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\title{\bf On a general $q$-identity}

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\author{Aimin Xu\thanks{Supported by the National Natural Science Foundation of China (grant
11201430), and the Ningbo Natural
Science Foundation.}\\
\small Institute of Mathematics \\[-0.8ex]
\small Zhejiang Wanli University\\[-0.8ex]
\small Ningbo 315100, China\\
\small\tt xuaimin1009@hotmail.com; xuaimin@zwu.edu.cn\\
%\and
%%Forgotten Second Author \qquad  Forgotten Third Author\\
%%\small School of Hard Knocks\\[-0.8ex]
%%\small University of Western Nowhere\\[-0.8ex]
%%\small Nowhere, Australasiaopia\\
%%\small\tt \{fsa,fta\}@uwn.edu.ao
}

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\date{\dateline{Dec 12, 2013}{Apr 24, 2014}\\
\small Mathematics Subject Classifications: 05A19, 11B65}

\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}
  In this paper, by means of the $q$-Rice formula we obtain a general $q$-identity which is a unified generalization of three kinds of identities.
  Some known results are special cases of ours. Meanwhile, some identities on $q$-generalized harmonic numbers are also derived.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} $q$-Rice formula; $q$-identity; $q$-generalized harmonic number; Cauchy's integral formula; Fa\`{a} di Bruno's formula
\end{abstract}

\section{Introduction}

Three kinds of identities will be introduced in this paper.

In the paper \cite{Van1982a}, Van Hamme gave the following identity
\begin{align}
  \sum_{k=1}^n(-1)^{k-1}\qbinomial{n}{k}\frac{q^{{k+1\choose 2}}}{1-q^k}=\sum_{k=1}^{n}\frac{q^k}{1-q^k}.
\end{align}
One of the generalizations of $(1.1)$ was given by Dilcher \cite{Dilcher1995a}:
\begin{align}
    \sum_{k=1}^n(-1)^{k-1}\qbinomial{n}{k}\frac{q^{{k\choose 2}+k\lambda}}{(1-q^k)^\lambda}=\sum_{1\leq \alpha_1\leq\cdots\leq \alpha_\lambda\leq n }\prod_{j=1}^\lambda\frac{q^{\alpha_j}}{1-q^{\alpha_j}}.
\end{align}
Prodinger \cite{Prodinger2001a} gave another generalization of $(1.1)$:
\begin{align}
    \sum_{k=0,\neq m}^n(-1)^{k-1}\qbinomial{n}{k}\frac{q^{{k+1\choose 2}}}{1-q^{k-m}}=(-1)^mq^{{m+1\choose 2}}\sum_{k=0,\neq m}^{n}\frac{q^k}{1-q^{k-m}},
\end{align}
where $0\leq m \leq n$. Many works have been devoted to the study of the generalizations of these identities. See for example \cite{Fu2003a,Fu2005a,Prodinger2004a,Zeng2005a}. Recently,
Guo and Zhang \cite{Guo2011a} made use of the Lagrange interpolation formula to give a generalization of Prodinger's identity $(1.3)$. They also gave a generalization of Dilcher's identity $(1.2)$. See Theorems $1.1$ and $1.2$ in \cite{Guo2011a}, respectively. Ismail and Stanton used the theory of basic hypergeometric functions to generalize Dilcher's identity. See Theorem $2.2$ in \cite{Ismail2012a}.

