} &=& U_{2(m_{2}-m_{1})k}W_{-m_{1} k}\overline{W}_{-m_{1} k},\\
		b_{2} &=& -U_{2m_{1}k}W_{(m_{2}-m_{1})k}\overline{W}_{(m_{2}-m_{1})k}.
\end{eqnarray*}
With $a_{0}$ as for Theorem \ref{thm2}, we have
\begin{theorem}\label{thm3}  Let $S_{6}^{1}$ be as defined in Section \ref{sec1}. Then
\begin{eqnarray*}
	a_{0}\left(S_{6}^{1}(n)-S_{6}^{1}(2)\right) & = & U_{k(n-2)}\left(b_{1}\Omega_{W}(0,m_{1})+b_{2}\Omega_{W}(m_{1},m_{2})
	\right.\nonumber\\ & & \left.
	{}-\Delta b_{1}\Omega_{\overline{W}}(0,m_{1})-\Delta b_{2}\Omega_{\overline{W}}(m_{1},m_{2})\right).
\end{eqnarray*}
\end{theorem}

For $1 \leq i \leq 2$ define $c_{i}=c_{i}(k,m_{1},m_{2})$ as
\begin{eqnarray*}
		c_{1} &=& U_{2(m_{2}-m_{1})k}W_{-m_{2} k}\overline{W}_{-m_{2} k},\\
		c_{2} &=& -U_{2m_{1}k}W_{0}\overline{W}_{0}.
\end{eqnarray*}
Then, with $a_{0}$ as for Theorem \ref{thm2}, we have
\begin{theorem}\label{thm4}  Let $S_{6}^{2}$ be as defined in Section \ref{sec1}. Then
\begin{eqnarray*}
	a_{0}\left(S_{6}^{2}(n)-S_{6}^{2}(2)\right) & = & U_{k(n-2)}\left(c_{1}\Omega_{W}(0,m_{1})+c_{2}\Omega_{W}(m_{1},m_{2})
	\right.\nonumber\\ & & \left.
	{}-\Delta c_{1}\Omega_{\overline{W}}(0,m_{1})-\Delta c_{2}\Omega_{\overline{W}}(m_{1},m_{2})\right).
\end{eqnarray*}
\end{theorem}

Let $k=1$, $m=0$, and $\left(m_{1},m_{2}\right)=(1,2)$. Take $W_{n}=F_{n+2}$. Then with the use of the identity $F_{n}L_{n}=F_{2n}$ to simplify the summand, and also
the right side, the result in Theorem \ref{thm3} becomes
\begin{eqnarray*}
	27720\sum_{i=1}^{n-1}\frac{F_{2(i+1)}}{F_{2(i+2)}F_{2(i+3)}F_{2(i+4)}} &=& 9+8F_{2(n-2)}\left(\frac{-55}{F_{2(n+2)}}+\frac{168}{F_{2(n+3)}}\right).
\end{eqnarray*}


\section{The summand has eight factors in the denominator}\label{sec4}

In this section we take $0<m_{1}< m_{2}< m_{3}$ to be integers.

Define the finite sum
\begin{equation*}
	S_{8}(k,m,n,m_{1},m_{2},m_{3})=\sum_{i=1}^{n-1}\frac{1}{P_{8}(W,\overline{W},k,m,i,0,m_{1},m_{2},m_{3})}.
\end{equation*}


 For $0 \leq i \leq 3$ define $a_{i}=a_{i}(k,m,m_{1},m_{2},m_{3})$ by
\begin{eqnarray*}
		a_{0} &=& 2 e_{W}^3 U_{2m_{1}k}U_{2m_{2}k}U_{2m_{3}k}U_{2(m_{3}-m_{2})k}U_{2(m_{3}-m_{1})k}U_{2(m_{2}-m_{1})k},\\
		a_{1} &=& (-1)^m U_{2(m_{3}-m_{2})k}U_{2(m_{3}-m_{1})k}U_{2(m_{2}-m_{1})k},\\
		a_{2} &=& (-1)^{m+1}U_{2m_{1}k}U_{2(m_{3}-m_{2})k}U_{2(m_{3}+m_{2}-m_{1})k},\\
		a_{3} &=& (-1)^m U_{2m_{1}k}U_{2m_{2}k}U_{2(m_{2}-m_{1})k}.
\end{eqnarray*}
Then
\begin{theorem}\label{thm5}  We have
\begin{eqnarray*}
	a_{0}\left(S_{8}(n)-S_{8}(2)\right) & = & U_{k(n-2)}\left(a_{1}\Omega_{W}(0,m_{1})+a_{2}\Omega_{W}(m_{1},m_{2})
	\right.\nonumber\\ & & \left.
	{}+a_{3}\Omega_{W}(m_{2},m_{3})-\Delta a_{1}\Omega_{\overline{W}}(0,m_{1})
		\right.\nonumber\\ & & \left.
	{}-\Delta a_{2}\Omega_{\overline{W}}(m_{1},m_{2})-\Delta a_{3}\Omega_{\overline{W}}(m_{2},m_{3})\right).
\end{eqnarray*}
\end{theorem}

