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	\begin{center}
	\vskip 1cm{\LARGE\bf On $k$-Fibonacci Brousseau Sums
	}
	\vskip 1cm
	\large
 	Prabha Sivaraman Nair\\
	Department of Mathematics\\ 
	Baby John Memorial Government College, Chavara\\ 
	Kerala 691583\\ India\\
	\href{mailto:prabhamaths@gmail.com }{\tt prabhamaths@gmail.com }\\ 
	\ \\
	Rejikumar Karunakaran\\ Department of Mathematics\\ 
	NSS College, Cherthala\\ 
	Kerala 688541\\ India\\
    \href{mailto:rkkmaths@gmail.com}{\tt rkkmaths@gmail.com}\\
	\end{center}
	
	\vskip .2 in
	\author{\\ Department of Mathematics\\ Baby John Memorial Government College, Chavara\\ Kerala, India\\\href{mailto:prabhamaths@gmail.com }{prabhamaths@gmail.com }\and K. Rejikumar\\ Department of Mathematics\\ NSS College, Cherthala, Alappuzha\\ Kerala, India\\\href{mailto:rkkmaths@gmail.com}{rkkmaths@gmail.com}}
\begin{abstract}
	Using elementary methods, we provide formulas for evaluating the Brousseau sum $\sum_{i=1}^{n} i^{p} F_{k,i}$ and the shifted Brousseau sum $\sum_{i=1}^{n} i^{p} F_{k,m+i}$ for all integers $m,p\geq0$, where $(F_{k,i})_{i\geq0}$ is the $k$-Fibonacci sequence defined by the two-term linear recurrence $F_{k,i}=kF_{k,i-1}+F_{k,i-2}$ for $i\geq2$ 
    with initial values $F_{k,0}=0$ and $F_{k,1}=1$. 
\end{abstract}

	\section{Introduction}
	
	In the second issue of {\it Fibonacci Quarterly} in 1963, Brousseau \cite{Brousseau1} proposed a problem to discover an expression for the Fibonacci sum of the form
	\[\sum_{i=1}^{n}i^3F_{i},\]
	where $F_{i}$ is the $i^{\rm th}$ Fibonacci number. In the following year, Erbacher and Fuchs \cite{Erbacher} gave a solution for this problem in terms of $F_{n+2}$ and $F_{n+3}$. Later, Ledin \cite{Ledin}, Brousseau \cite{ Brousseau2}, and Zeitlin \cite{Zeitlin} developed various methods to determine expressions for the Brousseau sum 
	\[\sum_{i=1}^{n} i^{p}F_{i},\] where $p$ is a non-negative integer. Ledin showed that the solution of Erbacher and Fuchs can be expressed in the form
	\begin{equation}\label{eqn.1}
		\sum_{i=1}^{n}i^3F_i=(n^3-6n^2+24n-50)F_{n+1}+(n^3-3n^2+15n-31)F_n+50.
	\end{equation}
	
	Recently, Ollerton \cite{Ollerton},
	Shannon \cite{Shannon}, Hendel \cite{Hendel}, and Adegoke \cite{Adegoke} derived expressions for such sums. Dresden \cite{Dresden} used a different technique to find the sum $\sum_{i=1}^{n} i^{p}F_{i} $ using just the binomial coefficients. 
	Motivated by this, we develop formulas for the sums \[\sum_{i=1}^{n} i^{p}F_{k,i},\hspace{1mm} \text{and} \hspace{1mm} \sum_{i=1}^{n}i^{p}F_{k,m+i},\]
	for integers $m,p\geq0$, where $F_{k,i}$ is the $i^{\rm th}$ $k$-Fibonacci number. For a positive integer $k$, the $k$-Fibonacci sequence (see \cite{FalconPlaza}) $(F_{k,n})_{n\geq0}$ is defined by the two-term linear recurrence  
	\begin{equation}
		\label{eqn.2}
		F_{k,n}=kF_{k,n-1}+F_{k,n-2}, 
	\end{equation}
	with initial values $F_{k,0}=0$ and $F_{k,1}=1$. The numbers $F_{k,n}$ are sometimes called the ``metallic'' or ``metallonacci'' numbers. The numbers $F_{1,n}$ are the ``regular'' Fibonacci numbers  and  $F_{2,n}$ are the Pell numbers $P_n$. The Pell numbers $F_{2,n}$  are sometimes called the {\it silver} Fibonacci numbers, and the numbers $F_{3,n}$  are sometimes called the {\it bronze} Fibonacci numbers. For $k=1, 2, 3$, and $4$, the numbers $F_{k,n}$ are \seqnum{A000045}, \seqnum{A000129}, \seqnum{A006190}, and \seqnum{A001076}, respectively, in the OEIS \cite{oeis}. The $k$-Fibonacci numbers can be extended to negative subscripts by 
	\begin{equation*}
		F_{k,n}=F_{k,n+2}-kF_{k,n+1}, \hspace{1mm} \text{for} \hspace{1mm} n<0.
	\end{equation*}
	
	The Lucas sequence $\bigl(L_n\bigr)_{n\geq0}$ and the Pell-Lucas sequence $\bigl(Q_n\bigr)_{n\geq0}$ are defined, respectively, by
	\[L_{0}=2, \ L_{1}=1, \  L_{n}=L_{n-1}+L_{n-2} \ \text{for} \ n\geq2,\]
	and
	\[Q_{0}=Q_{1}=2, \  Q_{n}=2Q_{n-1}+Q_{n-2} \ \text{for} \ n\geq2.\]
	The numbers $L_n$ and $Q_n$ are \seqnum{A000032} and \seqnum{A002203},
	respectively, in the OEIS. Falc\'on \cite{Falcon} proved that the $4$-Fibonacci numbers $F_{4,n}$ are just ${F_{3n}}/{F_{3}}$, the $11$-Fibonacci numbers $F_{11,n}$ are just ${F_{5n}}/{F_5}$, the $29$-Fibonacci numbers $F_{29,n}$ are just ${F_{7n}}/{F_{7}}$, and so on. In general, for all odd indexed Lucas numbers $L_{m}$, the $L_m$-Fibonacci numbers $F_{{L_m},n}$ are just
	${F_{mn}}/{F_m}$. The sequence $1, 4, 11, 29, 76, \ldots$ is \seqnum{A002878} in the OEIS. Falc\'on \cite{Falcon} also proved that the $14$-Fibonacci numbers $F_{14,n}$ are just ${P_{3n}}/{P_{3}}$, the $82$-Fibonacci numbers $F_{82,n}$ are just ${P_{5n}}/{P_{5}}$, the $478$-Fibonacci numbers $F_{478,n}$ are just ${P_{7n}}/{P_{7}}$, and so on. In general, for all odd indexed Pell-Lucas numbers $Q_{m}$, the $Q_{m}$-Fibonacci numbers are just ${P_{mn}}/{P_{m}}$. The sequence $2, 14, 82, 478, 2786, \ldots$ is \seqnum{A077444} in the OEIS.
	
Hendel proved that
\begin{equation}\label{eqn.3}
	2\cdot\sum_{i=1}^{n}i^3P_i=(n^3-3n^2+6n-7)P_{n+1}+(n^3+3n-3)P_n+7.
