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\title{\bf Some identities involving the partial sum\\ of $q$-binomial coefficients}

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\author{Bing He%\thanks{Supported by NASA grant ABC123.}
\\
\small Department of Mathematics, Shanghai Key Laboratory of PMMP \\[-0.8ex]
\small East China Normal University\\[-0.8ex]
\small 500 Dongchuan Road, Shanghai 200241,
 People's Republic of China\\
\small\tt yuhe001@foxmail.com\\
}

\date{\dateline{Feb 21, 2014}{Jul 21, 2014}{Jul 25, 2014}\\
\small Mathematics Subject Classifications: 05A10; 05A15}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
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\begin{abstract}
  We give some identities involving  sums of powers of the partial sum of $q$-binomial coefficients, which are $q$-analogues of Hirschhorn's identities [\emph{Discrete Math.} 159 (1996), 273--278] and  Zhang's  identity [\emph{Discrete Math.} 196 (1999), 291--298].

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} binomial coefficients, $q$-binomial coefficients,
$q$-binomial theorem
\end{abstract}

\section{Introduction}
In \cite{r2}, Calkin  proved the following  curious identity:
\begin{equation*}
  \sum_{k=0}^{n}\left(\sum_{j=0}^{k}{n \choose j}\right)^{3}=n\cdot 2^{3n-1}+2^{3n}-3n{2n \choose n}2^{n-2}.
\end{equation*}
Hirschhorn \cite{r3} established the following two identities on sums of powers of  binomial partial sums:
\begin{equation}\label{l1}
  \sum_{k=0}^{n}\sum_{j=0}^{k}{n \choose j}=n\cdot2^{n-1}+2^{n},
\end{equation}
and
\begin{equation}\label{l2}
  \sum_{k=0}^{n}\left(\sum_{j=0}^{k}{n \choose j}\right)^{2}=n\cdot 2^{2n-1}+2^{2n}-\frac{n}{2}{2n \choose n}.
\end{equation}
In \cite{r5}, Zhang  proved the following alternating form of \eqref{l2}:
\begin{equation}\label{l4}
  \sum_{k=0}^{n}(-1)^{k}\left(\sum_{j=0}^{k}{n \choose j}\right)^{2}=
  \begin{cases}
     1, &\text{if $n=0$,}\\
 2^{2n-1}, &\text{if $n$ is even and $n \neq 0 $,}\\
       -2^{2n-1}-(-1)^{(n-1)/2}{n-1 \choose (n-1)/2}, &\text{if $n$ is odd.}
   \end{cases}
\end{equation}
 Several generalizations are given in \cite{r8,r9,r10}.
Later, Guo \emph{et al.}~\cite{r11} gave the following $q$-identities:
 \begin{equation*}
   \sum_{k=0}^{2n}(-1)^{k}\left(\sum_{j=0}^{k}{2n \brack j}_{q}\right)^{2}=\left(\sum_{k=0}^{2n}{2n \brack k}_{q}\right)\left(\sum_{k=0}^{n}{2n \brack 2k}_{q}\right),
 \end{equation*}
 and
\begin{align*}
 \sum_{k=0}^{2n+1}(-1)^{k}\left(\sum_{j=0}^{k}{2n+1 \brack j}_{q}\right)^{2}&=-\left(\sum_{k=0}^{n}{2n+1 \brack 2k}_{q}\right)\left(\sum_{k=0}^{2n+1}{2n+1 \brack k}_{q}\right)\\
 &\quad -\sum_{k=0}^{n}(-1)^{k}{2n+1 \brack k}_{q}^{2}-2\!\!\sum\limits_{0\leq i < j \leq n}\!\!\!(-1)^{i}{2n+1 \brack i}_{q}{2n+1 \brack j}_{q}.
\end{align*}
Here and in what follows, ${n\brack k}_q$ is the $q$-binomial coefficient defined by
\[
{n\brack k}_q=
\begin{cases}
\displaystyle\frac{(q;q)_{n}}{(q;q)_{k}(q;q)_{n-k}}, &\text{if $0\leq k\leq n$},\\[5pt]
0,&\text{otherwise,}
\end{cases}
\]
where $(z;q)_n=(1-z)(1-zq)\cdots(1-zq^{n-1})$ is  the  $q$-shifted factorial for $n\geq 0$.