In the paper\cite {Diaz2007a}, D\'{\i}az-Barrero et al. obtained two identities involving rational sums:
\begin{align*}
  &\sum_{k=1}^n(-1)^{k-1}{n\choose k}{x+k\choose k}^{-1}\sum_{1\leq \alpha\leq \beta\leq k}\frac{1}{x^2+(\alpha+\beta)x+\alpha\beta}=\frac{n}{(x+n)^3},\\
  &\sum_{k=1}^n(-1)^{k-1}{n\choose k}{x+k\choose k}^{-1}\Bigg\{\sum_{\alpha=1}^k\frac{1}{(x+\alpha)^3}+\sum_{1\leq \alpha\leq \beta\leq k}\frac{1}{(x+\alpha)(x+\beta)(2x+\alpha+\beta)}\nonumber\\
  &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+\sum_{1\leq \alpha<\beta<\gamma\leq k}\frac{1}{(x+\alpha)(x+\beta)(x+\gamma)}\Bigg\}=\frac{n}{(x+n)^4}.
\end{align*}
Recently, Prodinger \cite{Prodinger2008a} made use of partial fraction decomposition and inverse pairs to present a more general formula:
\begin{align}
  \sum_{k=1}^n(-1)^{k-1}{n\choose k}{x+k\choose k}^{-1}\sum_{c_1+2c_2+\cdots=\lambda}\prod_{j\geq 1}\frac{s_{k,j}^{c_j}}{c_j! j^{c_j}}=\frac{n}{(x+n)^{\lambda+1}},
\end{align}
where $s_{k,j}=\sum_{\alpha=1}^k (x+\alpha)^{-j}$.
Almost at the same time, Chu and Yan \cite{Chu2008a} presented a generalization with multiple $\lambda$-fold sum:
\begin{align}
  \sum_{k=0}^n(-1)^k{n\choose k}{x+k\choose k}^{-1}\sum_{0\leq \alpha_1\leq\cdots\leq \alpha_\lambda\leq k}\prod_{j=1}^\lambda\frac{1}{x+\alpha_j}=\frac{x}{(x+n)^{\lambda+1}}.
\end{align}
A direct proof of $(1.5)$ can be found in Chu \cite{Chu2012a}.
More recently, Mansour et al. \cite{Mansour2013a} established a $q$-analog for the rational sum identity $(1.4)$:
\begin{align}
  \sum_{k=1}^n(-1)^{k-1}q^{{k\choose 2}-k(n-1)}\qbinomial{n}{k}{\qbinomial{x+k}{k}}^{-1}\sum_{c_1+2c_2+\cdots=\lambda}\prod_{j\geq 1}\frac{s_{k,j}(q)^{c_j}}{c_j! j^{c_j}}=\frac{q^{n\lambda}[n]_q}{[x+n]_q^{\lambda+1}},
\end{align}
where $s_{k,j}(q)=\sum_{\alpha=1}^k q^{j\alpha}[x+\alpha]_q^{-j}$. In particular, they gave a very nice bijective proof for the case $\lambda=1$.

In the recent paper \cite{Prodinger2013a},  Prodinger established an interesting identity involving harmonic numbers:
  \begin{align}
    &\sum_{k=0,\neq m}^n(-1)^{k-1}{n\choose k}{n+k\choose k}\frac{1}{(k-m)^\lambda}\nonumber\\
     &\quad\quad\quad =(-1)^m {n\choose m}{n+m\choose n}\sum_{c_1+2c_2+\cdots=\lambda}\frac{1}{c_1!c_2!\cdots}\prod_{j=1}^{\lambda}\left(\frac{\mathcal{H}_j}{j}\right)^{c_j},
  \end{align}
 where
   \begin{align*}
       \mathcal{H}_j=(-1)^{j-1}\left(H_{m+n}^{(j)}-2H_m^{(j)}\right)+H_{n-m}^{(j)},
      \end{align*}
      and $H_n^{(r)}$ are the generalized harmonic numbers defined by
      \begin{align*}
        H_{0}^{(r)}=0, \quad H_{n}^{(r)}=\sum_{k=1}^n\frac{1}{k^r} \quad for \quad n, r=1,2,\ldots.
      \end{align*}
      Mansour \cite{Mansour2012a} obtained a general rational sum to generalize this identity. He also obtained a $q$-analog of this result involving $q$-harmonic numbers.