\end{equation}
We can see the clear similarities between Eqs.~(\ref{eqn.1}) and (\ref{eqn.3}). For further clarity, we rewrite Eqs.~(\ref{eqn.1}) and (\ref{eqn.3}) as
\begin{equation}\label{eqn.4}
	\sum_{i=1}^{n}i^3F_i=(n^3-6n^2+24n-50)F_{n+1}+((n+1)^3-6(n+1)^2+24(n+1)-50)F_n+50,
\end{equation}
and
\begin{equation}\label{eqn.5}
	2\cdot\sum_{i=1}^{n}i^3P_i=(n^3-3n^2+6n-7)P_{n+1}+((n+1)^3-3(n+1)^2+6(n+1)-7)F_n+7.
\end{equation}
The identity \eqref{eqn.4} appears in the OEIS at \seqnum{A259546}.
If we use the $3$-Fibonacci numbers as another example, then we would have
\begin{equation}\label{eqn.6}
	3\cdot\sum_{i=1}^{n}i^3F_{3,i}=\bigg(n^3-2n^2+\frac{8}{3}n-\frac{22}{9}\bigg)F_{3,n+1}+\bigg((n+1)^3-2(n+1)^2+\frac{8}{3}(n+1)-\frac{22}{9}\bigg)F_{3,n}+\frac{22}{9}.
\end{equation}
Eqs.~(\ref{eqn.4}), (\ref{eqn.5}), and (\ref{eqn.6}) naturally suggest the following interesting generalization about the Brousseau sums of the $k$-Fibonacci numbers:
\begin{equation}\label{eqn.7}
	k\cdot\sum_{i=1}^{n}i^pF_{k,i}=(C_{k}^{(p)}(n))F_{k,n+1}+(C_{k}^{(p)}(n+1))F_{k,n}-C_{k}^{(p)}(0),
\end{equation}
	where $C_{k}^{(p)}(n)$ is a ``coefficient polynomial" in $n$ of degree $p$ with rational coefficients. The main task is to find an expression for the polynomial $C_{k}^{(p)}(n)$. We do this using some simple recursion formulas involving just binomial coefficients. We use a similar technique as that of Dresden. Throughout this paper, we assume $\binom{0}{0}=1$.
	%\section{\texorpdfstring{$k$}{pdfbookmark}-Fibonacci numbers and powers}
	\section{$k$-Fibonacci numbers and powers}
	We begin with the following set of identities, which are similar to those with the ``regular'' Fibonacci numbers \cite{Dresden}.	
	\begin{equation*}
		\begin{split}
			F_{k,n}&=n+\sum_{i=1}^{n}\bigg(ki-2\cdot\binom{1}{1}\bigg)F_{k,n-i},\\
			F_{k,n}&=n^2+\sum_{i=1}^{n}\bigg(ki^2-2\cdot\binom{2}{1}i\bigg)F_{k,n-i},\\
			F_{k,n}&=n^3+\sum_{i=1}^{n}\bigg(ki^3-2\cdot\binom{3}{1}i^2-2\cdot\binom{3}{3}\bigg)F_{k,n-i},\\
			F_{k,n}&=n^4+\sum_{i=1}^{n}\bigg(ki^4-2\cdot\binom{4}{1}i^3-2\cdot\binom{4}{3}i\bigg)F_{k,n-i},\\
			F_{k,n}&=n^5+\sum_{i=1}^{n}\bigg(ki^5-2\cdot\binom{5}{1}i^4-2\cdot \binom{5}{3}i^2-2\cdot\binom{5}{5}\bigg)F_{k,n-i},
		\end{split}
	\end{equation*}
	and so on.
Our first theorem involves the generalization of these formulas.
		\begin{theorem} For all integers $n, p\geq1$, we have
		\label{theo1}
		\begin{equation}
			\label{eqn.8}
			F_{k,n}=n^{p}+\sum_{i=1}^{n}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n-i}.
		\end{equation}
	\end{theorem}
	\begin{proof}
		We use induction on $n$. When $n=1$, the left-hand side of Eq.~(\ref{eqn.8}) is $F_{k,1}=1$ and the right-hand side is 
\begin{equation*}
\begin{split}
				&1+\sum_{i=1}^{1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,1-i}\\&=				1+\bigg(k-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}\bigg)F_{k,0}\\&=1+0\\&=1.
\end{split}
\end{equation*}
		When $n=2$, the left-hand side of Eq.~(\ref{eqn.8}) is $F_{k,2}=k$ and the right-hand side is 
		\begin{equation*}
			\begin{split}
				&2^{p}+\sum_{i=1}^{2}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,2-i}\\&=2^{p}+\bigg(k-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}\bigg)F_{k,1}+\bigg(k2^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}2^{p-2j+1}\bigg)F_{k,0}\\&=2^{p}+(k-2\cdot2^{p-1})+0\\&=k.
			\end{split}
		\end{equation*}
		Thus, Eq.~(\ref{eqn.8}) holds for $n=1$ and $n=2$. Now fix $n\geq2$. Assume that Eq.~(\ref{eqn.8}) holds for $n-1$ and $n$. Then
		\begin{equation}
			\label{eqn.9}
			F_{k,n}=n^{p}+\sum_{i=1}^{n}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n-i},
		\end{equation}
		and
		\begin{equation}
			\label{eqn.10}
			F_{k,n-1}=(n-1)^{p}+\sum_{i=1}^{n-1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n-1-i}.
		\end{equation}
		Since $F_{k,n-i}=0$ for $i=n$, Eq.~(\ref{eqn.9}) can be put in the form
		\begin{equation}
			\label{eqn.11}
			F_{k,n}=n^{p}+\sum_{i=1}^{n-1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n-i}.
		\end{equation}
		Multiplying Eq.~(\ref{eqn.11}) by $k$ and adding it to Eq.~(\ref{eqn.10}), we get
		\begin{equation*}
			\begin{split}
				kF_{k,n}+F_{k,n-1}=kn^{p}+(n-1)^{p}+\sum_{i=1}^{n-1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)(kF_{k,n-i}+F_{k,n-i-1}).
			\end{split}
		\end{equation*}
		 Using the $k$-Fibonacci recurrence (\ref{eqn.2}), this becomes
		\begin{equation}
			\label{eqn.12}
			\begin{split}
				F_{k,n+1}=kn^{p}+(n-1)^{p}+\sum_{i=1}^{n-1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n-i+1}.
			\end{split}
		\end{equation}
		Using the binomial expansion, we have the identity
		\begin{equation*}
			\begin{split}
				(n+1)^{p}-(n-1)^{p}&=2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}n^{p-2j+1},
			\end{split}
		\end{equation*}
		and hence
		\begin{equation}
			\label{eqn.13}
			\begin{split}
				(n-1)^{p}=(n+1)^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}n^{p-2j+1}.
			\end{split}
		\end{equation}
		Now using Eqs.~(\ref{eqn.12}) and (\ref{eqn.13}), we obtain
		\begin{equation*}
			\begin{split}
				F_{k,n+1}&
				=kn^{p}+(n+1)^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}n^{p-2j+1}\\&+\sum_{i=1}^{n-1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n+1-i}\\&
				=(n+1)^{p}+\bigg(kn^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}n^{p-2j+1}\bigg)\\&+\sum_{i=1}^{n-1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n+1-i}.
			\end{split}
		\end{equation*}
		Since $F_{k,1}=1$, we rewrite this as
		\begin{equation*}
			\begin{split}	
				F_{k,n+1}=(n+1)^{p}+\sum_{i=1}^{n}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n+1-i}.
			\end{split}
		\end{equation*}
		Since $F_{k,n+1-i}=0$ when $i=n+1$, we may simply add the corresponding term to the summand on the right-hand side of the above equation to get
		\begin{equation*}
			\begin{split}	
				F_{k,n+1}=(n+1)^{p}+\sum_{i=1}^{n+1}\bigg(ki^{p}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}\bigg)F_{k,n+1-i},
			\end{split}
		\end{equation*}
		and this completes the induction step.
