In this section, we give some applications of Theorem 1 and Corollary 2. We can obtain many old and new binomial identities.
Below we give an extension of binomial identity (7) and deduce some new binomial identities.
Theorem 4
Let
\(r, l, n \in\mathbb{N}\), \(m \in\mathbb{N}_{0}\), \(0 \leq m \leq n\); \(\theta> 0\), \(\beta\neq0\).
When
\(\theta\notin\{m+1, m+2,\ldots, n\}\)
and
\(r>0\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr) \\ &\qquad{}\times\Biggl( \biggl(\frac{\beta }{\theta} \biggr)^{r}+ \sum_{j=1}^{r-1}\sum _{1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta +s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta^{r}}{\theta(k_{1}-\theta)\cdots (k_{i}-\theta)(\theta+s_{1})\cdots(\theta+s_{l})} \Biggr). \end{aligned}$$
(10)
When
\(\theta\in\{m+1, m+2,\ldots, n\}\)
and
\(r>1\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \\ &\qquad{}\times \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr) \Biggl(\frac {\beta^{r}}{\theta^{r-1}}+\sum_{j=1}^{r-2} \sum_{1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta +s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-2}\sum_{i=1}^{n-m-1} \mathop{\sum_{k_{1},\ldots,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta ^{r}}{\theta(k_{1}-\theta)\cdots(k_{i}-\theta)(\theta+s_{1})\cdots (\theta+s_{l})} \Biggr). \end{aligned}$$
(11)
Proof
Taking \(f(z)= (\frac{\beta}{\theta+z} )^{r}\) in (8), and \(\mathcal{C}\) is a positively oriented, encloses the poles \(0, 1, 2,\ldots,n\) and no others, by means of the residues calculus, we obtain the following.
• When \(\theta\notin\{m+1, m+2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!} \int_{\mathcal{C}}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r}\,dz \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!} \mathop{\operatorname {Res}}\limits_{z=-\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(\theta+z)^{r-1}\bigr]\frac{(z+1+m)\cdots (z+n)\beta^{r}}{z(z-1)\cdots(z-n)} \\ &\quad= \frac{n!}{(n-m)!}\bigl[z^{r-1}\bigr] \Biggl(\prod _{k=1}^{n}\frac{1}{\theta +k} \Biggr) \Biggl(\prod _{k=m+1}^{n}(k-\theta) \Biggr) \frac{(1+\frac {z}{m+1-\theta})\cdots(1+\frac{z}{n-\theta})\beta^{r}}{\theta (1-\frac{z}{\theta})(1-\frac{z}{\theta+1})\cdots(1-\frac {z}{\theta+n})} \\ &\quad= \Biggl(\prod_{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr)\bigl[z^{r -l-1}\bigr] \Biggl(1+\sum _{i=1}^{n-m}\sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \frac{z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots (\theta+s_{l})} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr)\bigl[z^{r -l-1}\bigr] \Biggl(1+\sum _{i=1}^{n-m}\sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \frac{z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times \Biggl(\frac{\beta^{r}}{\theta^{l+1}}+\sum_{j=1}^{l} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac{\beta ^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \Biggr)\\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr) \Biggl( \biggl(\frac{\beta }{\theta} \biggr)^{r}+ \sum_{j=1}^{r-1}\sum _{1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta +s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta^{r}}{\theta(k_{1}-\theta)\cdots (k_{i}-\theta)(\theta+s_{1})\cdots(\theta+s_{l})} \Biggr). \end{aligned}$$
