\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand\BD{\mathrm{B}} \newcommand\SD{\mathrm{S}} \newcommand{\N}{{\mathbb N}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\C}{{\mathcal C}} \newcommand{\A}{{\mathcal A}} \newcommand{\F}{{\mathcal F}} \begin{center} \vskip 1cm{\LARGE\bf Factored Closed-form Expressions for the Sums of Cubes of Fibonacci and Lucas Numbers \vskip 1cm} \large Kunle Adegoke\\ Department of Physics and Engineering Physics\\ Obafemi Awolowo University\\ 220005 Ile-Ife\\ Nigeria\\ \href{mailto:adegoke00@gmail.com}{\tt adegoke00@gmail.com} \end{center} \vskip .2 in \begin{abstract} We obtain factored closed-form expressions for the sums of cubes of Fibonacci and Lucas numbers. \end{abstract} \section{Introduction} The Fibonacci numbers, $F_n$, and Lucas numbers, $L_n$, are defined, for \mbox{$n\in\Z$}, as usual, through the recurrence relations \mbox{$F_n=F_{n-1}+F_{n-2}$}, \mbox{$F_0=0$, $F_1=1$} and \mbox{$L_n=L_{n-1}+L_{n-2}$}, $L_0=2$, $L_1=1$, with $F_{-n}=(-1)^{n-1}F_n$ and $L_{-n}=(-1)^nL_n$. Clary and Hemenway~\cite{clary} derived the remarkable formulas \begin{equation}\label{equ.mnyq31m} 4\sum_{k=1}^nF_{2k}^3= \begin{cases} F_n^2L_{n+1}^2F_{n-1}L_{n+2}, & \text{if $n$ is even};\\ L_n^2F_{n+1}^2L_{n-1}F_{n+2}, & \text{if $n$ is odd}, \end{cases} \end{equation} and \begin{equation}\label{equ.gd1awb5} 8\sum_{k = 1}^n {F_{4k}^3 } = F_{2n}^2 F_{2n + 2}^2 (L_{4n + 2} + 6)\,. \end{equation} In this present paper we will derive the following corresponding Lucas counterparts of~\eqref{equ.mnyq31m} and~\eqref{equ.gd1awb5}: \begin{equation}\label{equ.aa5tal0} 4\sum_{k=1}^nL_{2k}^3= \begin{cases} 5F_nF_{n+1}(L_nL_{n+1}L_{2n+1}+16), & \text{if $n$ is even};\\ L_nL_{n+1}(5F_nF_{n+1}L_{2n+1}+16), & \text{if $n$ is odd}, \end{cases} \end{equation} and \begin{equation} 8\sum_{k = 1}^n {L_{4k}^3 } = F_{2n} L_{2n + 2} (5L_{2n}F_{2n + 2}F_{4n + 2} + 32)\,. \end{equation} In fact we will derive the following more general results: \begin{itemize} \item If $r$ is odd, then \begin{equation}\label{equ.ewu6n75} L_{3r}\sum_{k=1}^nF_{2rk}^3= \begin{cases} F_{rn}^2L_{rn+r}^2(L_{rn}F_{rn+r}+F_r), & \text{if $n$ is even};\\ L_{rn}^2F_{rn+r}^2(F_{rn}L_{rn+r}+F_r), & \text{if $n$ is odd}, \end{cases} \end{equation} and \begin{equation}\label{equ.bnbadtf} L_{3r}\sum_{k=1}^nL_{2rk}^3= \begin{cases} 5F_{rn}F_{rn+r}(L_{rn}L_{rn+r}L_{2rn+r}+4(L_{2r}+1)), & \text{if $n$ is even};\\ L_{rn}L_{rn+r}(5F_{rn}F_{rn+r}L_{2rn+r}+4(L_{2r}+1)), & \text{if $n$ is odd}. \end{cases} \end{equation} \item If $r$ is even, then \begin{equation}\label{equ.px681ib} F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } = F_{rn}^2 F_{rn + r}^2 (L_{rn} L_{rn + r} + L_r ) \end{equation} and \begin{equation} F_{3r} \sum_{k = 1}^n {L_{2rk}^3 } = F_{rn} L_{rn + r} (5L_{rn}F_{rn + r}F_{2rn + r} + 4(L_{2r} + 1))\,. \end{equation} \end{itemize} As variations on identities~\eqref{equ.ewu6n75} and~\eqref{equ.px681ib} we will prove \begin{itemize} \item If $r$ is odd, then \[ L_{3r}\sum_{k=1}^nF_{2rk}^3= \begin{cases} F_{rn}L_{rn+r}(L_{rn}F_{rn+r}F_{2rn+r}-2F_r^2), & \text{if $n$ is even};\\ L_{rn}F_{rn+r}(F_{rn}L_{rn+r}F_{2rn+r}-2F_r^2), & \text{if $n$ is odd}. \end{cases} \] \item