		This concludes the proof.
	\end{proof}
	\section{Convolutions}
	Applying Theorem \ref{theo1} along with the $k$-Fibonacci sum \cite[Proposition 8]{FalconPlaza} \[\sum_{i=0}^{n}F_{k,n}=\frac{1}{k}\bigl(F_{k,n+1}+F_{k,n}-1\bigr),\] we can recursively find the following convolution identities:
	\begin{equation*}
\begin{split}
	k\cdot\sum_{i=0}^{n}F_{k,n-i}&=F_{k,n+1}+F_{k,n}-1,\\
	k\cdot\sum_{i=0}^{n}iF_{k,n-i}&=\frac{1}{k}\bigl(2F_{k,n+1}+(-k+2)F_{k,n}-(kn+2)\bigr)\\&=\bigg(\frac{2}{k}\bigg)F_{k,n+1}+\bigg(\frac{-k+2}{k}\bigg)F_{k,n}-\frac{kn+2}{k},\\
	k\cdot\sum_{i=0}^{n}i^2F_{k,n-i}&=\frac{1}{k^2}\bigl(8F_{k,n+1}+(k^2-4k+8)F_{k,n}-(k^2n^2+4kn+8)\bigr)\\&=\bigg(\frac{8}{k^2}\bigg)F_{k,n+1}+\bigg(\frac{k^2-4k+8}{k^2}\bigg)F_{k,n}-\frac{k^2n^2+4kn+8}{k^2},\\
	k\cdot\sum_{i=0}^{n}i^3F_{k,n-i}&=\frac{1}{k^3}\bigl((2k^2+48)F_{k,n+1}+(-k^3+8k^2-24k+48)F_{k,n}\\&-(k^3n^3+6k^2n^2+24kn+2k^2+48)\bigr),\\&=\bigg(\frac{2k^2+48}{k^3}\bigg)F_{k,n+1}+\bigg(\frac{-k^3+8k^2-24k+48}{k^3}\bigg)F_{k,n}\\&-\frac{k^3n^3+6k^2n^2+24kn+2k^2+48}{k^3},\\
\end{split}
		\end{equation*}
	and so on. Each sum on the left-hand side of the above set of equations is a convolution of the powers of $i$ and the $k$-Fibonacci numbers. We define
	\begin{equation}
		\label{eqn.14}
		T_{k,n}^{(p)} = \begin{cases}
			\sum_{i=0}^{n} F_{k,n-i}, & \text{if $p=0$;}\\
			\sum_{i=0}^{n}i^{p} F_{k,n-i}, & \text{if $p\geq1$.}
		\end{cases}
	\end{equation}
	A pattern is evident in the above set of equations. Note that each equation is of the form
\begin{equation}\label{eqn.15}
		k\cdot T_{k,n}^{(p)}=\Phi_{k}^{(p)}(0)F_{k,n+1}+\Phi_{k}^{(p)}(-1)F_{k,n}-\Phi_{k}^{(p)}(n),
\end{equation}
	where $\Phi_{k}^{(p)}(n)$ is a polynomial in $n$ of degree $p$. 
	To find an explicit formula for $\Phi_{k}^{(p)}(n)$, we must define the sequence $\bigl(A_{k}^{(p)}\bigr)_{p\geq0}$ as follows:
\begin{definition}\label{def1} The sequence $\bigl(A_{k}^{(p)}\bigr)_{p\geq0}$ of numbers is defined by the recurrence
	\begin{equation}
		\label{eqn.16}
		A_{k}^{(p)} = \begin{cases}
			1, & \text{if $p=0$;}\\
			\frac{2}{k}\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}A_{k}^{(p-2j+1)}, & \text{if $p\geq1$.}
		\end{cases}
	\end{equation}
\end{definition}
The recurrence $A_{k}^{(p)}$ generates the sequence \[1, \frac{2}{k},  \frac{8}{k^2}, \frac{2k^2+48}{k^3}, \frac{32k^2+384}{k^4}, \ldots.\]
Note that these numbers are the coefficients of $F_{k,n+1}$ in the above set of equations. The numbers $A_{k}^{(p)}$ for $p=1, 2, 3$, and $4$ are given below:

\begin{enumerate}
	\item [\rm (i)] $A_{1}^{(p)}: 1, 2, 8, 50, 416, 4322, 53888, \ldots$. This sequence is \seqnum{A000557} in the OEIS \cite{oeis}.
	\item [\rm (ii)] $A_{2}^{(p)}: 1, 1, 2, 7, 32, 181, 1232, \ldots$. This sequence is \seqnum{A006154} in the OEIS \cite{oeis}.
	\item [\rm (iii)] $A_{3}^{(p)}: 1, \dfrac{2}{3}, \dfrac{8}{9}, \dfrac{22}{9}, \dfrac{224}{27}, \dfrac{2774}{81}, \dfrac{13952}{81}, \ldots$.
	\item [\rm (iv)] $A_{4}^{(p)}: 1, \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{5}{4}, \dfrac{7}{2}, \dfrac{47}{4}, \dfrac{197}{4}, \ldots$.
\end{enumerate}
From the last equation in the set of convolution identities given above, we recognize that
\begin{equation*}
	\begin{split}
		\Phi_{k}^{(3)}(n)&=n^3+\frac{6}{k}n^2+\frac{24}{k^2}n+\frac{2k^2+48}{k^3}\\&=1\cdot \binom{3}{0}n^3+\frac{2}{k}\cdot \binom{3}{1}n^2+\frac{8}{k^2}\cdot \binom{3}{2}n+\frac{2k^2+48}{k^3}\cdot\binom{3}{3},
	\end{split}
\end{equation*}
where $1, {2}/{k}, {8}/{k^2}, {(2k^2+48)}/{k^3}$ are the first four terms of the sequence $\bigl(A_{k}^{(p)}\bigr)_{p\geq0}$. With all this in mind, we make the following definition:
\begin{definition}\label{def2}
	For all integers $p\geq0$, we define the polynomial $\Phi_{k}^{(p)}(n)$  as
\begin{equation} \label{eqn.17}
		\Phi_{k}^{(p)}(n)=
	\begin{cases}
		A_{k}^{(p)}, & \text{if $n=0$;}\\
		\sum_{r=0}^{p}A_{k}^{(r)}\binom{p}{r}n^{p-r}, & \text{if $n\neq0$.}
	\end{cases}
\end{equation}
Rewriting Eq.~(\ref{eqn.15}) in terms of $A_{k}^{(p)}$, we have the following theorem:
\end{definition}
		\begin{theorem}\label{theo2} If $T_{k,n}^{(p)}$ is as defined in Eq.~(\ref{eqn.14}), then for all integers $p\geq0$, we have
		\begin{equation}\label{eqn.18}
			k\cdot T_{k,n}^{(p)}=A_{k}^{(p)}F_{k,n+1}+\sum_{r=0}^{p}A_{k}^{(r)}\binom{p}{r}\big((-1)^{p-r}F_{k,n}-n^{p-r}\big).
		\end{equation}
	\end{theorem}
	\begin{proof}
		We use induction on $p$. When $p=0$, the left-hand side of Eq.~(\ref{eqn.18}) is \begin{equation*}
				k\cdot T_{k,n}^{(0)}=	k\sum_{i=0}^{n} F_{k,n-i}=F_{k,n+1}+F_{k,n}-1,
		\end{equation*}
	and the right-hand side is
	\begin{equation*}
			A_{k}^{(0)}F_{k,n+1}+A_{k}^{(0)}(F_{k,n}-1)=\frac{1}{k}(F_{k,n+1}+F_{k,n}-1).