Motivated by these interesting work, by means of the $q$-Rice formula used in \cite{Prodinger2001a,Prodinger2004a}, we will establish a general $q$-identity which is a common generalization of those three kinds of identities introduced before.
\begin{thm}
Let $\lambda$ be any positive integer. For $0\leq m\leq n$ and $0\leq l\leq n+\lambda-1$, there holds
        \begin{align}
  &\sum_{k=0,\neq m}^n\qbinomial{n}{k}\frac{q^{(\lambda-1) k+m}(1-q^{k-m})(q/z;q)_k(zq^{-l};q)_{n-k+\lambda-1}}{(1-xq^{k-m})^{\lambda+1}}z^k\nonumber
  \\&\quad=-\frac{(q;q)_n(zq^{-l};q)_l(zxq^{-m};q)_{n-l+\lambda-1}}{(xq^{-m};q)_m(xq;q)_{n-m}}\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^{\lambda}\left(\frac{u_j}{j}\right)^{c_j},
      \end{align}
      where
     $\vec{c}!=c_1!c_2!\cdots c_{\lambda}!$, $\|\vec{c}\|=c_1+2c_2+\cdots+\lambda c_{\lambda}$ and
      \begin{align*}
        u_j=-\sum_{k=0}^{n-l+\lambda-2}\left(\frac{zq^k}{1-zxq^{k-m}}\right)^j+\sum_{k=0,\neq m}^{n}\left(\frac{q^k}{1-xq^{k-m}}\right)^j.
      \end{align*}
\end{thm}
This is a very general $q$-series sum identity involving five parameters $\lambda$, $l$, $m$, $x$ and $z$. It contains several known identities by choosing different parameters, which will be shown in the third section. By means of our identity, we will also obtain some identities on $q$-generalized harmonic numbers.


Throughout this paper, we will use the standard notation. For any real number $x$ and any integer $m$, define
\begin{align*}
  [x]_q=\frac{1-q^x}{1-q}, \quad (x;q)_\infty=\prod_{k=0}^{\infty}(1-x q^k), \quad (x;q)_m=\frac{(x;q)_\infty}{(x q^m;q)_{\infty}}.
  \end{align*}
  For any nonnegative integer $n$, define
  \begin{align*}
  [n]_q!=[1]_q[2]_q\cdots[n]_q, ~\qbinomial{n}{k}=\frac{[n]_q!}{[k]_q![n-k]_q!}.
    \end{align*}



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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem 1.1}

In the very interesting paper \cite{Prodinger2001a}, Prodinger introduced the following formula
\begin{align*}
  \sum_{k=1}^n(-1)^{k-1}q^{{k\choose 2}}\qbinomial{n}{k}f(q^{-k})=\frac{1}{2\pi i}\int_{\mathcal{C}}\frac{(q;q)_n}{(t;q)_{n+1}}f(t)dt,
\end{align*}
where $\mathcal{C}$ encircles the poles $q^{-1}$, $q^{-2}$,\ldots, $q^{-n}$ and no other.
It is a $q$-analog of Rice's formula \cite{Flajolet1995a,Szpankowski2001a}:
\begin{align*}
  \sum_{k=1}^n{n\choose k}(-1)^kf(k)=\frac{(-1)^n}{2\pi i}\int_{\mathcal{C}}\frac{n!}{t(t-1)\cdots(t-n)}f(t)dt,
\end{align*}
where $\mathcal{C}$ encircles the poles $1$, $2$,\ldots, $n$ and no other.
Indeed, by Cauchy's integral formula one is not hard to find that for any integer $m\in\{0,1,\ldots,n\}$ there holds
\begin{align}
  \sum_{k=0,\neq m}^n(-1)^{k-1}q^{{k\choose 2}}\qbinomial{n}{k}f(q^{-k})=(-1)^{n-1}\frac{(q;q)_n}{q^{{n+1\choose 2}}}\frac{1}{2\pi i}\int_{\mathcal{C}}\frac{f(t)dt}{\prod_{j=0}^n(t-q^{-j})},
\end{align}
where $\mathcal{C}$ encircles the poles $q^{-j}$, $j\in\{0,1,\ldots,n\}-\{m\}$ and no other. Prodinger first applied the $q$-analog of Rice's formula to prove many identities such as the identities of Van Hamme, Uchimura, Dilcher, Andrews-Crippa-Simon, and Fu-Lascoux, see \cite{Prodinger2001a,Prodinger2004a} and references therein. It was shown that this formula is a very powerful and useful tool.
Now, in this section we will use this important formula and present a proof of Theorem $1.1$.