The purpose of this paper is to study  $q$-analogues of \eqref{l1}--\eqref{l2} and establish a new $q$-version of \eqref{l4}.
Our main results may be stated as follows.
\begin{theorem}\label{t1}
  For any positive integer $n$ and any non-zero integer $m$, we have
  \begin{equation}\label{l5}
    \sum_{k=0}^{n}\sum_{j=0}^{k}{n \brack j}_{q}q^{mk+{j \choose 2}}=\frac{(-q^{m},q)_{n}-q^{m(n+1)}(-1,q)_{n}}{1-q^{m}},
  \end{equation}
  and
\begin{align}
   &\sum_{k=0}^{n}q^{-k}\left(\sum_{i=0}^{k}{n \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=0}^{k}{n \brack j}_{q}q^{{j \choose 2}+2(1-n)j}\right)\notag \\
       &\quad=\frac{\left((-q^{-1};q)_{n}-q^{-(n+1)}(-1;q)_{n}\right)(-q^{2(1-n)};q)_{n}}{1-q^{-1}}
-\sum_{i=0}^{n-1}\frac{1-q^{n-i}}{1-q}{2n \brack i}_{q}q^{{i \choose 2}-\frac{3n^{2}}{2}+\frac{n}{2}+1}.\label{l6}
\end{align}
\end{theorem}
\begin{theorem}
For any non-negative integer $n$, we have
\begin{align}
  &\sum_{k=0}^{2n+1}(-1)^{k}\left(\sum_{i=0}^{k}{2n+1 \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=0}^{k}{2n+1 \brack j}_{q}q^{2n-j+1 \choose 2}\right)\notag\\
       &\quad =-q^{2n^{2}+n}(-q^{-2n};q)_{4n+1}-\sum_{i=0}^{n}(-1)^{i}{2n+1 \brack i}_{q^{2}}q^{2{i \choose 2}},\label{l8}
\end{align}
and
     \begin{equation}\label{l9}
       \sum_{k=0}^{2n+2}(-1)^{k}\left(\sum_{i=0}^{k}{2n+2 \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{i=0}^{k}{2n+2 \brack i}_{q}q^{2n+2-i \choose 2}\right)=q^{2n^{2}+3n+1}(-q^{-1-2n};q)_{4n+3}.
     \end{equation}
\end{theorem}
Letting $q \rightarrow 1$ and using  L'H{\^ o}pital's rule and some familiar identities, we easily find that the identities \eqref{l5}--\eqref{l6} and \eqref{l8}--\eqref{l9} are $q$-analogues of \eqref{l1}--\eqref{l2} and \eqref{l4} respectively.

%Throughout the paper, we need the following results derived from the $q$-binomial theorem (see, for example, \cite{r1}).
%\begin{lem}Let $z$ and $q$ be complex numbers with $|z|<1$ and $|q|<1.$ Then
%  \begin{equation*}
%    (z,q)_{n}=\sum_{k=0}^{n}(-1)^{k}{n \brack k}_{q}q^{k \choose 2}z^{k}.
%  \end{equation*}
%  and
%  \begin{equation*}
%    \frac{1}{(z,q)_{n}}=\sum\limits_{i\geq 0}{n+i-1\brack i}_{q}z^{i}.
%  \end{equation*}
%\end{lem}
In Sections 2 and 3, we will give proofs of Theorems 1.1 and 1.2 respectively by using the $q$-binomial theorem and generating functions.

\section{Proof of Theorem 1.1}
 To give our proof of Theorem 1.1, we need to establish a result, which is a $q$-analogue of Chang and Shan's identity (see \cite{r12}).
\begin{lemma}
For any positive integer $n$, we have
  \begin{equation*}
     \sum_{k=0}^{n-1}q^{-k}\left(\sum_{i=0}^{k}{n \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=k+1}^{n}{n \brack j}_{q}q^{{j \choose 2}+2(1-n)j}\right)=\sum_{i=0}^{n-1}\frac{1-q^{n-i}}{1-q}{2n \brack i}_{q}q^{{i \choose 2}-\frac{3n^{2}}{2}+\frac{n}{2}+1}.
   \end{equation*}
\end{lemma}