	\end{equation*}
Thus, Eq.~(\ref{eqn.18}) holds for $p=0$.
	Now, fix $p\geq1$. Assume that Eq.~(\ref{eqn.18}) holds for all non-negative integers less than $p$. 
		First, we rewrite Eq.~(\ref{eqn.8}) in Theorem \ref{theo1} as
		\begin{equation*}
			\begin{split}	
				F_{k,n}&=n^{p}+k\sum_{i=1}^{n}i^{p}F_{k,n-i}-2\sum_{i=1}^{n}\sum_{j=1}^{\lceil\frac{p}{2}\rceil} \binom{p}{2j-1}i^{p-2j+1}F_{k,n-i}.
			\end{split}
		\end{equation*}
		By switching the order of summation in the double summand, we get
		\begin{equation*}
			\begin{split}
				F_{k,n}=n^{p}+k\sum_{i=1}^{n}i^{p}F_{k,n-i}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}\sum_{i=1}^{n} i^{p-2j+1}F_{k,n-i}.
			\end{split}
		\end{equation*}
		Since $i^{p}F_{k,n-i}=0$ for $i=0$, we can start the first summand at $i=0$ instead of at $i=1$, and thus we obtain
		\begin{equation}
			\label{eqn.19}
			\begin{split}
				F_{k,n}=n^{p}+k\sum_{i=0}^{n}i^{p}F_{k,n-i}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}\sum_{i=1}^{n} i^{p-2j+1}F_{k,n-i}.
			\end{split}
		\end{equation}
		If $ p$ is even, then Eq.~(\ref{eqn.19}) can be put in the form
		\begin{equation}
			\label{eqn.20}
			\begin{split}
				F_{k,n}=n^{p}+k\sum_{i=0}^{n}i^{p}F_{k,n-i}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}\sum_{i=0}^{n} i^{p-2j+1}F_{k,n-i},
			\end{split}
		\end{equation}
		because the term corresponding to $i=0$ in the  last summand is $0^{p-2j+1}F_{k,n}=0$ for all $j=1, 2, \ldots, {p}/{2}$.
		On the other hand, if $p$ is odd, then Eq.~(\ref{eqn.19}) can be written as 
		\begin{equation*}
			\begin{split}
				F_{k,n}=n^{p}+k\sum_{i=0}^{n}i^{p}F_{k,n-i}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}\sum_{i=0}^{n} i^{p-2j+1}F_{k,n-i}+2F_{k,n},
			\end{split}
		\end{equation*}
		because the term corresponding to $i=0$ in the last summation is $0^{p-2j+1}F_{k,n}=0$  for $j\neq {(p+1)}/{2}$ and  $F_{k,n}$ for $j={(p+1)}/{2}$. Therefore, if $p$ is odd, we have
		\begin{equation}
			\label{eqn.21}
			\begin{split}
				-F_{k,n}=n^{p}+k\sum_{i=0}^{n}i^{p}F_{k,n-i}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}\sum_{i=0}^{n} i^{p-2j+1}F_{k,n-i}.
			\end{split}
		\end{equation}
		Thus, from Eqs.~(\ref{eqn.20}) and (\ref{eqn.21}), we conclude that
		\begin{equation*}
			\begin{split}
				(-1)^{p}F_{k,n}=n^{p}+k\cdot T_{k,n}^{(p)}-2\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}T_{k,n}^{(p-2j+1)},
			\end{split}
		\end{equation*}
		and hence
		\begin{equation}
			\label{eqn.22}
			\begin{split}
					k\cdot T_{k,n}^{(p)}=(-1)^{p}F_{k,n}-n^{p}+\frac{2}{k}\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}k\cdot T_{k,n}^{(p-2j+1)}.
			\end{split}
		\end{equation}
		Now, by induction hypothesis, for $1\leq j\leq \lceil\frac{p}{2}\rceil$, we have
		\begin{equation*}
		\begin{split}
					k\cdot T_{k,n}^{(p-2j+1)}=&A_{k}^{(p-2j+1)}F_{k,n+1}+\sum_{r=0}^{p-2j+1}A_{k}^{(r)}\binom{p-2j+1}{r}\bigl((-1)^{p-2j+1-r}F_{k,n}-n^{p-2j+1-r}\bigr).
		\end{split}
		\end{equation*}
		Substituting this in Eq.~(\ref{eqn.22}), we get
		\begin{equation*}
			\begin{split}
					k\cdot T_{k,n}^{(p)}&=(-1)^{p}F_{k,n}-n^{p}+\frac{2}{k}\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}A_{k}^{(p-2j+1)}F_{k,n+1}\\&+\frac{2}{k}\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\binom{p}{2j-1}\sum_{r=0}^{p-2j+1}A_{k}^{(r)}\binom{p-2j+1}{r}\bigl((-1)^{p-2j+1-r}F_{k,n}-n^{p-2j+1-r}\bigr).
			\end{split}
		\end{equation*}
		Using the Definition \ref{def1}, this can be written as
		\begin{equation*}
		\begin{split}
				k\cdot T_{k,n}^{(p)}&=(-1)^pF_{k,n}-n^p+A_{k}^{(p)}F_{k,n+1}\\&+\frac{2}{k}\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\sum_{r=0}^{p-2j+1}A_{k}^{(r)}\binom{p}{2j-1}\binom{p-2j+1}{r}\bigl((-1)^{p-2j+1-r}F_{k,n}-n^{p-2j+1-r}\bigr).
		\end{split}
		\end{equation*}
			If we execute the change of variable $r'=2j+r-1$, this becomes
			\begin{equation*}
				\begin{split}
					k\cdot T_{k,n}^{(p)}&=(-1)^pF_{k,n}-n^p+A_{k}^{(p)}F_{k,n+1}\\&+	\frac{2}{k}\sum_{j=1}^{\lceil\frac{p}{2}\rceil}\sum_{r'=2j-1}^{p}A_{k}^{(r'-2j+1)}\binom{p}{2j-1}\binom{p-2j+1}{r'-2j+1}\bigl((-1)^{p-r'}F_{k,n}-n^{p-r'}).
				\end{split}
			\end{equation*}
		Now, by switching the order of summation, we obtain
		\begin{equation*}
			\begin{split}
			k\cdot T_{k,n}^{(p)}&=(-1)^pF_{k,n}-n^p+A_{k}^{(p)}F_{k,n+1}\\&+\frac{2}{k}\sum_{r'=1}^{p}\sum_{j=1}^{\lceil\frac{r'}{2}\rceil}A_{k}^{(r'-2j+1)}\binom{p}{2j-1}\binom{p-2j+1}{r'-2j+1}\bigl((-1)^{p-r'}F_{k,n}-n^{p-r'}\bigr).\end{split}
		\end{equation*}
		Using the well-known binomial identity $\binom{n}{r}=\binom{n}{n-r}$, we rewrite this as
				\begin{equation}\label{eqn.23}
			\begin{split}
					k\cdot T_{k,n}^{(p)}&=(-1)^pF_{k,n}-n^p+A_{k}^{(p)}F_{k,n+1}\\&+\frac{2}{k}\sum_{r'=1}^{p}\sum_{j=1}^{\lceil\frac{r'}{2}\rceil}A_{k}^{(r'-2j+1)}\binom{p}{p-2j+1}\binom{p-2j+1}{p-r'}\bigl((-1)^{p-r'}F_{k,n}-n^{p-r'}\bigr).