• When \(\theta\in\{m+1, m+2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!}\int_{\mathcal{C}}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r}\,dz \\ &\quad=\frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=-\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=-\theta}\frac{(z+1+m)\cdots (z+\theta-1)(z+\theta+1)\cdots(z+n-\theta)}{z(z-1)\cdots(z-n)} \frac{\beta^{r}}{(\theta+z)^{r-1}} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(\theta+z)^{r-2}\bigr]\frac{(z+1+m)\cdots (z+\theta-1)(z+\theta+1)\cdots(z+n-\theta)\beta^{r}}{z(z-1)\cdots (z-n)} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[z^{r-2}\bigr]\frac{(z+1+m-\theta)\cdots (z-1)(z+1)\cdots(z+n-\theta)\beta^{r}}{(z-\theta)(z-\theta -1)\cdots(z-n-\theta)} \\ &\quad= \frac{n!}{(n-m)!}\bigl[z^{r-2}\bigr] \Biggl(\prod _{k=1}^{n}\frac{1}{\theta +k} \Biggr) \Biggl(\prod _{k=m+1,k\neq\theta}^{n}(k-\theta) \Biggr)\\ &\qquad{}\times \frac{(1+\frac{z}{m+1-\theta})\cdots(1-z)(1+z)\cdots(1+\frac {z}{n-\theta})\beta^{r}}{\theta(1-\frac{z}{\theta})(1-\frac {z}{\theta+1})\cdots(1-\frac{z}{\theta+n})} \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr)\bigl[z^{r -l-2}\bigr] \\ &\qquad{}\times\Biggl(1+\sum _{i=1}^{n-m-1}\mathop{\sum_{k_{1},\ldots ,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \frac {z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots (\theta+s_{l})} \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr)\bigl[z^{r -l-2}\bigr] \\ &\qquad{}\times\Biggl(1+\sum _{i=1}^{n-m-1}\mathop{\sum_{k_{1},\ldots ,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \frac {z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times \Biggl(\frac{\beta^{r}}{\theta^{l+1}}+\sum_{j=1}^{l} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac{\beta ^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \Biggr) \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr)\\ &\qquad{}\times \Biggl(\frac {\beta^{r}}{\theta^{r-1}}+\sum_{j=1}^{r-2} \sum_{1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta +s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-2}\sum_{i=1}^{n-m-1} \mathop{\sum_{k_{1},\ldots,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta ^{r}}{\theta(k_{1}-\theta)\cdots(k_{i}-\theta)(\theta+s_{1})\cdots (\theta+s_{l})} \Biggr). \end{aligned}$$
This proof is complete. □
Setting \(r=1\) and \(\theta=\beta\) in (10) of Theorem 4 and \(r=2\) and \(\theta=\beta\) in (11) of Theorem 4, we have the following new binomial identity.
Corollary 5
Let
\(n \in\mathbb{N}\), \(m \in\mathbb{N}_{0}\), \(0 \leq m \leq n\); \(\theta> 0\).
When
\(\theta\notin\{m+1, m+2,\ldots, n\}\), we have
$$\begin{aligned} \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \frac{\theta}{\theta+k} = \Biggl(\prod _{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(\prod _{k=m+1}^{n}\frac{k-\theta}{k-m} \Biggr). \end{aligned}$$
(12)
When
\(\theta\in\{m+1, m+2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\theta}{\theta+k} \biggr)^{2} \\ &\quad=\frac{\theta}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta +k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr). \end{aligned}$$
(13)
Corollary 6
(see [4])
For
\(r, n \in\mathbb{N}\); \(\theta> 0\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta }{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod _{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(1+\sum_{j=1}^{r-1}\sum _{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac {\theta^{j}}{(\theta+s_{1})(\theta+s_{2})\cdots(\theta +s_{j})} \Biggr). \end{aligned}$$
(14)
Proof
Take \(m=n\), \(\beta= \theta\) in (10) of Theorem 4, it is a must that \(\theta\notin\{m+1, m+2,\ldots, n\}\), and define that \(\prod_{k=n+1}^{n}=1\), \(\sum_{i=1}^{0}=0\), then Corollary 6 follows. □
Setting \(r=0\) in (14), defining the empty sum \(\sum_{j=1}^{-1}=-1\), we have the following familiar formula.
Corollary 7
For all
\(n \in\mathbb{N}\)
and
\(\theta> 0\),
$$ \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}=0. $$
(15)
Setting \(r=1\) in (14), defining the empty sum \(\sum_{j=1}^{0}=0\), we have the following.
Corollary 8
(see [4])
For all
\(n \in\mathbb{N}\)
and
\(\theta> 0\),
$$ \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\frac{\theta}{\theta+k} =\prod_{k=1}^{n} \frac{k}{\theta+k}. $$
(16)
Setting \(m=2\) in (14), we have the following.