If $r$ is even, then \[ 5F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } = F_{rn} F_{rn + r}(L_{rn} L_{rn + r} L_{2rn + r} - 2L_r^2)\,. \] \end{itemize} \section{Required identities and preliminary results} \subsection{Telescoping summation identity} The following telescoping summation identity is a special case of more general identities proved by Adegoke~\cite{adegoke}. \begin{lemma}\label{finall} If $f(k)$ is a real sequence and $m$, $q$ and $n$ are positive integers, then \[ \sum_{k = 1}^n {\left[ {f(mk + mq) - f(mk)} \right]} = \sum_{k = 1}^q {f(mk + mn)} - \sum_{k = 1}^q {f(mk)}\,. \] \end{lemma} \subsection{First-power Fibonacci summation identities} \begin{lemma}\label{thm.esyph89} If $r$ and $n$ are integers, then \begin{enumerate} \item[(i)] If $r$ is even, then \[ F_r\sum_{k=1}^nF_{2rk}=F_{rn}F_{rn+r}\,. \] \item[(ii)] If $r$ is odd, then \[ L_r\sum_{k=1}^nF_{2rk}= \begin{cases} F_{rn}L_{rn+r}, & \text{if $n$ is even};\\ L_{rn}F_{rn+r}, & \text{if $n$ is odd}. \end{cases} \] \end{enumerate} \end{lemma} \begin{proof} Setting $v=2r$ and $u=2rk$ in the identity \begin{equation} L_{u + v} - (-1)^vL_{u-v}=5F_uF_v\label{equ.q79u3q4} \end{equation} gives \begin{equation}\label{equ.jv8ztfh} L_{2rk + 2r} - L_{2rk - 2r} = 5F_{2r} F_{2rk}\,. \end{equation} Taking \mbox{$f(k)=L_{k-2r}$}, $q=2$ and $m=2r$ in Lemma~\ref{finall} and employing identity~\eqref{equ.jv8ztfh} we have \begin{equation}\label{equ.w8cpa7u} \begin{split} 5F_{2r} \sum_{k = 1}^n {F_{2rk} } &= \sum_{k = 1}^2 {L_{2rk + 2rn - 2r} } - \sum_{k = 1}^2 {L_{2rk - 2r} }\\ &= L_{2rn + 2r} + L_{2rn} - L_{2r} - 2\,. \end{split} \end{equation} If $r$ is even, then on account of the identity \begin{equation} L_{u + v} + (-1)^vL_{u-v}=L_uL_v \label{equ.ztsb3uk},\\ \end{equation} we have \[ L_{2rn+2r}+L_{2rn}=L_rL_{2rn+r},\quad L_{2r}+2=L_r^2\,, \] and since \begin{equation}\label{equ.gv7zoe2} F_{2u}=F_uL_u\,, \end{equation} identity~\eqref{equ.w8cpa7u} now becomes \begin{equation} \begin{split} 5F_{r} \sum_{k = 1}^n {F_{2rk} } &= L_{2rn + r} - L_r\\ &=5F_{rn}F_{rn+r}\,,\quad\mbox{by~\eqref{equ.q79u3q4}}\,, \end{split} \end{equation} that is, \[ F_{r} \sum_{k = 1}^n {F_{2rk} }=F_{rn}F_{rn+r},\quad\mbox{$r$ even}\,, \] and the first part of Lemma~\ref{thm.esyph89} is proved. If $r$ is odd, then on account of the identities~\eqref{equ.q79u3q4} and~\eqref{equ.ztsb3uk}, we have \[ L_{2rn+2r}+L_{2rn}=5F_rF_{2rn+r},\quad L_{2r}+2=5F_r^2\,, \] and identity~\eqref{equ.w8cpa7u} reduces to \begin{align*} L_r\sum_{k=1}^nF_{2rk} &=F_{2rn+r}-F_r\\ &= \begin{cases} F_{rn}L_{rn+r}, & \text{if $n$ is even};\\ L_{rn}F_{rn+r}, & \text{if $n$ is odd,} \end{cases} \end{align*} and the second part of Lemma~\ref{thm.esyph89} is proved. In the last stage of the above derivation we made use of the identities \begin{equation}\label{equ.yb05ue2} F_{u + v} - (-1)^vF_{u-v}=F_vL_u \end{equation} and \begin{equation}\label{equ.jjkzwa0} F_{u + v} + (-1)^vF_{u-v}=L_vF_u\,. \end{equation} \end{proof} \subsection{First-power Lucas summation identities} \begin{lemma}\label{thm.hnf079l} If $r$ and $n$ are integers, then \begin{enumerate} \item[(i)] If $r$ is even, then \[ F_r\sum_{k=1}^nL_{2rk}=F_{rn}L_{rn+r}\,. \] \item[(ii)] If $r$ is odd, then \[ L_r\sum_{k=1}^nL_{2rk}= \begin{cases} 5F_{rn}F_{rn+r}, & \text{if $n$ is even};\\ L_{rn}L_{rn+r}, & \text{if $n$ is odd}. \end{cases} \] \end{enumerate} \end{lemma} \begin{proof} Setting $v=2r$ and $u=2rk$ in the identity~\eqref{equ.yb05ue2} gives \begin{equation}\label{equ.gimtkiy} F_{2rk + 2r} - F_{2rk - 2r} = F_{2r} L_{2rk}\,. \end{equation} Taking \mbox{$f(k)=F_{k-2r}$}, $q=2$ and $m=2r$ in Lemma~\ref{finall} and employing identity~\eqref{equ.gimtkiy} we have \begin{equation}\label{equ.jhbv3g7} \begin{split} F_{2r} \sum_{k = 1}^n {L_{2rk} } &= \sum_{k = 1}^2 {F_{2rk + 2rn - 2r} } - \sum_{k = 1}^2 {F_{2rk - 2r} }\\ &= F_{2rn+2r}+F_{2rn}-F_{2r}\,. \end{split} \end{equation} If $r$ is even, then choosing $v=r$ and $u=2rn+r$ in identity~\eqref{equ.jjkzwa0} gives \begin{equation} F_{2rn+2r}+F_{2rn}=L_rF_{2rn+r} \end{equation} and, on account of identity~\eqref{equ.gv7zoe2}, the identity~\eqref{equ.jhbv3g7} reduces to \[ \begin{split} F_r\sum_{k=1}^nL_{2rk}&=F_{2rn+r}-F_r\\ &=F_{rn+r+rn}-F_{rn+r-rn}\\ &=F_{rn}L_{rn+r}\,,\quad\mbox{by identity~\eqref{equ.yb05ue2},} \end{split} \] and the first part of Lemma~\ref{thm.hnf079l} is proved. If $r$ is odd, then choosing $v=r$ and $u=2rn+r$ in identity~\eqref{equ.yb05ue2} gives \begin{equation} F_{2rn+2r}+F_{2rn}=F_rL_{2rn+r} \end{equation} and, again on account of identity~\eqref{equ.gv7zoe2}, the identity~\eqref{equ.jhbv3g7} now reduces to \[ \begin{split} L_r\sum_{k=1}^nL_{2rk}&=L_{2rn+r}-L_r\\ &=L_{rn+r+rn}-L_{rn+r-rn}\\ &= \begin{cases} 5F_{rn}F_{rn+r}, & \text{if $n$ is even};\\ L_{rn}L_{rn+r}, & \text{if $n$ is odd}, \end{cases} \end{split} \] where in the last step we used the identities~\eqref{equ.q79u3q4} and~\eqref{equ.ztsb3uk}. \end{proof} \subsection{Other identities} \begin{lemma}\label{thm.w3ql3gi} If $r$ and $n$ are integers, then \[ \frac{F_{3rn}F_{3rn+3r}}{F_{rn}F_{rn+r}}=L_{rn}L_{rn+r}L_{2rn+r}+L_{2r}+(-1)^{r-1}. \] \end{lemma} \begin{proof} Using the identity Clary~\cite[Eq.~(36)]{clary}, or Dresel~\cite[Eq.~(3.3)]{dresel}, namely, \begin{equation}\label{equ.oyvdke0} F_{3u}=5F_u^3+3(-1)^uF_u\,, \end{equation} we have \begin{equation}\label{equ.gs4xocq} \begin{split} \frac{{F_{3rn} F_{3rn + 3r} }}{{F_{rn} F_{rn + r} }} &= (5F_{rn}^2 + 3( - 1)^{rn} )(5F_{rn + r}^2 + 3( - 1)^{rn + r} )\\ &= (L_{rn}^2 - ( - 1)^{rn} )(L_{rn + r}^2 - ( - 1)^{rn + r} )\\ &= L_{rn}^2 L_{rn + r}^2 - ( - 1)^{rn + r} L_{rn}^2 - ( - 1)^{rn} L_{rn + r}^2 + ( - 1)^r\,, \end{split} \end{equation} where we have also made use of the identity \begin{equation}\label{equ.mpb5qh0} 5F_u^2-L_u^2=(-1)^{u-1}4\,. \end{equation} Now, \[ \begin{split} L_{rn}^2 L_{rn + r}^2&= L_{rn} L_{rn + r} (L_{rn} L_{rn + r} )\\ &= L_{rn} L_{rn + r} (L_{2rn + r} + ( - 1)^{rn} L_r )\quad\mbox{by~\eqref{equ.ztsb3uk}}\\ &= L_{rn} L_{rn + r} L_{2rn + r} + ( - 1)^{rn} L_{rn} L_{rn + r} L_r \,. \end{split} \] Therefore \[ \frac{{F_{3rn} F_{3rn + 3r} }}{{F_{rn} F_{rn + r} }} = L_{rn} L_{rn + r} L_{2rn + r} + ( - 1)^{rn} L_{rn + r} (L_{rn}L_r - L_{rn + r} ) - ( - 1)^{rn + r} L_{rn}^2 + ( - 1)^r\,. \] But \[ \begin{split} &( - 1)^{rn} L_{rn + r} (L_{rn} L_r - L_{rn + r} )\\ &= ( - 1)^{rn} L_{rn + r} (L_{rn + r} + ( - 1)^r L_{rn - r} - L_{rn + r} ),\quad\mbox{by~\eqref{equ.ztsb3uk}}\\ & = ( - 1)^{rn + r} L_{rn + r} L_{rn - r}\\ &= ( - 1)^{rn + r} (L_{2rn} + ( - 1)^{rn - r} L_{2r} ),\quad\mbox{again by~\eqref{equ.ztsb3uk}}\\ &= ( - 1)^{rn + r} L_{2rn} + L_{2r}\,. \end{split} \] Thus \begin{align*} \frac{{F_{3rn} F_{3rn + 3r} }}{{F_{rn} F_{rn + r} }} &= L_{rn} L_{rn + r} L_{2rn + r} + ( - 1)^{rn + r} L_{2rn} + L_{2r} - ( - 1)^{rn + r} L_{rn}^2 + ( - 1)^r \\ &= L_{rn} L_{rn + r} L_{2rn + r} + ( - 1)^{rn + r} (L_{2rn} - L_{rn}^2 ) + L_{2r} + ( - 1)^r\,. \end{align*} Finally, using the identity \begin{equation}\label{equ.fpjkos2} L_{2u}=L_u^2+(-1)^{u-1}2\,, \end{equation} obtained by setting $v=u$ in identity~\eqref{equ.ztsb3uk}, we have the statement of the Lemma. \end{proof} \begin{lemma}\label{thm.cdcw5y2} If $r$ and $n$ are integers, then \[ \frac{L_{3rn}L_{3rn+3r}}{L_{rn}L_{rn+r}}=5F_{rn}F_{rn+r}L_{2rn+r}+L_{2r}+(-1)^{r-1}. \] \end{lemma} \begin{proof} Using the following identity, of Dresel~\cite[Eq.~(1.6)]{dresel} \begin{equation}\label{equ.aiunttr} L_{3u}=L_u^3-3(-1)^uL_u\,, \end{equation} we have \[ \begin{split} \frac{{L_{3rn} L_{3rn + 3r} }}{{L_{rn} L_{rn + r} }} &= (L_{rn}^2 - 3( - 1)^{rn} )(L_{rn + r}^2 - 3( - 1)^{rn + r} )\\ &= (5F_{rn}^2 + ( - 1)^{rn} )(5F_{rn + r}^2 + ( - 1)^{rn + r} ),\quad\mbox{by~\eqref{equ.mpb5qh0}}\\ &= 25F_{rn}^2 F_{rn + r}^2 + ( - 1)^{rn + r} 5F_{rn}^2 + ( - 1)^{rn} 5F_{rn + r}^2 + ( - 1)^r\,, \end{split} \] and the rest of the calculation then proceeds as in the proof of Lemma~\ref{thm.w3ql3gi}, the basic required identities now being~\eqref{equ.q79u3q4}, \eqref{equ.jjkzwa0} and the identity \begin{equation}\label{equ.gr4yegj} L_{2u}=5F_u^2+(-1)^u2, \end{equation} obtained by setting $v=u$ in identity~\eqref{equ.q79u3q4}. \end{proof} \begin{lemma}\label{thm.ip29mgz} If $r$ and $n$ are integers, then \[ \frac{L_{3rn}F_{3rn+3r}}{L_{rn}F_{rn+r}}=5F_{rn}L_{rn+r}F_{2rn+r}+L_{2r}+(-1)^r. \] \end{lemma} \begin{lemma}\label{thm.c1bi001} If $r$ and $n$ are integers, then \[ \frac{F_{3rn}L_{3rn+3r}}{F_{rn}L_{rn+r}}=5L_{rn}F_{rn+r}F_{2rn+r}+L_{2r}+(-1)^r. \] \end{lemma} Different but equivalent versions of Lemmas~\ref{thm.w3ql3gi}--\ref{thm.c1bi001} are given below: \begin{lemma}\label{thm.c6tpo1r} If $r$ and $n$ are integers, then \[ \frac{{F_{3rn} F_{3rn + 3r} }}{{F_{rn} F_{rn + r} }} = L_{2rn + r}^2 + ( - 1)^{nr} L_{rn + r}^2 + ( - 1)^{(n - 1)r} L_{rn}^2 + L_r^2 + ( - 1)^{r - 1} 7\,. \] \end{lemma} \begin{proof} The proof is similar to that of Lemma~\ref{thm.w3ql3gi}, but here we use \begin{align*} L_{rn}^2 L_{rn + r}^2 &= (L_{2rn + r} + ( - 1)^{rn} L_r )^2\\ &= L_{2rn + r}^2 + L_r^2 + 2( - 1)^{rn} ( L_r L_{2rn + r})\\ &= L_{2rn + r}^2 + L_r^2 + 2( - 1)^{rn} ( L_{2rn + 2r} + ( - 1)^r L_{2rn} )\\ &= L_{2rn + r}^2 + L_r^2 + 2( - 1)^{rn} ( L_{rn + r}^2 + ( - 1)^{rn + r - 1} 2 + ( - 1)^r (L_{rn}^2 + ( - 1)^{rn - 1} 2)), \end{align*} and substitute in~\eqref{equ.gs4xocq}. \end{proof} \begin{lemma}\label{thm.da8g4ih} If $r$ and $n$ are integers, then \[ \frac{{L_{3rn} L_{3rn + 3r} }}{{L_{rn} L_{rn + r} }} = L_{2rn + r}^2 + ( - 1)^{nr-1} L_{rn + r}^2 - ( - 1)^{(n - 1)r} L_{rn}^2 + L_r^2 + ( - 1)^r\,. \] \end{lemma} \begin{lemma}\label{thm.kquo1a5} If $r$ and $n$ are