			\end{split}
		\end{equation}
		Next, using the binomial identity \cite[Identity 134, p.\ 67]{Benjamin} $\binom{p}{q}\binom{q}{r}=\binom{p}{r}\binom{p-r}{p-q}$, we have 
\begin{equation*}
	\begin{split}
				\binom{p}{p-2j-1}\binom{p-2j+1}{p-r'}=\binom{p}{p-r'}\binom{r'}{2j-1}=\binom{p}{r'}\binom{r'}{2j-1}.
	\end{split}
\end{equation*}
When we substitute this in Eq.~(\ref{eqn.23}), we obtain
		\begin{equation*}
			\begin{split}
					k\cdot T_{k,n}^{(p)}&=(-1)^pF_{k,n}-n^p+A_{k}^{(p)}F_{k,n+1}\\&+\frac{2}{k}\sum_{r'=1}^{p}\sum_{j=1}^{\lceil\frac{r'}{2}\rceil}A_{k}^{(r'-2j+1)}\binom{p}{r'}\binom{r'}{2j-1}\bigl((-1)^{p-r'}F_{n,k}-n^{p-r'}\bigr)\\&=(-1)^pF_{k,n}-n^p+A_{k}^{(p)}F_{k,n+1}\\&+\sum_{r'=1}^{p}\bigg(\frac{2}{k}\sum_{j=1}^{\lceil\frac{r'}{2}\rceil}\binom{r'}{2j-1}A_{k}^{(r'-2j+1)}\bigg)\binom{p}{r'}\bigl((-1)^{p-r'}F_{n,k}-n^{p-r'}\bigr).
			\end{split}
		\end{equation*}
		Now, using Definition \ref{def1}, this becomes
				\begin{equation*}
			\begin{split}
				k\cdot T_{k,n}^{(p)}=A_{k}^{(p)}F_{k,n+1}+\bigl((-1)^pF_{k,n}-n^p\bigr)+\sum_{r'=1}^{p}A_{k}^{(r')}\binom{p}{r'}\bigl((-1)^{p-r'}F_{n,k}-n^{p-r'}\bigr)).
			\end{split}
					\end{equation*}
		Since $A_{k}^{(r')}=1$ at $r'=0$, we conclude that
			\begin{equation*}
			\begin{split}
				k\cdot T_{k,n}^{(p)}
				=A_{k}^{(p)}F_{k,n+1}+\sum_{r'=0}^{p}A_{k}^{(r')}\binom{p}{r'}\bigl((-1)^{p-r'}F_{n,k}-n^{p-r'}\bigr).
			\end{split}
		\end{equation*}
		Hence, by induction, Eq.~(\ref{eqn.18}) holds for all $p\geq0$. This completes the proof.
	\end{proof}
\section{The Brousseau sums}
Let us begin this section with finding the identities about the Brousseau sums $\sum_{i=0}^{n}i^pF_{k,i}$ for $p=1,2,3, \ldots$. Consider the case $p=1$.
\begin{equation*}
	\begin{split}
		k\cdot\sum_{i=0}^{n}iF_{k,i}&
		=k\cdot\sum_{i=0}^{n}(n-i)F_{k,n-i}\\&
		=nk\cdot\sum_{i=0}^{n}F_{k,n-i}-k\cdot\sum_{i=0}^{n}iF_{k,n-i}\\&
		=n\bigl(F_{k,n+1}+F_{k,n}-1\bigr)-\bigg(\frac{2}{k}F_{k,n+1}+\frac{-k+2}{k}F_{k,n}-\frac{kn+2}{k}\bigg)\\&
		=\frac{1}{k}\bigl((kn-2)F_{k,n+1}+(kn+k-2)F_{k,n}+2\bigr).
	\end{split}
\end{equation*}
Proceeding like this, we get the following set of identities:
	\begin{equation}\label{Brousseausums}
	\begin{split}
		k\cdot \sum_{i=0}^{n}F_{k,i}&=F_{k,n+1}+F_{k,n}-1,\\
		k\cdot \sum_{i=0}^{n}iF_{k,i}&=\frac{1}{k}\bigl((kn-2)F_{k,n+1}+(kn+k-2)F_{k,n}+2\bigr)\\&
		=\frac{1}{k}\bigl((kn-2)F_{k,n+1}+(k(n+1)-2)F_{k,n}+2\bigr),\\
		k\cdot \sum_{i=0}^{n}i^2F_{k,i}&=\frac{1}{k^2}\bigl(\bigl(k^2n^{2}-4kn+8\bigr)F_{k,n+1}+\bigl(k^2n^2+2k(k-2)n+k^2-4k+8\bigr)F_{k,n}-8\bigr)\\
		&=\frac{1}{k^2}\bigl((k^2n^{2}-4kn+8)F_{k,n+1}+\bigl(k^2(n+1)^2-4k(n+1)+8\bigr)F_{k,n}-8\bigr),\\
		k\cdot \sum_{i=0}^{n}i^3F_{k,i}&=\frac{1}{k^3}\bigl(\bigl(k^3n^3-6k^2n^2+24kn-2k^2-48\bigr)F_{k,n+1}\\&+\bigl(k^3n^3+3k^2(k-2)n^2 		+3k(k^2-4k+8)n+k^3-8k^2+24k-48\bigr)F_{k,n}\\&+2k^2+48\bigr)\\&
=\frac{1}{k^3}\bigl(\bigl(k^3n^3-6k^2n^2+24kn-2k^2-48\bigr)F_{k,n+1}\\&+\bigl(k^3(n+1)^3-6k^2(n+1)^2+24k(n+1)-2k^2-48\bigr)F_{k,n}+2k^2+48\bigr),
	\end{split}
	\end{equation}
	and so on.
	As we expected, these equations also follow a pattern. 	If we define the sums $S_{k,n}^{(p)}$ as 
	\begin{equation}
		\label{eqn.24}
		S_{k,n}^{(p)} = \begin{cases}
			\sum_{i=0}^{n}F_{k,i}, & \text{if $p=0$;}\\
			\sum_{i=0}^{n}i^{p} F_{k,i}, & \text{if $p\geq1$,}
		\end{cases}
	\end{equation}
	then each equation is of the form
	\[k\cdot S_{k,n}^{(p)}=C_{k}^{(p)}(n)F_{k,n+1}+C_{k}^{(p)}(n+1)F_{k,n}-C_{k}^{(p)}(0),\]
	where $C_{k}^{(p)}(n)$ is a polynomial in $n$ of degree $p$.
	Let us try to investigate the rule of formation of the coefficients of this polynomial, $C_{k}^{(p)}(n)$. From the last equation in Eq.~(\ref{Brousseausums}), we identify that
\begin{equation*}
	\begin{split}
			C_{k}^{(3)}(n)&
			=\frac{1}{k^3}(k^3n^3-6k^2n^2+24kn-2k^2-48)\\&
			=n^3-\bigg(\frac{6}{k}\bigg)n^2+\bigg(\frac{24}{k^2}\bigg)n-\frac{2k^2+48}{k^3}\\&
			=1\cdot \binom{3}{0} n^3-\frac{2}{k}\cdot\binom{3}{1}n^2+\frac{8}{k^2}\cdot\binom{3}{2}n-\frac{2k^2+48}{k^3}\cdot\binom{3}{3},
	\end{split}
\end{equation*}
where the numbers $1$, ${2}/{k}$, ${8}/{k^2}$, ${(2k^2+48)}/{k^3}$ are the first four terms of the sequence $\bigl(A_{k}^{(p)}\bigr)_{p\geq0}$.