{\bf Proof of Theorem 1.1.} By simple calculations we have
\begin{align*}
  &\sum_{k=0,\neq m}^n\qbinomial{n}{k}\frac{q^{(\lambda-1) k}(1-q^{k-m})(q/z;q)_k(zq^{-l};q)_{n-k+\lambda-1}}{(1-xq^{k-m})^{\lambda+1}}z^k\nonumber\\
  =&\sum_{k=0,\neq m}^n(-1)^kq^{{k\choose 2}+k\lambda}\qbinomial{n}{k}\frac{1-q^{k-m}}{(1-xq^{k-m})^{\lambda+1}}(zq^{-k};q)_k(zq^{-l};q)_{n-k+\lambda-1}\nonumber\\
  =&\sum_{k=0,\neq m}^n(-1)^{k}q^{{k\choose 2}+k\lambda}\qbinomial{n}{k}\frac{1-q^{k-m}}{(1-xq^{k-m})^{\lambda+1}}\frac{(zq^{-k};q)_{\infty}}{(z;q)_{\infty}}\frac{(zq^{-l};q)_{\infty}}{(zq^{n+\lambda-1-l-k};q)_{\infty}}\nonumber\\
  =&(zq^{-l};q)_l\sum_{k=0,\neq m}^n(-1)^{k}q^{{k\choose 2}+k\lambda}\qbinomial{n}{k}\frac{1-q^{k-m}}{(1-xq^{k-m})^{\lambda+1}}(zq^{-k};q)_{n-l+\lambda-1}.
  \end{align*}
 Thus, by the $q$-Rice formula $(2.1)$ there holds
  \begin{align}
    &\sum_{k=0,\neq m}^n\qbinomial{n}{k}\frac{q^{(\lambda-1) k}(1-q^{k-m})(q/z;q)_k(zq^{-l};q)_{n-k+\lambda-1}}{(1-xq^{k-m})^{\lambda+1}}z^k\nonumber\\
    =&(-1)^n(zq^{-l};q)_l\frac{(q;q)_n}{q^{{n+1\choose 2}}}\frac{1}{2\pi i}\int_{\mathcal{C}}\frac{(zt;q)_{n-l+\lambda-1}dt}{(t-xq^{-m})^{\lambda+1}\prod_{k=0,\neq m}^n(t-q^{-k})},
  \end{align}
  where $\mathcal{C}$ (positively oriented) encloses the poles $q^{-j},~j\in \{0,1,\ldots,n\}-\{m\}$ and no other.
  It is obvious that
  \begin{align}
    \frac{1}{2\pi i}\int_{\mathcal{C}}\frac{(zt;q)_{n-l+\lambda-1}dt}{(t-xq^{-m})^{\lambda+1}\prod_{k=0,\neq m}^n(t-q^{-k})}=-\frac{1}{2\pi i}\int_{\mathcal{C}'}\frac{(zt;q)_{n-l+\lambda-1}dt}{(t-xq^{-m})^{\lambda+1}\prod_{k=0,\neq m}^n(t-q^{-k})},
  \end{align}
   where $\mathcal{C}'$ (positively oriented) encloses the pole $xq^{-m}$. By Cauchy's integral formula, there holds
   \begin{align}
     \frac{1}{2\pi i}\int_{\mathcal{C}'}\frac{(zt;q)_{n-l+\lambda-1}}{(t-xq^{-m})^{\lambda+1}\prod_{k=0,\neq m}^n(t-q^{-k})}=\frac{1}{\lambda !}\frac{d^\lambda}{dt^{\lambda}}\frac{(zt;q)_{n-l+\lambda-1}}{\prod_{k=0,\neq m}^n(t-q^{-k})}\bigg|_{t=xq^{-m}}.
   \end{align}
   Applying Fa\`{a} di Bruno's formula \cite{Comtet1974a} yields
   \begin{align}
     \frac{d^\lambda}{dt^{\lambda}}\frac{(zt;q)_{n-l+\lambda-1}}{\prod_{k=0,\neq m}^n(t-q^{-k})}\bigg|_{t=xq^{-m}}=&\frac{d^\lambda}{dt^{\lambda}}e^{\sum_{k=0}^{n-l+\lambda-2}\log(1-ztq^k)-\sum_{k=0,\neq m}^{n}\log(t-q^{-k})}\bigg|_{t=xq^{-m}}\nonumber\\
     =&\frac{(zxq^{-m};q)_{n-l+\lambda-1}}{\prod_{k=0,\neq m}^n(xq^{-m}-q^{-k})}\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^{\lambda}\left(\frac{u_j}{j}\right)^{c_j}.
   \end{align}
    From (2.2), (2.3),(2.4) and (2.5), the desired result is obtained.