integers, then \[ \frac{{L_{3rn} F_{3rn + 3r} }}{{L_{rn} F_{rn + r} }} = 5F_{2rn + r}^2 + ( - 1)^{nr - 1} 5F_{rn + r}^2 + ( - 1)^{(n - 1)r} 5F_{rn}^2 + 5F_r^2 + ( - 1)^r 3\,. \] \end{lemma} \begin{lemma}\label{thm.powhazn} If $r$ and $n$ are integers, then \[ \frac{{F_{3rn} L_{3rn + 3r} }}{{F_{rn} L_{rn + r} }} = 5F_{2rn + r}^2 + ( - 1)^{nr} 5F_{rn + r}^2 - ( - 1)^{(n - 1)r} 5F_{rn}^2 + 5F_r^2 + ( - 1)^r 3\,. \] \end{lemma} \section{Main results} \subsection{Sums of cubes of Fibonacci numbers} \begin{theorem}\label{thm.antrd3y} If $r$ and $n$ are integers such that $r$ is odd, then \[ L_{3r}\sum_{k=1}^nF_{2rk}^3= \begin{cases} F_{rn}L_{rn+r}(L_{rn}F_{rn+r}F_{2rn+r}-2F_r^2), & \text{if $n$ is even};\\ L_{rn}F_{rn+r}(F_{rn}L_{rn+r}F_{2rn+r}-2F_r^2), & \text{if $n$ is odd}. \end{cases} \] \end{theorem} \begin{proof} Setting $u=2rk$ in identity~\eqref{equ.oyvdke0} and summing, we have \[ 5\sum_{k = 1}^n {F_{2rk}^3 } = \sum_{k = 1}^n {F_{6rk} } - 3\sum_{k = 1}^n {F_{2rk} }\,, \] so that, \begin{equation}\label{equ.dpiq8yz} \begin{split} 5L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } &= L_{3r} \sum_{k = 1}^n {F_{6rk} } - 3\frac{{L_{3r} }}{{L_r }}L_r \sum_{k = 1}^n {F_{2rk} }\\ &= L_{3r} \sum_{k = 1}^n {F_{6rk} } - 3(L_r^2 + 3)L_r \sum_{k = 1}^n {F_{2rk} }\,. \end{split} \end{equation} %%numbered list \begin{itemize} \item If $n$ is even, then, by Lemma~\ref{thm.esyph89}, identity~\eqref{equ.dpiq8yz} can be written as \[ 5L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } = F_{3rn} L_{3rn + 3r} - 3(L_r^2 + 3)F_{rn} L_{rn + r}\,, \] so that \[ \begin{split} \frac{{5L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{F_{rn} L_{rn + r} }} &= \frac{{F_{3rn} L_{3rn + 3r} }}{{F_{rn} L_{rn + r} }} - 3(L_r^2 + 3)\\ &= 5L_{rn} F_{rn + r} F_{2rn + r} + L_{2r} - 1 - 3L_r^2 - 9,\quad\mbox{by Lemma~\ref{thm.c1bi001}}\\ &= 5L_{rn} F_{rn + r} F_{2rn + r} - 10F_r^2,\quad\mbox{by \eqref{equ.mpb5qh0} and \eqref{equ.fpjkos2}}\,. \end{split} \] \item If $n$ is odd, then, by Lemma~\ref{thm.esyph89}, we have \[ 5L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } = L_{3rn} F_{3rn + 3r} - 3(L_r^2 + 3)L_{rn} F_{rn + r}\,, \] so that \[ \begin{split} \frac{{5L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{L_{rn} F_{rn + r} }} &= \frac{{L_{3rn} F_{3rn + 3r} }}{{L_{rn} F_{rn + r} }} - 3(L_r^2 + 3)\\ &= 5F_{rn} L_{rn + r} F_{2rn + r} + L_{2r} - 1 - 3L_r^2 - 9,\quad\mbox{by Lemma~\ref{thm.c1bi001}}\\ &= 5F_{rn} L_{rn + r} F_{2rn + r} - 10F_r^2,\quad\mbox{by \eqref{equ.mpb5qh0} and \eqref{equ.fpjkos2}}\,. \end{split} \] \end{itemize} \end{proof} \begin{theorem} If $r$ and $n$ are integers such that $r$ is even, then \[ 5F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } = F_{rn} F_{rn + r}(L_{rn} L_{rn + r} L_{2rn + r} - 2L_r^2)\,. \] \end{theorem} \begin{proof} \[ \begin{split} 5F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } &= F_{3r} \sum_{k = 1}^n {F_{6rk} } - 3\frac{{F_{3r} }}{{F_r }}F_r \sum_{k = 1}^n {F_{2rk} }\\ &= F_{3rn} F_{3rn + 3r} - 3(5F_r^2 + 3)F_{rn} F_{rn + r},\\ & \mbox{by Lemma \ref{thm.esyph89} and identity \eqref{equ.oyvdke0}}\,, \end{split} \] so that \[ \begin{split} \frac{{5F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{F_{rn} F_{rn + r} }} &= \frac{{F_{3rn} F_{3rn + 3r} }}{{F_{rn} F_{rn + r} }} - 3(5F_r^2 + 3)\\ &= L_{rn} L_{rn + r} L_{2rn + r} + L_{2r} - 1 - 15F_r^2 - 9\\ &\mbox{(by Lemma \ref{thm.w3ql3gi} and identity \eqref{equ.oyvdke0})}\,,\\ &= L_{rn} L_{rn + r} L_{2rn + r} - 2L_r^2,\quad\mbox{by \eqref{equ.mpb5qh0}, \eqref{equ.fpjkos2} and \eqref{equ.gr4yegj}}\,. \end{split} \] \end{proof} \begin{theorem} If $r$ and $n$ are integers such that $r$ is odd, then \[ L_{3r}\sum_{k=1}^nF_{2rk}^3= \begin{cases} F_{rn}^2L_{rn+r}^2(L_{rn}F_{rn+r}+F_r), & \text{if $n$ is even};\\ L_{rn}^2F_{rn+r}^2(F_{rn}L_{rn+r}+F_r), & \text{if $n$ is odd}. \end{cases} \] \end{theorem} \begin{proof} %%numbered list \begin{itemize} \item If $n$ is even, then from Lemma~\ref{thm.esyph89} and identity~\eqref{equ.dpiq8yz} we have \[ \begin{split} \frac{{5L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{F_{rn} L_{rn + r} }} &= \frac{{F_{3rn} L_{3rn + 3r} }}{{F_{rn} L_{rn + r} }} - 3(L_r^2 + 3)\\ &= 5F_{2rn + r}^2 + 5F_{rn + r}^2 + 5F_{rn}^2 + 5F_r^2 - 3 - 3L_r^2 - 9,\,\mbox{by Lemma~\ref{thm.powhazn}}\\ &= 5F_{2rn + r}^2 + 5F_{rn + r}^2 + 5F_{rn}^2 - 10F_r^2\quad\mbox{by identity~\eqref{equ.mpb5qh0}}\,, \end{split} \] so that \[ \begin{split} \frac{{L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{F_{rn} L_{rn + r} }} &= F_{2rn + r}^2 + F_{rn + r}^2 + F_{rn}^2 - 2F_r^2\\ &= (F_{2rn + r}^2 - F_r^2 ) + (F_{rn + r}^2+ F_{rn}^2) - F_r^2\,. \end{split} \] Using the following identity, derived by Howard~\cite{howard}, \begin{equation} F_u^2 + (-1)^{u+v-1}F_v^2=F_{u-v}F_{u+v}\,,\label{equ.wwhhpdy} \end{equation} we have \[ \begin{split} \frac{{L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{F_{rn} L_{rn + r} }}&= F_{2rn} F_{2rn + 2r} + F_r F_{2rn + r} - F_r^2\\ &= F_{2rn} F_{2rn + 2r} + F_r (F_{2rn + r} - F_r )\\ &= F_{2rn} F_{2rn + 2r} + F_r F_{rn} L_{rn + r},\,\mbox{by identity~\eqref{equ.yb05ue2}}\\ &= F_{rn} L_{rn + r} L_{rn} F_{rn + r} + F_r F_{rn} L_{rn + r}\\ &= F_{rn} L_{rn + r} (L_{rn} F_{rn + r} + F_r )\,. \end{split} \] \item If $n$ is odd, then from Lemma~\ref{thm.esyph89} and identity~\eqref{equ.dpiq8yz} we have \[ \begin{split} \frac{{5L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{L_{rn} F_{rn + r} }} &= \frac{{L_{3rn} F_{3rn + 3r} }}{{L_{rn} F_{rn + r} }} - 3(L_r^2 + 3)\\ &= 5F_{2rn + r}^2 + 5F_{rn + r}^2 + 5F_{rn}^2 + 5F_r^2 - 3 - 3L_r^2 - 9,\,\mbox{by Lemma~\ref{thm.kquo1a5}}\\ &= 5F_{2rn + r}^2 + 5F_{rn + r}^2 + 5F_{rn}^2 - 10F_r^2\quad\mbox{by identity~\eqref{equ.mpb5qh0}}\,, \end{split} \] so that \[ \begin{split} \frac{{L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{L_{rn} F_{rn + r} }} &= F_{2rn + r}^2 + F_{rn + r}^2 + F_{rn}^2 - 2F_r^2\\ &= (F_{2rn + r}^2 - F_r^2 ) + (F_{rn + r}^2+ F_{rn}^2) - F_r^2\,. \end{split} \] Using identity~\eqref{equ.wwhhpdy}, we have \[ \begin{split} \frac{{L_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{L_{rn} F_{rn + r} }}&= F_{2rn} F_{2rn + 2r} + F_r F_{2rn + r} - F_r^2\\ &= F_{2rn} F_{2rn + 2r} + F_r (F_{2rn + r} - F_r )\\ &= F_{2rn} F_{2rn + 2r} + F_r L_{rn} F_{rn + r},\,\mbox{by identity~\eqref{equ.jjkzwa0}}\\ &= F_{rn} L_{rn + r} L_{rn} F_{rn + r} + F_r L_{rn} F_{rn + r}\\ &= L_{rn} F_{rn + r} (F_{rn} L_{rn + r} + F_r )\,. \end{split} \] \end{itemize} \end{proof} \begin{theorem} If $r$ and $n$ are integers such that $r$ is even, then \begin{equation} F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } = F_{rn}^2 F_{rn + r}^2 (L_{rn} L_{rn + r} + L_r )\,. \end{equation} \end{theorem} \begin{proof} \[ \begin{split} 5F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } &= F_{3r} \sum_{k = 1}^n {F_{6rk} } - 3\frac{{F_{3r} }}{{F_r }}F_r \sum_{k = 1}^n {F_{2rk} }\\ &= F_{3rn} F_{3rn + 3r} - 3(5F_r^2 + 3)F_{rn} F_{rn + r}\,, \end{split} \] so that \[ \begin{split} \frac{{5F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{F_{rn} F_{rn + r} }} &= \frac{{F_{3rn} F_{3rn + 3r} }}{{F_{rn} F_{rn + r} }} - 3(5F_r^2 + 3)\\ &=L_{2rn+r}^2+L_{rn+r}^2+L_{rn}^2+L_r^2-7 - 15F_r^2 - 9,\quad\mbox{by Lemma~\ref{thm.c6tpo1r}}\\ &=L_{2rn+r}^2+L_{rn+r}^2-2L_r^2+5F_{rn}^2,\quad\mbox{by \eqref{equ.mpb5qh0}}\\ &=(L_{2rn+r}^2-L_r^2)+(L_{rn+r}^2-L_r^2)+5F_{rn}^2\,. \end{split} \] Using the identity (derived by Howard~\cite{howard}) \begin{equation} L_u^2 + (-1)^{u+v-1}L_v^2=5F_{u-v}F_{u+v}\,,\label{equ.wodrq78} \end{equation} we see that \begin{equation} L_{2rn + r}^2 - L_r^2 = 5F_{2rn} F_{2rn + 2r} = 5F_{rn} F_{rn + r} L_{rn} L_{rn + r} \end{equation} and \begin{equation} L_{rn + r}^2 - L_r^2 = 5F_{rn} F_{rn + 2r}\,. \end{equation} Thus, \[ \begin{split} \frac{{F_{3r} \sum_{k = 1}^n {F_{2rk}^3 } }}{{F_{rn} F_{rn + r} }} &= F_{rn} F_{rn + r} L_{rn} L_{rn + r} + F_{rn} F_{rn + 2r} + F_{rn}^2\\ &= F_{rn} F_{rn + r} L_{rn} L_{rn + r} + F_{rn} (F_{rn} + F_{rn + 2r} )\\ &= F_{rn} F_{rn + r} L_{rn} L_{rn + r} + F_{rn} F_{rn + r} L_r,\quad\mbox{by identity \eqref{equ.jjkzwa0}}\\ &= F_{rn} F_{rn + r} (L_{rn} L_{rn + r} + L_r )\,. \end{split} \] \end{proof} \subsection{Sums of cubes of Lucas numbers} \begin{theorem}\label{thm.htq3o7h} If $r$ and $n$ are integers such that $r$ is odd, then \[ L_{3r}\sum_{k=1}^nL_{2rk}^3= \begin{cases} 5F_{rn}F_{rn+r}(L_{rn}L_{rn+r}L_{2rn+r}+4(L_{2r}+1)), & \text{if $n$ is even};\\ L_{rn}L_{rn+r}(5F_{rn}F_{rn+r}L_{2rn+r}+4(L_{2r}+1)), & \text{if $n$ is odd}. \end{cases} \] \end{theorem} \begin{proof} Using identity~\eqref{equ.aiunttr} with $u=2rk$, we have \[ \sum_{k = 1}^n {L_{2rk}^3 } = \sum_{k = 1}^n {L_{6rk} } + 3\sum_{k = 1}^n {L_{2rk} }\,, \] so that \[ \begin{split} L_{3r} \sum_{k = 1}^n {L_{2rk}^3 } &= L_{3r} \sum_{k = 1}^n {L_{6rk} } + 3\frac{{L_{3r} }}{{L_r }}L_r \sum_{k = 1}^n {L_{2rk} }\\ &= L_{3r} \sum_{k = 1}^n {L_{6rk} } + 3(L_r^2 + 3)L_r \sum_{k = 1}^n {L_{2rk} },\quad\mbox{by \eqref{equ.aiunttr}}\,.\\ \end{split} \] %%numbered list \begin{itemize} \item If $n$ is even, then by Lemma~\ref{thm.hnf079l} we have \begin{equation} L_{3r}\sum_{k = 1}^n {L_{2rk}^3 } = 5F_{3rn} F_{3rn + 3r} + 3(L_r^2 + 3)5F_{rn} F_{rn + r}\,, \end{equation} so that \[ \begin{split} \frac{{L_{3r}\sum_{k = 1}^n {L_{2rk}^3 } }}{{5F_{rn} F_{rn + r} }} &= \frac{{F_{3rn} F_{3rn + 3r} }}{{F_{rn} F_{rn + r} }} + 3(L_r^2 + 3)\\ &= L_{rn} L_{rn + r} L_{2rn + r} + L_{2r} + 1 + 3L_r^2 + 9,\quad\mbox{by Lemma \ref{thm.w3ql3gi}}\\ &= L_{rn} L_{rn + r} L_{2rn + r} + 4(L_{2r} + 1),\quad\mbox{by \eqref{equ.fpjkos2}}\,. \end{split} \] \item If $n$ is odd, then by