Let $k, \ell \leq n$ be non-negative integers. Does the following identity simplify?
\begin{align}
\sum_{j = 0}^{k} \binom{k}{j} \binom{j + n -\ell + 1}{n} = \binom{n - \ell + 1}{n} \phantom1_{2}\mathsf{F}_{1}(-k,n - \ell + 2, 2- \ell; -1)
\end{align}
where $\!\!\! \phantom1_{2}\mathsf{F}_{1}$ is a hypergeometric function. That is, does the right side have another representation in terms of simple functions given that $k,\ell$ and $n$ are non-negative integers?
Answer
Let us denote:
\begin{equation}
S_k^{n,l} := \sum\limits_{j=0}^k
\left( \begin{array}{c} k \\ j \end{array} \right)
\left( \begin{array}{c} n-l+1+j \\ n \end{array} \right)
\end{equation}
Then we have:
\begin{eqnarray}
S_k^{n,l} &:=& \left. \frac{d^n}{n! d x^n} x^{n-l+1} \cdot \left(1+x\right)^k \right|_{x=1} \\
&=&
\frac{1}{n!} \sum\limits_{p=0}^n
\left(\begin{array}{c} n \\ p \end{array} \right) (n-l+1)_{(p)} k_{(n-p)} 2^{k-n+p} \\
&=&2^{k-n}
\left( \begin{array}{c} k \\ n \end{array} \right) {}_2F_1\left[4-n,-n,1+k-n;2\right]
\end{eqnarray}
The last result, that in terms of the hypergeometric function is not particularily useful when $n$, $l$, $k$ are integers. However the result above is useful for example if $k$ is large.
No comments:
Post a Comment