\begin{rmk}
  Actually, careful checking the proof of Theorem $1.1$, one can find that Theorem $1.1$ still holds for $\lambda=0$ if in this case we assume the sum of the right hand side of $(1.8)$ is equal to $1$. This implies that for $0\leq l\leq n-1$ there holds
  \begin{align*}
  &\sum_{k=0,\neq m}^n\qbinomial{n}{k}\frac{(1-q^{m-k})(q/z;q)_k(zq^{-l};q)_{n-k-1}}{1-xq^{k-m}}z^k
  =\frac{(q;q)_n(zq^{-l};q)_l(zxq^{-m};q)_{n-l-1}}{(xq^{-m};q)_m(xq;q)_{n-m}}.
      \end{align*}
\end{rmk}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Consequences of Theorem 1.1}

Theorem $1.1$ can help us to find some new identities or retrieve some well known identities.

Let $\lambda=1$ and $x=1$. $(1.8)$ reduces to the following identity.
\begin{cor}
 For $0\leq m\leq n$ and $0\leq l\leq n$, there holds
\begin{align}
  &\sum_{k=0,\neq m}^n\qbinomial{n}{k}\frac{(q/z;q)_k(zq^{-l};q)_{n-k}}{1-q^{k-m}}z^k\nonumber\\
  &=(-1)^{m-1}q^{{m\choose 2}}\qbinomial{n}{m}(zq^{-l};q)_l(zq^{-m};q)_{n-l}\left\{-\sum_{k=0}^{n-l-1}\frac{zq^k}{1-zq^{k-m}}+\sum_{k=0,\neq m}^n\frac{q^k}{1-q^{k-m}}\right\}.
\end{align}
\end{cor}
Guo and Zhang \cite{Guo2011a} made use of Lagrange interpolation formula to obtain this identity which generalizes the identity $(1.3)$ due to Prodinger.
It is obvious that $(3.1)$ reduces to $(1.3)$ when $l=0$ and $z\rightarrow 0$.

  Let $x=1$, $l=\lambda-1$ and $z=q^{-n}$ in $(1.8)$. We have
  \begin{cor}
  Let $\lambda$ be any nonnegative integer. For $0\leq m\leq n$, there holds
    \begin{align*}
      &\sum_{k=0,\neq m}^n(-1)^{k-1}\qbinomial{n}{k}\qbinomial{n+k}{k}\frac{q^{{k\choose 2}+(\lambda-n)k}}{(1-q^{k-m})^\lambda}\nonumber\\
      &\quad\quad\quad=(-1)^m q^{{m\choose 2}-nm}\qbinomial{n}{m}\qbinomial{n+m}{n}\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^{\lambda}\left(\frac{\mathcal{H}_j(q)}{j}\right)^{c_j},
    \end{align*}
    where
     \begin{align*}
       \mathcal{H}_j(q)=-\sum_{k=0}^{n-1}\left(\frac{q^{k-n}}{1-q^{k-m-n}}\right)^j+\sum_{k=0,\neq m}^{n}\left(\frac{q^k}{1-q^{k-m}}\right)^j.
      \end{align*}
  \end{cor}
  This identity is a $q$-analog of Prodinger's identity $(1.7)$.
An alternative form of this $q$-identity was presented in \cite{Mansour2012a}.