With this in mind, we define the {\it coefficient polynomial}, $C_{k}^{(p)}(n)$, in $n$ of degree $p$ as follows:
\begin{definition}\label{def5} For all integers $p\geq0$, we define
	\begin{equation}\label{eqn.25}
	C_{k}^{(p)}(n)=\begin{cases}
		(-1)^pA(p), & \text{if $n=0$;}\\
		\sum_{r=0}^{p}(-1)^{r}A_{k}^{(r)}\binom{p}{r}n^{p-r}, & \text{if $n\neq0$.}
	\end{cases}
\end{equation}
\end{definition}
It should be noted that, for $k>2$, the polynomial $C_{k}^{(p)}(n)$ generally doesn't have integer coefficients. From Eqs.~(\ref{eqn.17}) and (\ref{eqn.25}), it is clear that $\Phi_{k}^{(p)}(n)=(-1)^pC_{k}^{(p)}(-n)$. Consequently, we may rewrite Eq.~(\ref{eqn.15}) as 
\[k\cdot T_{k,n}^{(p)}=(-1)^{p}\bigl(C_{k}^{(p)}(0)F_{k,n+1}+C_{k}^{(p)}(1)F_{k,n}-C_{k}^{(p)}(-n)\bigr).\]
The central result (\ref{eqn.7}) about the Brousseau sums of the $k$-Fibonacci numbers can now be established.
	\begin{theorem}
		\label{theo5}
	If $S_{k,n}^{(p)}$ is defined as in Eq.~(\ref{eqn.24}), then 	for all $p\geq0$, we have
		\begin{equation}
			\label{eqn.26}
			\begin{split}
				k\cdot S_{k,n}^{(p)}=C_{k}^{(p)}(n)F_{k,n+1}+C_{k}^{(p)}(n+1)F_{k,n}-C_{k}^{(p)}(0),
			\end{split}
		\end{equation}
		where $C_{k}^{(p)}(n)$ is the ``coefficient polynomial'' as defined in Eq.~(\ref{eqn.25}).
	\end{theorem}
	\begin{proof}
		If $p=0$, then the left-hand side of Eq.~(\ref{eqn.26}) is \begin{equation*}
			k\cdot S_{k,n}^{(p)}=k\cdot\sum_{i=0}^{n}F_{k,i}=F_{k,n+1}+F_{k,n}-1,
		\end{equation*} and the right-hand side is
\begin{equation*}
\begin{split}
				C_{k}^{(0)}(n)F_{k,n+1}+C_{k}^{(0)}(n+1)F_{k,n}-C_{k}^{(0)}(0)&=A_{k}^{(0)}F_{k,n+1}+A_{k}^{(0)}F_{k,n}-1\\&=F_{k,n+1}+F_{k,n}-1.
\end{split}
\end{equation*} 
Thus, Eq.~(\ref{eqn.26}) holds for $p=0$.
		Now fix $p\geq1$. Then, using the binomial expansion, we have
		\begin{equation*}
			\begin{split}
				S_{k,n}^{(p)}&
				=\sum_{i=0}^{n}i^{p}F_{k,i}\\&
				=\sum_{i=0}^{n}(n-i)^{p}F_{k,n-i}\\&=\sum_{i=0}^{n}\bigg(\sum_{r=0}^{p}\binom{p}{r}n^{p-r}(-i)^{r}\bigg)F_{k,n-i}\\&
				=\sum_{r=0}^{p}(-1)^{r}\binom{p}{r}n^{p-r}\bigg(\sum_{i=0}^{n}i^{r}F_{k,n-i}\bigg)\\&
				=\sum_{r=0}^{p}(-1)^{r}\binom{p}{r}n^{p-r} T_{k,n}^{(r)}.
			\end{split}
		\end{equation*}
	Thus,
\begin{equation}
	\label{eqn.27}
		k\cdot S_{k,n}^{(p)}=\sum_{r=0}^{p}(-1)^{r}\binom{p}{r}n^{p-r}\bigg(k\cdot T_{k,n}^{(r)}\bigg).
\end{equation}
		Now, using Theorem \ref{theo2}, we have 
		\begin{equation*}
			k\cdot T_{k,n}^{(r)}=A_{k}^{(r)}F_{k,n+1}+\sum_{j=0}^{r}A_{k}^{(j)}\binom{r}{j}\bigl((-1)^{r-j}F_{k,n}-n^{r-j}\bigr).
		\end{equation*}
	When we substitute this in Equation (\ref{eqn.27}), we obtain
		\begin{equation*}
			\begin{split}
				k\cdot S_{k,n}^{(p)}&=\sum_{r=0}^{p}(-1)^{r}\binom{p}{r}n^{p-r}\bigg(A_{k}^{(r)}F_{k,n+1}+\sum_{j=0}^{r}A_{k}^{(j)}\binom{r}{j}\bigl((-1)^{r-j}F_{k,n}-n^{r-j}\bigr)\bigg)\\&=\bigg(\sum_{r=0}^{p}(-1)^{r}A_{k}^{(r)}\binom{p}{r}n^{p-r}\bigg)F_{k,n+1}\\&+\sum_{r=0}^{p}\sum_{j=0}^{r}(-1)^{r}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}\bigl((-1)^{r-j}F_{k,n}-n^{r-j}\bigr)n^{p-r}\\&=C_{k}^{(p)}(n)F_{k,n+1}+\sum_{r=0}^{p}\sum_{j=0}^{r}(-1)^{r}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}\bigl((-1)^{r-j}F_{k,n}-n^{r-j}\bigr)n^{p-r}.
			\end{split}
		\end{equation*}
	By switching the order of summation, this becomes
	\begin{equation*}
		\begin{split}
			k\cdot S_{k,n}^{(p)}&=C_{k}^{(p)}(n)F_{k,n+1}+\sum_{j=0}^{p}\sum_{r=j}^{p}(-1)^{j}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}n^{p-r}F_{k,n}\\&-\sum_{j=0}^{p}\sum_{r=j}^{p}(-1)^{r}A_{k}^{(j)}\binom{p}{r}\binom{r}{j}n^{p-j}\\&
			=C_{k}^{(p)}(n)F_{k,n+1}+\sum_{j=0}^{p}(-1)^{j}A_{k}^{(j)}\bigg(\sum_{r=j}^{p}\binom{p}{r}\binom{r}{j}n^{p-r}\bigg)F_{k,n}\\&-\sum_{j=0}^{p}A_{k}^{(j)}n^{p-j}\bigg(\sum_{r=j}^{p}\binom{p}{r}\binom{r}{j}(-1)^{r}\bigg)
			\\&
			=C_{k}^{(p)}(n)F_{k,n+1}+\bigg(\sum_{j=0}^{p}(-1)^{j}A_{k}^{(j)}\binom{p}{j}(n+1)^{p-j}\bigg)F_{k,n}-(-1)^{p}A_{k}^{(p)},
		\end{split}
	\end{equation*}
	where the last equality follows from the binomial identities
	\[\sum_{r=j}^{p}\binom{p}{r}\binom{r}{j}n^{p-r}=\binom{p}{j}(n+1)^{p-j},\]

and

	\[\sum_{r=j}^{p}\binom{p}{r}\binom{r}{j}(-1)^{r}= \begin{cases}
		0, & \text{if $j\neq p$;}\\
		(-1)^{p}, & \text{if $j=p$,}
	\end{cases}\]
	from Gould's
	collection \cite[Identities 3.118, 3.119, p.\ 36]{Gould}.