Lemma~\ref{thm.hnf079l} we have \begin{equation} L_{3r}\sum_{k = 1}^n {L_{2rk}^3 } = L_{3rn} L_{3rn + 3r} + 3(L_r^2 + 3)L_{rn} L_{rn + r}\,, \end{equation} so that \[ \begin{split} \frac{{L_{3r}\sum_{k = 1}^n {L_{2rk}^3 } }}{{L_{rn} L_{rn + r} }} &= \frac{{L_{3rn} L_{3rn + 3r} }}{{L_{rn} L_{rn + r} }} + 3(L_r^2 + 3)\\ &= 5F_{rn} F_{rn + r} L_{2rn + r} + L_{2r} + 1 + 3L_r^2 + 9,\quad\mbox{by Lemma \ref{thm.cdcw5y2}}\\ &= 5F_{rn} F_{rn + r} L_{2rn + r} + 4(L_{2r} + 1),\quad\mbox{by \eqref{equ.fpjkos2}}\,. \end{split} \] \end{itemize} \end{proof} \begin{theorem}\label{thm.qa1rrs8} If $r$ and $n$ are integers such that $r$ is even, then \[ F_{3r} \sum_{k = 1}^n {L_{2rk}^3 } = F_{rn} L_{rn + r} (5L_{rn}F_{rn + r}F_{2rn + r} + 4(L_{2r} + 1))\,. \] \end{theorem} \begin{proof} \[ \begin{split} F_{3r} \sum_{k = 1}^n {L_{2rk}^3 } &= F_{3r} \sum_{k = 1}^n {L_{6rk} } + 3\frac{{F_{3r} }}{{F_r }}F_r \sum_{k = 1}^n {L_{2rk} }\\ &= F_{3r} \sum_{k = 1}^n {L_{6rk} } + 3(5F_r^2 + 3)F_r \sum_{k = 1}^n {L_{2rk} },\quad\mbox{by identity \eqref{equ.oyvdke0}}\\ &= F_{3rn} L_{3rn + 3r} + 3(5F_r^2 + 3)F_{rn} L_{rn + r},\quad\mbox{by Lemma \ref{thm.hnf079l}}\,. \end{split} \] Thus, \[ \begin{split} \frac{{F_{3r} \sum_{k = 1}^n {L_{2rk}^3 } }}{{F_{rn} L_{rn + r} }} &= \frac{{F_{3rn} L_{3rn + 3r} }}{{F_{rn} L_{rn + r} }} + 3(5F_r^2 + 3)\\ &= 5L_{rn} F_{rn + r} F_{2rn + r} + L_{2r} + 1 + 15F_r^2 + 9,\quad\mbox{by Lemma \ref{thm.c1bi001}}\\ &= 5L_{rn} F_{rn + r} F_{2rn + r} + 4(L_{2r} + 1),\quad\mbox{by \eqref{equ.fpjkos2} and \eqref{equ.gr4yegj}}\,. \end{split} \] \end{proof} \section{Acknowledgment} Thanks are due to the referee for a careful reading and for pointing out a couple of typographical errors. \begin{thebibliography}{99} \bibitem{adegoke} K. Adegoke, Generalizations of the reciprocal Fibonacci-Lucas sums of Brousseau, \emph{J. Integer Seq.} {\bf 21} (2018), \htmladdnormallink{Article 18.1.6}{http://www.cs.uwaterloo.ca/journals/JIS/VOL21/Adegoke/ade3.html}. \bibitem{clary} S. Clary and P. D. Hemenway, On sums of cubes of Fibonacci numbers, in G. E. Bergum, A. N. Philippou and A. F. Horadam, eds., \emph{Applications of Fibonacci Numbers}, Kluwer Academic Publishers, 1993, pp. 123--136. \bibitem{dresel} L. A. G. Dresel, Transformations of Fibonacci-Lucas identities, in G. E. Bergum, A. N. Philippou and A. F. Horadam, eds., \emph{Applications of Fibonacci Numbers}, Kluwer Academic Publishers, 1993, pp. 169--184. \bibitem{howard} F. T. Howard, The sum of the squares of two generalized Fibonacci numbers, \emph{Fibonacci Quart.} {\bf 41} (2003), 80--84. \bibitem{melham2000} R. S. Melham, Alternating sums of fourth powers of Fibonacci and Lucas numbers, \emph{Fibonacci Quart.} {\bf 38} (2000), 254--259. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B37. \noindent \emph{Keywords: } sum of cubes, Fibonacci number, Lucas number, telescoping summation. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000032} and \seqnum{A000045}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received February 11 2018; revised versions received March 8 2018; July 1 2018; July 3 2018. Published in {\it Journal of Integer Sequences}, August 22 2018. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}