We have found that as the number of products in the denominator of the summand increases it becomes more difficult to write down the closed
form of the corresponding finite sum. The same is true as the number of products in the numerator of the summand increases. Indeed, to find the closed form
for the finite sum $T_{8}$, defined below, we needed to specialize the values of $m_{1}$, $m_{2}$, and $m_{3}$ in a manner that we soon make clear.

Define the sum
\begin{equation*}
	T_{8}(k,m,n,m_{1},m_{2},m_{3})=\sum_{i=1}^{n-1}\frac{P_{4}\left(U,V,k,m,i,m_{1},m_{2}\right)}{P_{8}(W,\overline{W},k,m,i,0,m_{1},m_{2},m_{3})}.
\end{equation*}
Here, and in the sequel, we take $g\geq 1$ to be an integer.
For $0\leq i\leq 3$ define the quantities $b_{i}=b_{i}(g,k,m)$ as follows:
\begin{eqnarray*}
		b_{0} &=& 2 e_{W}^5 U_{2 g k}U_{4g k}U_{6 g k}, \\
    b_{1} &=& (-1)^m W_{-2 g k}W_{-g k}\overline{W}_{-2 g k}\overline{W}_{-g k},\\
		b_{2} &=& (-1)^{m+1} V_{2 g k}W_{-g k}W_{g k}\overline{W}_{-g k}\overline{W}_{g k},\\
		b_{3} &=& (-1)^m W_{g k}W_{2gk}\overline{W}_{g k}\overline{W}_{2 g k}.
\end{eqnarray*}
We now state our next theorem.
\begin{theorem}\label{thm6}  Let $(m_{1},m_{2},m_{3})=(g,2g,3g)$, so that $0$, $m_{1}$, $m_{2}$, and $m_{3}$ form an arithmetic progression. Then
\begin{eqnarray*}
	b_{0}\left(T_{8}(n)-T_{8}(2)\right) & = & U_{k(n-2)}\left(b_{1}\Omega_{W}(0,g)+b_{2}\Omega_{W}(g,2g)
	\right.\nonumber\\ & & \left.
	{}+b_{3}\Omega_{W}(2g,3g)-\Delta b_{1}\Omega_{\overline{W}}(0,g)
		\right.\nonumber\\ & & \left.
	{}-\Delta b_{2}\Omega_{\overline{W}}(g,2g)-\Delta b_{3}\Omega_{\overline{W}}(2g,3g)\right).
\end{eqnarray*}
\end{theorem}

Let $k=1$, $m=0$, and $g=1$. Then for $W_{n}=F_{n+3}$ the result in Theorem \ref{thm6} becomes
\begin{eqnarray*}
	&& 167207040\sum_{i=1}^{n-1}\frac{F_{2(i+1)}F_{2(i+2)}}{F_{2(i+3)}F_{2(i+4)}F_{2(i+5)}F_{2(i+6)}} \\
	&& =64+63F_{2(n-2)}\left(\frac{6032}{F_{2(n+3)}}-\frac{145145}{F_{2(n+4)}}+\frac{338800}{F_{2(n+5)}}\right).
\end{eqnarray*}

We attempted to find the closed forms for certain finite sums analogous to $T_{8}$, but without success. Firstly, we considered $T_{8}$ as defined above
but with the $m_{i}$ defined as different multiples of $g$, such as $(m_{1}, m_{2}, m_{3})=(2g,3g,5g)$. Secondly, in the definition of $T_{8}$, we
replaced $P_{4}\left(U,V,k,m,m_{1},m_{2}\right)$ by $P_{4}\left(U,V,k,m,0,m_{2}\right)$, and by $P_{4}\left(U,V,k,m,m_{1},m_{3}\right)$. In attempting
to find the corresponding closed forms in each of these two cases we set $(m_{1},m_{2},m_{3})=(g,2g,3g)$. Put simply, there seems to be a fine line
between success and failure.

\section{The summand has ten factors in the denominator}\label{sec5}


In this section we take $0<m_{1}< m_{2}< m_{3}< m_{4}$ to be integers.