\begin{proof}
According to the $q$-binomial theorem (see \cite{r1}), we have  for all complex numbers  $z$ and $q$ with $|z|<1$ and $|q|<1,$ there holds
\begin{equation}\label{e2-1}
    (z,q)_{n}=\sum_{k=0}^{n}(-1)^{k}{n \brack k}_{q}q^{k \choose 2}z^{k}
  \end{equation}
  and
  \begin{equation*}
    \frac{1}{(z,q)_{n}}=\sum\limits_{i\geq 0}{n+i-1\brack i}_{q}z^{i}.
  \end{equation*}
It follows that
\begin{equation*}
  (-z;q)_{n}\frac{1}{1-z}=\left(\sum_{i=0}^{n}{n \brack i}_{q}q^{{i \choose 2}}z^{i}\right)
  \left(\sum_{i=0}^{\infty}z^{i}\right),
\end{equation*}
\begin{equation*}
  (-z q^{n};q)_{n}\frac{1}{1-zq}=\left(\sum_{i=0}^{n}{n \brack i}_{q}q^{{i \choose 2}+n i}z^{i}\right)
  \left(\sum_{i=0}^{\infty}q^{ i}z^{i}\right),
\end{equation*}
and
\begin{equation*}
  (-z;q)_{2n}\frac{1}{(z;q)_{2}}=\left(\sum_{i=0}^{2n}{2n \brack i}_{q}q^{{i \choose 2}}z^{i}\right)\left(\sum_{i=0}^{\infty}{ 1+i \brack i}_{q}z^{i}\right).
\end{equation*}
Therefore, for any non-negetive integer $k$ with $k\leq n-1$, the coefficient of $z^{k}$ in $(-z;q)_{n}\frac{1}{1-z}$ is
\begin{equation*}
  \sum_{i=0}^{k}{n \brack i}_{q}q^{{i \choose 2}},
\end{equation*}
the coefficient of $z^{n-k-1}$ in $(-z q^{n};q)_{n}\frac{1}{(1-zq}$ is
\begin{equation*}
  \sum_{i=k+1}^{n}{n \brack i}_{q}q^{{n-i \choose 2}+n(n-i)+i-k-1}
\end{equation*}
and
the coefficient of $z^{n-1}$ in $(-z;q)_{2n}\frac{1}{(z;q)_{2}}$ is
\begin{equation*}
  \sum_{i=0}^{n-1}{2n \brack i}_{q}\frac{1-q^{n-i}}{1-q}q^{{i \choose 2}}.
\end{equation*}
Using the fact $$(-z;q)_{n}\frac{1}{1-z}\cdot (-z q^{n};q)_{n}\frac{1}{1-zq}=(-z;q)_{2n}\frac{1}{(z;q)_{2}},$$ equating the coefficients of $z^{n-1}$ and after some simplifications, we obtain Lemma 2.1.
\end{proof}
\begin{flushleft}
\emph{Proof of Theorem 1.1.} We first prove \eqref{l5}.
\end{flushleft}
\begin{align*}
        \sum_{k=0}^{n}\sum_{j=0}^{k}{n \brack j}_{q}q^{mk+{j \choose 2}}&=\sum_{j=0}^{n}{n \brack j}_{q}q^{j \choose 2}\sum_{k=j}^{n}q^{mk} \\
        &=\frac{\sum_{j=0}^{n}{n \brack j}_{q}q^{{j \choose 2}+mj}-q^{m(n+1)}\sum_{j=0}^{n}{n \brack j}_{q}q^{j \choose 2}}{1-q^{m}}\\
        &=\frac{(-q^{m},q)_{n}-q^{m(n+1)}(-1,q)_{n}}{1-q^{m}},
       \end{align*}
       where in the last step, we have used \eqref{e2-1}.