Corollary 9
(see [4])
For all
\(n \in\mathbb{N}\)
and
\(\theta> 0\),
$$ \sum_{k=0}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{2} = \Biggl(\prod _{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(1+\sum_{k=1}^{n}\frac{\theta}{\theta+k} \Biggr). $$
(17)
Corollary 10
Let
\(r, l, n \in\mathbb{N}\)
and
\(\theta> 0\).
When
\(\theta\notin\{1, 2,\ldots, n\}\),
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k-\theta}{\theta+k} \Biggr) \Biggl(1+\sum_{j=1}^{r-1} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac{\theta^{j}}{(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n} \sum_{1\leq k_{1}<\cdots<k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta^{r-1}}{(k_{1}-\theta)\cdots(k_{i}-\theta )(\theta+s_{1})\cdots(\theta+s_{l})} \Biggr). \end{aligned}$$
(18)
When
\(\theta\in\{1, 2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{1}{\theta+k} \Biggr) \Biggl(\prod_{k=1,k \neq\theta}^{n}(k- \theta) \Biggr) \Biggl(\theta+\sum_{j=1}^{r-2} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac {\theta^{j}}{(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-2}\sum_{i=1}^{n-1} \mathop{\sum_{k_{1},\ldots,k_{i}\neq\theta}}_{ 1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta ^{r-1}}{(k_{1}-\theta)\cdots(k_{i}-\theta)(\theta+s_{1})\cdots (\theta+s_{l})} \Biggr). \end{aligned}$$
(19)
Proof
Take \(m=0\), \(\beta= \theta\) in Theorem 4, then Corollary 10 follows. □
Setting \(r=0\) in (18) and defining the suitable empty sum, we have the following familiar formula.
Corollary 11
([7])
For all
\(n \in\mathbb{N}\)
and
\(\theta> 0\),
$$ \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k}= (-1)^{n}. $$
(20)
Setting \(r=1\) and \(\theta=\beta\) in (18) and defining the empty sum \(\sum_{j=1}^{0}=0\), we have the following new binomial identity.
Corollary 12
For all
\(n \in\mathbb{N}\)
and
\(\theta> 0\),
$$ \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k}\frac{\theta}{\theta+k}= \Biggl(\prod _{k=1}^{n}\frac {k-\theta}{\theta+k} \Biggr). $$
(21)
Setting \(r=2\) and \(\theta=\beta\) in (18) and defining the empty sum \(\sum_{j=1}^{0}=0\), we have the following new binomial identity.
Corollary 13
For all
\(n \in\mathbb{N}\)
and
\(\theta> 0\),
$$ \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{2}= \Biggl(\prod_{k=1}^{n}\frac{k-\theta}{\theta+k} \Biggr) \Biggl(1+\sum_{k=1}^{n} \frac{\theta}{\theta+k} \Biggr). $$
(22)
Theorem 14
For
\(r,l,n \in\mathbb{N}\), \(0 \leq m \leq n\), \(\beta\neq0\).
When
\(\theta\notin\{0,1, 2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \Biggl(- \biggl(-\frac{\beta}{\theta} \biggr)^{r}+\sum _{j=1}^{r-1} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac {\beta^{r}}{\theta(k_{1}+\theta)\cdots(k_{i}+\theta)(s_{1}-\theta )\cdots(s_{l}-\theta)} \Biggr). \end{aligned}$$
(23)
When
\(\theta\in\{0,1, 2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta } \Biggl( \biggl(- \frac{\beta}{\theta} \biggr)^{r}+\sum_{j=1}^{r} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \mathop{\sum_{s_{1},\ldots,s_{l}\neq\theta}}_{ 0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta^{r}}{(k_{1}+\theta )\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots(s_{l}-\theta)} \Biggr). \end{aligned}$$
(24)
Proof
By means of Theorem 1 and the residues calculus, we obtain the following.