Define the sum
\begin{equation*}
	S_{10}(k,m,n,m_{1},m_{2},m_{3},m_{4})=\sum_{i=1}^{n-1}\frac{P_{6}\left(U,V,k,m,i,m_{1},m_{2},m_{3}\right)}{P_{10}(W,\overline{W},k,m,i,0,m_{1},m_{2},m_{3},m_{4})}.
\end{equation*}
For $0\leq i\leq 4$ define the
quantities $a_{i}=a_{i}(g,k,m)$ as follows:
\begin{eqnarray*}
		a_{0} &=& 2 e_{W}^7 U_{2 g k}U_{4 g k}U_{6 g k}U_{8 g k}, \\
    a_{1} &=& (-1)^{m+1} W_{-3g k}W_{-2g k}W_{-g k}\overline{W}_{-3g k}\overline{W}_{-2g k}\overline{W}_{-g k},\\
		a_{2} &=& (-1)^m \left(V_{4 g k}+1\right)W_{-2 g k}W_{-g k}W_{g k}\overline{W}_{-2 g k}\overline{W}_{-g k}\overline{W}_{g k},\\
		a_{3} &=& (-1)^{m+1} \left(V_{4 g k}+1\right)W_{-g k}W_{g k}W_{2 g k}\overline{W}_{-g k}\overline{W}_{g k}\overline{W}_{2 g k},\\
		a_{4} &=& (-1)^m W_{g k}W_{2g k}W_{3g k}\overline{W}_{g k}\overline{W}_{2 g k}\overline{W}_{3 g k}.
\end{eqnarray*}
Once again, to discover our next result we needed the $m_{i}$ to take on special values.
\begin{theorem}\label{thm7}  Let $(m_{1},m_{2},m_{3},m_{4})=(g,2g,3g,4g)$, so that $0$, $m_{1}$, $m_{2}$, $m_{3}$, and $m_{4}$ form an arithmetic progression. Then
\begin{eqnarray*}
	a_{0}\left(S_{10}(n)-S_{10}(2)\right) & = & U_{k(n-2)}\left(a_{1}\Omega_{W}(0,g)+a_{2}\Omega_{W}(g,2g)
	\right.\nonumber\\ & & \left.
	{}+a_{3}\Omega_{W}(2g,3g)+a_{4}\Omega_{W}(3g,4g)
		\right.\nonumber\\ & & \left.
	{}-\Delta a_{1}\Omega_{\overline{W}}(0,g)-\Delta a_{2}\Omega_{\overline{W}}(g,2g)
	\right.\nonumber\\ & & \left.
	{}-\Delta a_{3}\Omega_{\overline{W}}(2g,3g)-\Delta a_{4}\Omega_{\overline{W}}(3g,4g)\right).
\end{eqnarray*}
\end{theorem}

We have found that the most interesting examples of our results occur when we take $W_{n}=F_{n+c}$ for some non-negative integer $c$. Accordingly, as an instance of
Theorem \ref{thm7} let $k=1$, $m=0$, and $g=1$. Then for $W_{n}=F_{n+5}$ Theorem \ref{thm7} yields
\begin{eqnarray*}
	&& b_{0}\sum_{i=1}^{n-1}\frac{F_{2(i+1)}F_{2(i+2)}F_{2(i+3)}}{F_{2(i+5)}F_{2(i+6)}F_{2(i+7)}F_{2(i+8)}F_{2(i+9)}} \\
	&& = 7+144F_{2(n-2)}\left(\frac{b_{1}}{F_{2(n+5)}}+\frac{b_{2}}{F_{2(n+6)}}+\frac{b_{3}}{F_{2(n+7)}}+\frac{b_{4}}{F_{2(n+8)}}\right),
\end{eqnarray*}
where
\begin{eqnarray*}
		b_{0} &=& 13009146630480,\\
		b_{1} &=& -239632085,\\
		b_{2} &=& 35147981440,\\
		b_{3} &=& -632668766730,\\
		b_{4} &=& 1419740509642.
\end{eqnarray*}

We also considered variants of $S_{10}$ that we obtained in ways similar to those described in the paragraph at the end of Section \ref{sec4}. However, in each case
we were unable to find the closed form of the corresponding finite sum. Furthermore, we considered summands with only two factors in the numerator. Once again, in each
case, we were unable to find the closed form of the corresponding finite sum. Theorem \ref{thm7} is the only result of its kind (i.e., where the summand has ten factors in the denominator, and
where the $m_{i}$ are multiples of a positive integer parameter g) that we could find. We have discovered closed forms for such sums where
the $m_{i}$ are {\it specific} integers, but we refrain from giving these sums.


\section{The summand has twelve factors in the denominator}\label{sec6}


In this section we take $0<m_{1}< m_{2}< m_{3}< m_{4}< m_{5}$ to be integers.