We next show \eqref{l6}. By \eqref{e2-1}, we have
      \begin{equation*}
        \sum_{j=0}^{n}{n \brack j}_{q}q^{{j \choose 2}+2(1-n)j}=(-q^{2(1-n)};q)_{n},
      \end{equation*}
      and taking $m=-1$ in \eqref{l5}, we obtain
      \begin{equation*}
         \sum_{k=0}^{n}q^{-k}\sum_{j=0}^{k}{n \brack j}_{q}q^{j \choose 2}=\frac{(-q^{-1},q)_{n}-q^{-(n+1)}(-1,q)_{n}}{1-q^{-1}}.
      \end{equation*}
  Hence, by Lemma 2.1, we get
\begin{align*}
&\sum_{k=0}^{n}q^{-k}\left(\sum_{i=0}^{k}{n \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=0}^{k}{n \brack j}_{q}q^{{j \choose 2}+2(1-n)j}\right)\\                                                                                                                                                    &=\sum_{k=0}^{n}q^{-k}\left(\sum_{i=0}^{k}{n \brack i}_{q}q^{i \choose 2}\right)\left((-q^{2(1-n)};q)_{n}-\sum_{j=k+1}^{n}{n \brack j}_{q}q^{{j \choose 2}+2(1-n)j}\right)\\
&=(-q^{2(1-n)};q)_{n}\sum_{k=0}^{n}q^{-k}\left(\sum_{i=0}^{k}{n \brack i}_{q}q^{i \choose 2}\right)\\
&\qquad -\sum_{k=0}^{n-1}q^{-k}\left(\sum_{i=0}^{k}{n \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=k+1}^{n}{n \brack j}_{q}q^{{j \choose 2}+2(1-n)j}\right)\\
&=\frac{\left((-q^{-1};q)_{n}-q^{-(n+1)}(-1;q)_{n}\right)(-q^{2(1-n)},q)_{n}}{1-q^{-1}}-\sum_{i=0}^{n-1}\frac{1-q^{n-i}}{1-q}{2n \brack i}_{q}q^{{i \choose 2}-\frac{3n^{2}}{2}+\frac{n}{2}+1}.
\end{align*}
%%%%%%%%%%%%%%%
\section{Proof of Theorem 1.2}
In order to prove the Theorem 1.2, we need the following result, which gives a $q$-analogue of alternating sums of Chang and Shan's identity.
\begin{lemma}
  For any non-negative integer $n$, we have
     \begin{equation*}
       \sum_{k=0}^{2n}(-1)^{k}\left(\sum_{i=0}^{k}{2n+1 \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=k+1}^{2n+1}{2n+1 \brack j}_{q}q^{2n-j+1 \choose 2}\right)=\sum_{i=0}^{n}(-1)^{i}{2n+1 \brack i}_{q^{2}}q^{2{i \choose 2}}.
     \end{equation*}
\end{lemma}
\begin{proof}
By \eqref{e2-1}, we find that
\begin{equation*}
  (z;q)_{n}\frac{1}{1+z}=\left(\sum_{i=0}^{n}{n \brack i}_{q}q^{{i \choose 2}}(-z)^{i}\right)
  \left(\sum_{i=0}^{\infty}(-z)^{i}\right),
\end{equation*}
\begin{equation*}
  (-z;q)_{n}\frac{1}{1-z}=\left(\sum_{i=0}^{n}{n \brack i}_{q}q^{i \choose 2}z^{i}\right)
  \left(\sum_{i=0}^{\infty}z^{i}\right),
\end{equation*}
and
\begin{equation*}
  (z^{2};q^{2})_{n}\frac{1}{1-z^{2}}=\left(\sum_{i=0}^{n}{n \brack i}_{q^{2}}q^{2{i \choose 2}}(-1)^{i}z^{2i}\right)\left(\sum_{i=0}^{\infty}z^{2i}\right).
\end{equation*}
Therefore, for any non-negetive integer $k$ with $k\leq n-1$, the coefficient of $z^{k}$ in $(z;q)_{n}\frac{1}{1+z}$ is
\begin{equation*}
  (-1)^{k}\sum_{i=0}^{k}{n \brack i}_{q}q^{i \choose 2},
\end{equation*}
the coefficient of $z^{n-k-1}$ in $(-z;q)_{n}\frac{1}{1-z}$ is
\begin{equation*}
  \sum_{i=k+1}^{n}{n \brack i}_{q}q^{n-i \choose 2}
\end{equation*}
and
the coefficient of $z^{n-1}$ in $(z^{2};q^{2})_{n}\frac{1}{1-z^{2}}$ is
\begin{equation*}
  \sum_{i=0}^{(n-1)/2}(-1)^{i}{n \brack i}_{q^{2}}q^{2{i \choose 2}}[2|(n-1)],
\end{equation*}
where $[2|n]$ is defined by
\begin{equation*}
  [2|n]=
  \begin{cases}
     1, &\text{if $2|n$,}\\
 0, &\text{otherwise.}
   \end{cases}
\end{equation*}
Using the fact $$(-z;q)_{n}\frac{1}{1-z}\cdot (z;q)_{n}\frac{1}{1+z}=(z^{2};q^{2})_{n}\frac{1}{1-z^{2}},$$ equating the coefficients of $z^{n-1}$ and after some simplifications, we obtain Lemma 3.1.
\end{proof}
\begin{flushleft}
\emph{Proof of Theorem 1.2.} We first prove \eqref{l8}. By \eqref{e2-1}, we have
\end{flushleft}
\begin{align}
      \sum_{k=0}^{n}(-1)^k\sum_{j=0}^{k}{n \brack j}_{q}q^{j \choose 2}
   &=\sum_{j=0}^{n}{n \brack j}_{q}q^{{j \choose 2}}\sum_{k=j}^{n}(-1)^k\notag\\
   &=\frac{1}{2}\sum_{j=0}^{n}{n \brack j}_{q}q^{{j \choose 2}}((-1)^{j}-(-1)^{n+1})\notag\\
   &=\frac{(-1)^{n}}{2}(-1,q)_{n},  \label{e3-1}
     \end{align}
     and
\begin{equation*}
  \sum_{j=0}^{2n+1}{2n+1 \brack j}_{q}q^{2n-j+1 \choose 2}=q^{2n^{2}+n}(-q^{-2n};q)_{2n+1}.
\end{equation*}
Replacing $n$ by $2n+1$ in \eqref{e3-1}, we obtain
\begin{equation*}
  \sum_{k=0}^{2n+1}(-1)^{k}\left(\sum_{i=0}^{k}{2n+1 \brack i}_{q}q^{i \choose 2}\right)=-(-q;q)_{2n}.
\end{equation*}
Hence, by Lemma 3.1, we get
\begin{align*}
   & \sum_{k=0}^{2n+1}(-1)^{k}\left(\sum_{i=0}^{k}{2n+1 \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=0}^{k}{2n+1 \brack j}_{q}q^{2n-j+1 \choose 2}\right) \\
   &=\sum_{k=0}^{2n+1}(-1)^{k}\left(\sum_{i=0}^{k}{2n+1 \brack i}_{q}q^{i \choose 2}\right)\left(q^{2n^{2}+n}(-q^{-2n};q)_{2n+1}-\sum_{j=k+1}^{2n+1}{2n+1 \brack j}_{q}q^{2n-j+1 \choose 2}\right)\\
   &=-q^{2n^{2}+n}(-q^{-2n};q)_{4n+1}-\sum_{k=0}^{2n}(-1)^{k}\left(\sum_{i=0}^{k}{2n+1 \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{j=k+1}^{2n+1}{2n+1 \brack j}_{q}q^{2n-j+1 \choose 2}\right)\\
   &=-q^{2n^{2}+n}(-q^{-2n};q)_{4n+1}-\sum_{i=0}^{n}(-1)^{i}{2n+1 \brack i}_{q^{2}}q^{2{i \choose 2}}.
\end{align*}