For $m=0$, $l=0$ and $z\rightarrow 0$ in (1.8), the following identity is true.
\begin{cor}
Let $\lambda$ be any nonnegative integer. There holds
\begin{align}
  \sum_{k=1}^n\qbinomial{n}{k}(-1)^{k-1}q^{{k\choose 2}+\lambda k}\frac{1-q^k}{(1-xq^k)^{\lambda+1}}
  =\frac{(q;q)_n}{(xq;q)_n}\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^{\lambda}\left(\frac{1}{j}{\sum_{k=1}^n\frac{q^{jk}}{(1-xq^k)^j}}\right)^{c_j}.
\end{align}
\end{cor}
It is clear that
  \begin{align}
    \prod_{j=1}^n\frac{1}{1-x_jt}=\prod_{j=1}^n\sum_{k\geq 0}(x_j t)^k=\sum_{\lambda\geq 0}t^\lambda\sum_{1\leq \alpha_1\leq\cdots\leq \alpha_\lambda\leq n}\prod_{j=1}^\lambda x_{\alpha_j}.
  \end{align}
Since
  \begin{align*}
    \frac{d^\lambda}{dt^\lambda}\prod_{j=1}^n\frac{1}{1-x_jt}\bigg|_{t=0}=\frac{d^\lambda}{dt^\lambda}e^{-\sum_{j=1}^n \log(1-x_jt)}\bigg|_{t=0},
  \end{align*}
  we apply Fa\`{a} di Bruno's formula \cite{Comtet1974a} to obtain
   \begin{align}
     \frac{d^\lambda}{dt^\lambda}\prod_{j=1}^n\frac{1}{1-x_jt}\bigg|_{t=0}=\sum_{\|\vec{c}\|=\lambda}\frac{\lambda!}{\vec{c}! }\prod_{j=1}^\lambda\left(\frac{\sum_{k=1}^nx_k^j}{j}\right)^{c_j}.
   \end{align}
      Comparing $(3.3)$ with $(3.4)$, there holds
  \begin{align*}
    \sum_{1\leq \alpha_1\leq\cdots\leq \alpha_\lambda\leq n}\prod_{j=1}^\lambda x_{\alpha_j}=\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}! }\prod_{j=1}^\lambda\left(\frac{\sum_{k=1}^nx_k^j}{j}\right)^{c_j}.
  \end{align*}
Therefore, $(3.2)$ can be rewritten as
 \begin{align*}
  \sum_{k=1}^n\qbinomial{n}{k}(-1)^{k-1}q^{{k\choose 2}+\lambda k}\frac{1-q^k}{(1-xq^k)^{\lambda+1}}
  =\frac{(q;q)_n}{(xq;q)_n}\sum_{1\leq \alpha_1\leq\cdots\leq \alpha_\lambda\leq n}\prod_{j=1}^\lambda \frac{q^{\alpha_j}}{1-xq^{\alpha_j}} ,
\end{align*}
or
 \begin{align}
  \sum_{k=1}^n\qbinomial{n}{k}(-1)^{k-1}q^{{k\choose 2}+\lambda k}\frac{1-q^k}{(1-xq^k)^{\lambda+1}}
  =\frac{(q;q)_n}{(xq;q)_n}\sum_{|\vec{b}|=\lambda}\prod_{j=1}^n \left(\frac{q^{j}}{1-xq^{j}}\right)^{b_j},
\end{align}
where $|\vec{b}|=b_1+b_2+\cdots+b_n$.
 By the theory of basic hypergeometric functions Ismail and Stanton \cite{Ismail2012a} found Eq. $(3.5)$ which reduces to the Dilcher identity \cite{Dilcher1995a} when $x=1$.

 In fact, it has been recently pointed out in \cite{Guo2014a} that the Ismail-Stanton
 result $(3.5)$ is the $i=1$ (with $m=\lambda+1$) case of following formula due to Zeng \cite{Zeng2005a}:
 \begin{align*}
   \sum_{k=i}^n(-1)^{k-i}\qbinomial{n}{k}\qbinomial{k}{i}\frac{q^{{k-i\choose 2}+km}}{(1-zq^k)^m}=\frac{q^i(q;q)_{i-1}(q;q)_n}{(q;q)_i(zq;q)_n}h_{m-1}\left(\frac{q^i}{1-zq^i},\ldots,\frac{q^n}{1-zq^n}\right),
 \end{align*}
where $1\leq i\leq n$ and $h_k(x_1,\ldots,x_n)$ is the $k$th homogeneous symmetric polynomial in $x_1,x_2,\ldots,x_n$ defined by
\begin{align*}
  h_k(x_1,\ldots,x_n)=\sum_{1\leq i_1\leq\cdots\leq i_k\leq n}x_{i_1}\cdots x_{i_k}=\sum_{|\vec{b}|=k}x_1^{b_1}\cdots x_n^{b_n}.
\end{align*}
This more general formula can not follow from Theorem $1.1$ and it can be viewed as a different generalization of the Ismail-Stanton result $(3.5)$.