Define the sum
\begin{equation*}
	S_{12}(k,m,n,m_{1},m_{2},m_{3},m_{4},m_{5})=\sum_{i=1}^{n-1}\frac{1}{P_{12}(W,\overline{W},k,m,i,0,m_{1},m_{2},m_{3},m_{4},m_{5})}.
\end{equation*}
 For $0\leq i\leq 5$ we define the quantities $a_{i}=a_{i}(g,k,m)$ as
\begin{eqnarray*}
		a_{0} &=& 2 e_{W}^5 U_{2 g k}U_{4 g k}U_{8 g k}U_{10 g k}U_{12 g k}, \\
    a_{1} &=& a_{5}=(-1)^m,\\
		a_{2} &=& a_{4}=(-1)^{m+1}\left(V_{8 g k}+V_{4 g k}-1\right),\\
		a_{3} &=& (-1)^m \left(V_{12 g k}-V_{8 g k}+2\right).
\end{eqnarray*}
In our next theorem 0, $m_{1}$, $m_{2}$, $m_{3}$, $m_{4}$, and $m_{5}$ are not required to form an arithmetic progression.
\begin{theorem}\label{thm8}  Setting $(m_{1},m_{2},m_{3},m_{4},m_{5})=(g,2g,4g,5g,6g)$ we have
\begin{eqnarray*}
	a_{0}\left(S_{12}(n)-S_{12}(2)\right) & = & U_{k(n-2)}\left(a_{1}\Omega_{W}(0,g)+a_{2}\Omega_{W}(g,2g)
	\right.\nonumber\\ & & \left.
	{}+a_{3}\Omega_{W}(2g,4g)+a_{4}\Omega_{W}(4g,5g)
		\right.\nonumber\\ & & \left.
	{}+a_{5}\Omega_{W}(5g,6g)-\Delta a_{1}\Omega_{\overline{W}}(0,g)
	\right.\nonumber\\ & & \left.
	{}-\Delta a_{2}\Omega_{\overline{W}}(g,2g)-\Delta a_{3}\Omega_{\overline{W}}(2g,4g)
		\right.\nonumber\\ & & \left.
	{}-\Delta a_{4}\Omega_{\overline{W}}(4g,5g)-\Delta a_{5}\Omega_{\overline{W}}(5g,6g)\right).
\end{eqnarray*}
\end{theorem}

We also managed to find a closed form for $S_{12}$ if $(m_{1},m_{2},m_{3},m_{4},m_{5})=(g,2g,3g,4g,5g)$, but we refrain from giving this result here.

To indicate other types of results that are possible, define the sum
\begin{equation*}
		T_{12}(k,m,n,m_{1},m_{2},m_{3},m_{4},m_{5}) = \sum_{i=1}^{n-1}\frac{P_{8}\left(U,V,k,m,i,m_{1},m_{2},m_{3},m_{4}\right)}{P_{12}(W,\overline{W},k,m,i,0,m_{1},m_{2},m_{3},m_{4},m_{5})}.
\end{equation*}
We managed to find a closed form for $T_{12}$ under the assumption that the $m_{i}$ are certain multiples of $g$. One such instance is for $(m_{1},m_{2},m_{3},m_{4},m_{5})$=$(g,2g,3g,4g,5g)$.
Another instance is for $(m_{1},m_{2},m_{3},m_{4},m_{5})$=$(2g,3g,4g,5g,7g)$. We have also discovered other results of a similar nature that we do not present here.


\section{Concluding comments}\label{sec7}

Earlier we stated that we chose to limit the scope of this paper to finite sums that have four, or six, or eight, or ten, or twelve products
in the denominator of the summand. We have, however, discovered closed forms for finite sums (with the structure described in the introduction) that have fourteen, or
sixteen, or eighteen, or twenty products in the denominator of the summand. The possibilities seem limitless.

\begin{thebibliography}{9}

\bibitem{mel1}
R.~S.~Melham, Finite sums that involve reciprocals of products of generalized Fibonacci numbers, {\it Integers} \textbf{13} (2013), \#A40.
Available at \newline \url{http://www.integers-ejcnt.org/vol13.html}.

\bibitem{mel2}
R.~S.~Melham, More on finite sums that involve reciprocals of products of generalized Fibonacci numbers, {\it Integers} \textbf{14} (2014), \#A4.
Available at \newline \url{http://www.integers-ejcnt.org/vol14.html}.

\end{thebibliography}


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\noindent {\it 2010 Mathematics Subject Classification:} Primary 11B39; Secondary  11B37.

\noindent {\it Keywords:} reciprocal summation, Fibonacci number, Lucas number.


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

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\vspace*{+.1in}
\noindent
Received February 4 2014;
revised version received April 16 2014. 
Published in {\it Journal of Integer Sequences}, May 12 2014.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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