We next show \eqref{l9}. By \eqref{e2-1}, we have
\begin{equation*}
  \sum_{j=0}^{2n}{2n \brack j}_{q}q^{2n-j \choose 2}=q^{2n^{2}-n}(-q^{1-2n};q)_{2n},
\end{equation*}
and replacing $n$ by $2n$ in \eqref{e3-1}, we obtain
\begin{equation*}
  \sum_{k=0}^{2n}(-1)^{k}\left(\sum_{i=0}^{k}{2n \brack i}_{q}q^{i \choose 2}\right)=(-q;q)_{2n-1}.
  \end{equation*}
  Hence, by the fact
     \begin{equation*}
     \sum_{k=0}^{2n-1}(-1)^{k}\left(\sum_{i=0}^{k}{2n \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{i=k+1}^{2n}{2n \brack i}_{q}q^{2n-i \choose 2}\right)=0
     \end{equation*}
     which follows easily from the substitution $k\rightarrow 2n-1-k,$
 we have
  \begin{align*}
     & \sum_{k=0}^{2n}(-1)^{k}\left(\sum_{i=0}^{k}{2n \brack i}_{q}q^{i \choose 2}\right)\left(\sum_{i=0}^{k}{2n \brack i}_{q}q^{2n-i \choose 2}\right) \\
     &=\sum_{k=0}^{2n}(-1)^{k}\left(\sum_{i=0}^{k}{2n \brack i}_{q}q^{i \choose 2}\right)\left(q^{2n^{2}-n}(-q^{1-2n};q)_{2n}-\sum_{i=k+1}^{2n}{2n \brack i}_{q}q^{2n-i \choose 2}\right)\\
     &=q^{2n^{2}-n}(-q^{1-2n};q)_{4n-1}.
  \end{align*}

\bigskip
\subsection*{Acknowledgement} I would like to thank the referee for his/her helpful comments.

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\end{document}