		Thus, we conclude that
		\begin{equation*}
		k\cdot S_{k,n}^{(p)}=C_{k}^{(p)}(n)F_{k,n+1}+C_{k}^{(p)}(n+1)F_{k,n}-C_{k}^{(p)}(0).
		\end{equation*}
	\end{proof}
	\begin{example}\label{eg:7}
		Setting $p=2$ in Eq.~(\ref{eqn.26}) we get
		\begin{equation*}
		\begin{split}
		k\cdot \sum_{i=1}^{n}i^{2}F_{k,i}=\bigg(n^{2}-\frac{4}{k}n+\frac{8}{k^2}\bigg)F_{k,n+1}+\bigg((n+1)^2-\frac{4}{k}(n+1)+\frac{8}{k^2}\bigg)F_{k,n}-\frac{8}{k^2}.
			\end{split}
		\end{equation*}
		In particular, when $k=11$ this becomes
		 		\begin{equation*}
		 	\begin{split}
		 		11\cdot \sum_{i=1}^{n}i^{2}F_{11,i}=\bigg(n^{2}-\frac{4}{11}n+\frac{8}{121}\bigg)F_{11,n+1}+\bigg((n+1)^2-\frac{4}{11}(n+1)+\frac{8}{121}\bigg)F_{11,n}-\frac{8}{121}.
		 	\end{split}
		 \end{equation*}
		 Since $F_{11,i}={F_{5i}}/{F_5}={F_{5i}}/{5}$ (see \cite{Falcon}), we obtain
		 	\begin{equation*}
		 	\begin{split}
		 		11\cdot \sum_{i=1}^{n}i^{2}F_{5i}=\bigg(n^{2}-\frac{4}{11}n+\frac{8}{121}\bigg)F_{5n+5}+\bigg((n+1)^2-\frac{4}{11}(n+1)+\frac{8}{121}\bigg)F_{5n}-\frac{5\cdot8}{121},
		 	\end{split}
		 \end{equation*}
		 which gives the identity about the Brousseau sums of the sequence $\bigl(F_{5i}\bigr)_{i\geq1}$.
		 	\end{example}
		 		\begin{example}\label{eg:8}
		 		Setting $p=2$ and $k=14$ in Eq.~(\ref{eqn.26}) we get
		 		\begin{equation*}
		 			\begin{split}
		 				14\cdot \sum_{i=1}^{n}i^{2}F_{14,i}=\bigg(n^{2}-\frac{2}{7}n+\frac{2}{49}\bigg)F_{14,n+1}+\bigg((n+1)^2-\frac{2}{7}(n+1)+\frac{2}{49}\bigg)F_{14,n}-\frac{2}{49}.
		 			\end{split}
		 		\end{equation*}
		 		Since $F_{14,i}={P_{3i}}/{P_3}={P_{3i}}/{5}$ (see \cite{Falcon}), we obtain
		 		\begin{equation*}
		 			\begin{split}
		 				14\cdot \sum_{i=1}^{n}i^{2}P_{3i}=\bigg(n^{2}-\frac{2}{7}n+\frac{2}{49}\bigg)P_{3n+3}+\bigg((n+1)^2-\frac{2}{7}(n+1)+\frac{2}{49}\bigg)P_{3n}-\frac{5\cdot2}{49},
		 			\end{split}
		 		\end{equation*}
		 		which gives the identity about the Brousseau sums of the sequence $\bigl(P_{3i}\bigr)_{i\geq1}$.
		 	\end{example}
		 Examples \ref{eg:7} and \ref{eg:8} suggest  two interesting identities Eqs.~(\ref{eqn.28}) and (\ref{eqn.30}). Eq.~(\ref{eqn.28}) is about the Brousseau sums of every $m^{\rm th}$ Fibonacci number, and Eq.~(\ref{eqn.30}) is that of every $m^{\rm th}$ Pell number, when $m$ is odd.
	\begin{corollary}
		\label{coro9} Let $m\geq1$ be an odd integer. Then for all integers $p\geq0$, the following identity holds:
		\begin{equation}\label{eqn.28}
			\begin{split}
				L_{m}\cdot\sum_{i=1}^{n}i^{p}F_{mi}=C_{L_{m}}^{(p)}(n)F_{m(n+1)}+C_{L_{m}}^{(p)}(n+1)F_{mn}-C_{L_{m}}^{(p)}(0)F_{m},
			\end{split}
		\end{equation}
		where	$L_{m}$ is the $m^{th}$ Lucas number.
	\end{corollary}
	\begin{proof}
		Setting $k=L_{m}$ in Eq.~(\ref{eqn.26}) yields
		\begin{equation}
			\begin{split}\label{eqn.29}
				L_{m}\cdot\sum_{i=1}^{n}i^{p}F_{L_{m},i}=C_{L_{m}}^{(p)}(n)F_{L_{m},n+1}+C_{{L_{m}}}^{(p)}(n+1)F_{L_{m},n}-C_{L_{m}}^{(p)}(0).
			\end{split}
		\end{equation}
		If $m$ is odd, then we have (see \cite{Falcon})
		\begin{equation*}
		F_{L_{m},i}=\dfrac{F_{mi}}{F_m}.
		\end{equation*}
		Applying this in Eq.~(\ref{eqn.29}) and multiplying through by $F_{m}$, we get Eq.~(\ref{eqn.28}).
	\end{proof}
	\begin{corollary} \label{coro10}Let $m\geq1$ be an odd number.  Then for all integers $p\geq0$, the following identity holds:
	\begin{equation}\label{eqn.30}
		\begin{split}
			Q_{m}\cdot\sum_{i=1}^{n}i^{p}P_{mi}=C_{Q_{m}}^{(p)}(n)P_{m(n+1)}+C_{Q_{m}}^{(p)}(n+1)P_{mn}-C_{Q_{m}}^{(p)}(0)P_{m},
		\end{split}
	\end{equation}
	where	$Q_{m}$ is the $m^{th}$ Pell-Lucas number.
	\end{corollary}
	\begin{proof} The proof is similar to the proof of Corollary (\ref{coro9}) by using the fact that	(see \cite{Falcon})
	\begin{equation*}
		F_{Q_{m},i}=\dfrac{P_{mi}}{P_m},
	\end{equation*}
	when $m$ is odd.
 \end{proof}
	\section{Shifted Brousseau sums}
	In this section, we find the formula for the shifted Brousseau sums 
	\[\sum_{i=1}^{n}i^{p}F_{k,m+i},\]
	for all integers $m,p\geq0$. For example, if we take $p=1$, then
	\begin{equation*}
\begin{split}
			k\cdot\sum_{i=1}^{n}iF_{k,m+i}&
			=k\cdot\sum_{i=m+1}^{m+n}(i-m)F_{k,i}\\&
			=k\cdot\sum_{i=m+1}^{m+n}iF_{k,i}-mk\cdot\sum_{i=m+1}^{m+n}F_{k,i}\\&
			=k\bigg(\sum_{i=0}^{m+n}iF_{k,i}-\sum_{i=0}^{m}iF_{k,i}\bigg)-mk\bigg(\sum_{i=0}^{m+n}F_{k,i}-\sum_{i=0}^{m}F_{k,i}\bigg).