Since
\begin{align*}
  \qbinomial{n}{k}=\qbinomial{n-1}{k-1}\frac{1-q^n}{1-q^k},
\end{align*}
Eq. $(3.2)$ can be rewritten as
\begin{align*}
  \sum_{k=1}^n(-1)^{k-1}\qbinomial{n-1}{k-1}q^{{k\choose 2}+\lambda k}\frac{1}{(1-xq^k)^{\lambda+1}}=\frac{(q;q)_{n-1}}{(xq;q)_n}\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^{\lambda}\left(\frac{1}{j}{\sum_{\alpha=1}^n\frac{q^{j\alpha}}{(1-xq^\alpha)^j}}\right)^{c_j}.
\end{align*}
Using the $q$-inverse pair formula \cite{Goulden1983a}
  \begin{align*}
    f_n=\sum_{k=1}^n(-1)^kq^{{k\choose 2}}\qbinomial{n-1}{k-1}g_k \Leftrightarrow g_n=\sum_{k=1}^n(-1)^kq^{{k\choose 2}-k(n-1)}\qbinomial{n-1}{k-1}f_k,
  \end{align*}
  we obtain the inverse of $(3.2)$
  \begin{align*}
    &\sum_{k=1}^n(-1)^{k-1}q^{{k\choose 2}-k(n-1)}\qbinomial{n-1}{k-1}\frac{(q;q)_{k-1}}{(xq;q)_k}\nonumber\\
    &\quad\quad\quad\quad\times\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^{\lambda}\left(\frac{1}{j}\sum_{\alpha=1}^k\frac{q^{j\alpha}}{(1-xq^\alpha)^j}\right)^{c_j}=\frac{q^{n\lambda}}{(1-xq^n)^{\lambda+1}}.
  \end{align*}
Replacing $x$ by $q^x$, we rediscover an identity due to Mansour et al. \cite{Mansour2013a}:
\begin{cor}
Let $\lambda$ be any nonnegative integer. There holds
  \begin{align}
    &\sum_{k=1}^n(-1)^{k-1}q^{{k\choose 2}-k(n-1)}\qbinomial{n}{k}\qbinomial{x+k}{k}^{-1}\nonumber\\
    &\quad\quad\quad\quad\times\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^{\lambda}\left(\frac{1}{j}{\sum_{\alpha=1}^k\frac{q^{j\alpha}}{[x+\alpha]_q^j}}\right)^{c_j}=\frac{q^{n\lambda}[n]_q}{[x+n]_q^{\lambda+1}}.
  \end{align}
\end{cor}
This identity is a $q$-analog for the rational sum identity $(1.4)$ due to Prodinger.
If we further replace $n$ by $n+1$ and $x$ by $x-1$ in $(3.6)$, then a $q$-analog of Chu-Yan's identity $(1.5)$ is derived:
\begin{cor}
Let $\lambda$ be any nonnegative integer. There holds
\begin{align*}
  \sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-kn}\qbinomial{n}{k}\qbinomial{x+k}{k}^{-1}\sum_{0\leq \alpha_1\leq\cdots\leq \alpha_\lambda\leq k}\prod_{j=1}^\lambda\frac{q^{\alpha_j}}{[x+\alpha_j]_q}=\frac{q^{n(\lambda+1)}[x]_q}{[x+n]_q^{\lambda+1}}.
  \end{align*}
 \end{cor}