• When \(\theta\notin\{0,1, 2,\ldots, n\}\), taking \(f(z)= (\frac{\beta}{z-\theta} )^{r}\) in (8), and \(\mathcal{C}\) is a positively oriented, encloses the poles \(0, 1, 2,\ldots,n\) and no others.
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!}\int_{\mathcal{C}}\frac {(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta } \biggr)^{r}\,dz\\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac{(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)}\frac{\beta^{r}}{(z-\theta)^{r}} \\ &\quad=\frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(z-\theta)^{r-1}\bigr]\frac {(z+1+m)(z+2+m)\cdots(z+n)\beta^{r}}{z(z-1)\cdots(z-n)} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[z^{r-1}\bigr]\frac{(z+1+m+\theta )(z+2+m+\theta)\cdots(z+n+\theta)\beta^{r}}{(z+\theta)(z+\theta -1)\cdots(z+\theta-n)} \\ &\quad= (-1)^{n+1}\frac{n!(\theta-n-1){!}(m+1+\theta)_{n-m}}{(n-m)!\theta {!}}\bigl[z^{r-1}\bigr] \frac{(1+\frac{z}{m+1+\theta})\cdots(1+\frac {z}{n+\theta})\beta^{r}}{(1+\frac{z}{\theta})(1+\frac{z}{\theta -1})\cdots(1+\frac{z}{\theta-n})} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \bigl[z^{r-l-1}\bigr] \Biggl(1+\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{l}-\theta)} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \bigl[z^{r-l-1}\bigr] \Biggl(1+\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times \Biggl((-1)^{l}\frac{\beta^{r}}{\theta^{l+1}}+\sum _{j=1}^{l} \ \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{j}-\theta)} \Biggr) \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \Biggl(- \biggl(-\frac{\beta}{\theta} \biggr)^{r}+\sum _{j=1}^{r-1}\ \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac {\beta^{r}}{\theta(k_{1}+\theta)\cdots(k_{i}+\theta)(s_{1}-\theta )\cdots(s_{l}-\theta)} \Biggr). \end{aligned}$$
• When \(\theta\in\{0,1, 2,\ldots, n\}\), taking \(f(z)= (\frac{\beta}{z-\theta} )^{r}\) in (8), and \(\mathcal {C}\) is a positively oriented, encloses the poles \(0, 1, \ldots,\theta -1,\theta+1,\ldots,n\) and no others.
$$\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!}\int_{\mathcal{C}}\frac {(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta } \biggr)^{r}\,dz \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac{(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-\theta+1)(z-\theta-1)\cdots (z-n)}\frac{\beta^{r}}{(z-\theta)^{r+1}} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(z-\theta)^{r}\bigr]\frac {(z+1+m)(z+2+m)\cdots(z+n)\beta^{r}}{z(z-1)\cdots(z-\theta +1)(z-\theta-1)\cdots(z-n)} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[z^{r}\bigr]\frac{(z+1+m+\theta )(z+2+m+\theta)\cdots(z+n+\theta)\beta^{r}}{(z+\theta)(z+\theta -1)\cdots(z+1)(z-1)\cdots(z+\theta-n)} \\ &\quad= (-1)^{\theta+1}\frac{n!(m+1+\theta)_{n-m}}{(n-m)!\theta {!}(n-\theta){!}}\bigl[z^{r}\bigr] \frac{(1+\frac{z}{m+1+\theta})\cdots (1+\frac{z}{n+\theta})\beta^{r}}{(1+\frac{z}{\theta})(1+\frac {z}{\theta-1})\cdots(1+z)(1-z)\cdots(1-\frac{z}{n-\theta})} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta }\bigl[z^{r-l} \bigr] \Biggl(1+\sum_{i=1}^{n-m}\sum _{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times\mathop{\sum_{s_{1},\ldots,s_{l}\neq\theta}}_{ 0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta ^{r}}{(s_{1}-\theta)\cdots(s_{l}-\theta)} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta }\bigl[z^{r-l} \bigr] \Biggl(1+\sum_{i=1}^{n-m}\sum _{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times \Biggl(\frac{\beta^{r}}{(-\theta)^{l}}+\sum_{j=1}^{l} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\beta^{r}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \Biggr) \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta } \Biggl( \biggl(- \frac{\beta}{\theta} \biggr)^{r}+\sum_{j=1}^{r} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \mathop{\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}}_{s_{1},\ldots,s_{l}\neq\theta} \frac{\beta^{r}}{(k_{1}+\theta )\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots(s_{l}-\theta)} \Biggr). \end{aligned}$$
This proof is complete. □
Corollary 15
For
\(r,n \in\mathbb{N}\).