\end{split}
	\end{equation*}
	Now, using the first two identities in Eq.~(\ref{Brousseausums}), we have
		\begin{equation*}
		\begin{split}
			k\cdot\sum_{i=1}^{n}iF_{k,m+i}&
			=\frac{1}{k}\bigl((k(m+n)-2)F_{k,m+n+1}+(k(m+n+1)-2)F_{k,m+n}-(km-2)F_{k,m+1}\\&-(k(m+1)-2)F_{k,m}\bigr)-m\bigl(F_{k,m+n+1}+F_{k,m+n}-F_{k,m+1}-F_{k,m}\bigr)\\&
			=\frac{1}{k}\bigl((kn-2)F_{k,m+n+1}+(k(n+1)-2)F_{k,m+n}+2F_{k,m+1}+(-k+2)F_{k,m}\bigr).
		\end{split}
	\end{equation*}
	We generalize this identity for all integers $p\geq0$ in the next theorem.
	\begin{theorem}
		For all integers $m, p\geq0$, we have
		\begin{equation}\label{eqn.32}
		k\cdot\sum_{i=1}^{n}i^{p}F_{k,m+i}=C_{k}^{(p)}(n)F_{k,m+n+1}+C_{k}^{(p)}(n+1)F_{k,m+n}-C_{k}^{(p)}(0)F_{k,m+1}-C_{k}^{(p)}(1)F_{k,m}.
		\end{equation}
	\end{theorem}
	\begin{proof}
		Since $F_{k,0}=0$ and $F_{k,1}=1$, the case $m=0$ follows from Theorem \ref{theo5}. Now fix $m\ge1$. Then, using the binomial expansion, we have
		\begin{equation*}
			\begin{split}
				\sum_{i=1}^{n}i^{p}F_{k,m+i}&
				=\sum_{i=m+1}^{m+n}(i-m)^{p}F_{k,i}\\&
				=\sum_{i=m+1}^{m+n}\sum_{j=0}^{p}\binom{p}{j}i^{p-j}(-m)^{j}F_{k,i}\\&
				=\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}\sum_{i=m+1}^{m+n}i^{p-j}F_{k,i}\\&
				=\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}\bigl(S_{k,m+n}^{(p-j)}-S_{k,m}^{(p-j)}\bigr).
			\end{split}
		\end{equation*}
	Thus, 
	\begin{equation*}
			k\cdot\sum_{i=1}^{n}i^{p}F_{k,m+i}=\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}\bigl(k\cdot S_{k,m+n}^{(p-j)}-k\cdot S_{k,m}^{(p-j)} \bigr).
	\end{equation*}
		Now, applying Theorem \ref{theo5}, this becomes
	\begin{equation}\label{eqn.33}
	\begin{split}
					k\cdot\sum_{i=1}^{n}i^{p}F_{k,m+i}&=\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}\bigg(C_{k}^{(p-j)}(m+n)F_{k,m+n+1}+C_{k}^{(p-j)}(m+n+1)F_{k,m+n}\\&-C_{k}^{(p-j)}(m)F_{k,m+1}-C_{k}^{(p-j)}(m+1)F_{k,m}\bigg)
	\end{split}
\end{equation}
	Consider
	\begin{equation*}
		\begin{split}
			\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}C_{k}^{(p-j)}(m+n)=\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}\sum_{r=0}^{p-j}(-1)^rA_{k}^{(r)}\binom{p-j}{r}		(m+n)^{p-j-r}.
		\end{split}
	\end{equation*}
	By switching the order of summation, this becomes
		\begin{equation}\label{eqn.34}
		\begin{split}
			\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}C_{k}^{(p-j)}(m+n)=\sum_{r=0}^{p}\sum_{j=0}^{p-r}(-1)^rA_{k}^{(r)}\binom{p}{j}\binom{p-j}{r}(m+n)^{p-j-r}(-m)^{j}.
		\end{split}
	\end{equation}
	Next, we use the binomial identity \cite[Identity 134, p.\ 67]{Benjamin} to get 
	\begin{equation*}
		\binom{p}{j}\binom{p-j}{r}=	\binom{p}{p-j}\binom{p-j}{r}=	\binom{p}{r}\binom{p-r}{j}.
	\end{equation*}
	Substituting this in Eq.~(\ref{eqn.34}), we obtain
			\begin{equation}\label{eqn.35}
		\begin{split}
			\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}C_{k}^{(p-j)}(m+n)&
			=\sum_{r=0}^{p}(-1)^{r}A_{k}^{(r)}\binom{p}{r}\bigg(\sum_{j=0}^{p-r}\binom{p-r}{j}(m+n)^{p-r-j}(-m)^{j}\bigg)\\&
			=\sum_{r=0}^{p}(-1)^{r}A_{k}^{(r)}\binom{p}{r}n^{p-r}\\&
			=C_{k}^{(p)}(n).
		\end{split}
	\end{equation}
	Similarly, we can show that
				\begin{equation}\label{eqn.36}
		\begin{split}
			\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}C_{k}^{(p-j)}(m+n+1)=C_{k}^{(p)}(n+1),
		\end{split}
	\end{equation}
				\begin{equation}\label{eqn.37}
		\begin{split}
			\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}C_{k}^{(p-j)}(m)=C_{k}^{(p)}(0),
		\end{split}
	\end{equation}
	and
	\begin{equation}\label{eqn.38}
		\begin{split}
			\sum_{j=0}^{p}\binom{p}{j}(-m)^{j}C_{k}^{(p-j)}(m+1)=C_{k}^{(p)}(1).
		\end{split}
	\end{equation}
	Thus, Eq.~(\ref{eqn.32}) follows by substituting Eq.~(\ref{eqn.35}) through Eq.~(\ref{eqn.38}) in Eq.~(\ref{eqn.33}).
	\end{proof}
	\section{Conclusion}
	 While all the results presented above assume that $k$ is a positive integer, there is no reason not to extend them to nonzero real numbers as well. The only drawback is that the numbers $F_{k,n}$ are not necessarily integers. The $k$-Fibonacci numbers are  just the Fibonacci polynomials $F_n(x)$ (see \cite{Bicknell}) calculated at $x=k$. Hence, we strongly believe that all the above results are still valid if we allow non-integer values of $k$. For example, we can have the identity
	 \begin{equation*}
	 	\sqrt{2}\cdot \sum_{i=1}^{n}i^2F_i\bigl(\sqrt{2}\bigr)=\bigl(n^2-2\sqrt{2}n+4\bigr)F_{n+1}\bigl(\sqrt{2}\bigr)+\bigl((n+1)^2-2\sqrt{2}(n+1)+4\bigr)F_{n}\bigl(\sqrt{2}\bigr)-4.
	 \end{equation*}
	\section{Acknowledgment}
	The authors would like to thank the anonymous referee for valuable comments and
	suggestions for improving the quality of the original version of this manuscript.
	
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\noindent 2020 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B37, 11B65, 11B83.

\noindent \emph{Keywords}: Brousseau sum, Fibonacci number, Fibonacci polynomial,  $k-$Fibonacci number, binomial coefficient, metallic number.

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\noindent 
(Concerned with sequences 
\seqnum{A000032}, 
\seqnum{A000045},  
\seqnum{A000129}, 
\seqnum{A000557}, 
\seqnum{A001076}, 
\seqnum{A002203}, 
\seqnum{A002878}, 
\seqnum{A006154}, 
\seqnum{A006190} 
\seqnum{A077444}, and 
\seqnum{A259546}.)
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\vspace*{+.1in}
\noindent
Received October 11 2023;
revised versions received February 23 2024; July 17 2024.
Published in {\it Journal of Integer Sequences}, July 17 2024.

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