 Let the generalized $q$-harmonic numbers
\begin{align*}
H_0^{(r)}(q)=0, \quad  H_n^{(r)}(q)=\sum_{k=1}^nq^{rk}[k]^{-r},\quad n\geq 1.
\end{align*}
Recently, the $q$-generalized harmonic number sums have been useful in studying Feynman diagram contributions an relations among special functions \cite{Coffey2008a}.
 Taking $x=0$ in $(3.6)$, we have the following identities on $q$-generalized harmonic numbers:
 \begin{cor}
For $\lambda\geq 1$, there holds
\begin{align*}
  \sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-(k+\lambda)n}\qbinomial{n}{k}\sum_{\|\vec{c}\|=\lambda}\frac{1}{\vec{c}!}\prod_{j=1}^\lambda\left(\frac{H_k^{(j)}(q)}{j}\right)^{c_j}=-\frac{1}{[n]_q^\lambda}.
\end{align*}
\end{cor}
\noindent The first few cases are listed as follows.
\begin{align*}
  &\sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-(k+1)n}\qbinomial{n}{k}H_k(q)=-\frac{1}{[n]_q},\\
  &\sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-(k+2)n}\qbinomial{n}{k}\left(\left(H_k(q)\right)^2+H_k^{(2)}(q)\right)=-\frac{2}{[n]_q^2},\\
  &\sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-(k+3)n}\qbinomial{n}{k}\left(\left(H_k(q)\right)^3+3H_k(q)H_k^{(2)}(q)+2H_k^{(3)}(q)\right)=-\frac{6}{[n]_q^3},\\
  &\sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-(k+4)n}\qbinomial{n}{k}\bigg(\left(H_k(q)\right)^4+6\left(H_k(q)\right)^2H_k^{(2)}(q)+3\left(H_k^{(2)}(q)\right)^2\nonumber\\
  &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+8H_k(q)H_k^{(3)}(q)+6H_k^{(4)}(q)\bigg)=-\frac{24}{[n]_q^4},\\
  &\sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-(k+5)n}\qbinomial{n}{k}\bigg(\left(H_k(q)\right)^5+10\left(H_k(q)\right)^3H_k^{(2)}(q)+15H_k(q)\left(H_k^{(2)}(q)\right)^2\nonumber\\
  &+20\left(H_k(q)\right)^2H_k^{(3)}(q)+20H_k^{(2)}(q)H_k^{(3)}(q)+30H_k(q)H_k^{(4)}(q)+24H_k^{(5)}(q)\bigg)=-\frac{120}{[n]_q^5}
\end{align*}
These identities are $q$-analogs of generalized harmonic number identities which were presented in \cite{Wang2013a}:
\begin{align}
  &\sum_{k=0}^n(-1)^k{n\choose k}H_k=-\frac{1}{n},\\
  &\sum_{k=0}^n(-1)^k{n\choose k}\left(H_k^2+H_k^{(2)}\right)=-\frac{2}{n^2},\\
  &\sum_{k=0}^n(-1)^k{n\choose k}\left(H_k^3+3H_kH_k^{(2)}+2H_k^{(3)}\right)=-\frac{6}{n^3},\\
  &\sum_{k=0}^n(-1)^k{n\choose k}\bigg(H_k^4+6H_k^2H_k^{(2)}+3\left(H_k^{(2)}\right)^2+8H_kH_k^{(3)}+6H_k^{(4)}\bigg)=-\frac{24}{n^4},\\
  &\sum_{k=0}^n(-1)^k{n\choose k}\bigg(H_k^5+10H_k^3H_k^{(2)}+15H_k\left(H_k^{(2)}\right)^2\nonumber\\
  &\quad\quad\quad\quad\quad+20H_k^2H_k^{(3)}+20H_k^{(2)}H_k^{(3)}+30H_kH_k^{(4)}+24H_k^{(5)}\bigg)=-\frac{120}{n^5}
\end{align}
It is worth noticing that starting from
\begin{align*}
  \sum_{k=0}^n(-1)^kq^{{k+1\choose 2}-kn}\qbinomial{n}{k}\frac{(q;q)_k}{(xq;q)_k}=\frac{q^{n}(1-x)}{1-xq^n}
  \end{align*}
  and taking the $j$th derivative of both sides at $x=1$, we can also arrive at Corollary $3.6$. Wang and Jia \cite{Wang2013a} applied the Newton-Andrews method to some well known identities and found many interesting identities on harmonic numbers which include the identities from $(3.7)$ to $(3.11)$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
We thank the anonymous referee for his/her careful reading of our manuscript and very helpful comments.

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