When
\(\theta\notin\{0,1, 2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta }{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{n}\binom{\theta-1}{n}^{-1} \Biggl(1-\sum _{j=1}^{r-1} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\theta ^{j}}{(s_{1}-\theta)\cdots(s_{j}-\theta)} \Biggr). \end{aligned}$$
(25)
When
\(\theta\in\{0,1, 2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta} \Biggl((-1)^{r}+\sum _{j=1}^{r} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq \cdots\leq s_{j}\leq n} \frac{\theta^{j}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \Biggr). \end{aligned}$$
(26)
Proof
Take \(m=n\), \(\beta= \theta\) in Theorem 14 and define that \(\prod_{k=n+1}^{n}=1\), \(\sum_{i=1}^{0}=0\), then Corollary 15 follows. □
Corollary 16
For
\(r,l,n \in\mathbb{N}\).
When
\(\theta\notin\{0,1, 2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{\theta} \Biggl( (-1 )^{r+1}+\sum_{j=1}^{r-1} \sum _{1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\theta^{j}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n} \sum_{1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta ^{r-1}}{(k_{1}+\theta)\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots (s_{l}-\theta)} \Biggr). \end{aligned}$$
(27)
When
\(\theta\in\{0,1, 2,\ldots, n\}\), we have
$$\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \binom{n+k}{k} \biggl(\frac{\theta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{\theta} \\ &\qquad{}\times\Biggl( (-1 )^{r}+\sum_{j=1}^{r} \mathop{\sum _{s_{1},\ldots ,s_{j}\neq\theta}}_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac {\theta^{j}}{(s_{1}-\theta)\cdots(s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r}\sum_{i=1}^{n} \sum_{1\leq k_{1}<\cdots< k_{i}\leq n} \mathop{\sum_{s_{1},\ldots,s_{l}\neq\theta}}_{ 0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta^{r}}{(k_{1}+\theta )\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots(s_{l}-\theta)} \Biggr). \end{aligned}$$
(28)
Proof
Take \(m=0\), \(\beta= \theta\) in Theorem 14 and define that \(\prod_{k=n+1}^{n}=1\), \(\sum_{i=1}^{0}=0\), then Corollary 16 follows. □
Remark 17
By applying Bell polynomials (3) and (4), we can deduce all the results of [18] and [19]. Our results are another representation of the corresponding results in [18] and [19]. For example, compare formula (24) with \(m=0\), \(\theta=M\), \(\beta=1\) and the following formula [19]: For \(0\leq M\leq n\),
$$\begin{aligned} &\sum_{0\leq k \leq n,k \neq M}(-1)^{k-1} \binom{n}{k} \binom{n+k}{k}\frac{1}{(k-M)^{r}} \\ &\quad=(-1)^{M}\binom{n}{M}\binom{M+n}{M}\sum _{m_{1}+2m_{2}+3m_{3}+\cdots=r}\frac{1}{m_{1}!m_{2}!m_{3}! \cdots} \biggl(\frac {x_{1}}{1} \biggr)^{m_{1}} \biggl(\frac{x_{2}}{2} \biggr)^{m_{2}} \biggl( \frac {x_{3}}{3} \biggr)^{m_{3}}\cdots, \end{aligned}$$
(29)
where
$$\begin{aligned} x_{k}=(-1)^{k-1}H_{M+n}^{(k)}-2H_{M}^{(k)}+H_{n-M}^{(k)},\quad H_{n}^{(r)}=\sum_{k=1}^{n} \frac{1}{k^{r}},\ n,r \in\mathbb{N}. \end{aligned}$$
Obviously, formula (24) includes the above formula (29) and the main result of [18], i.e., formula (24) is an extension of them.