How to prove the combinatorial identity $\sum_{k=\ell}^{n}\binom{2n-k-1}{n-1}k2^k=2^\ell n\binom{2n-\ell}{n}$ for $n\ge\ell\ge0$? With the aid of the simple identity
\begin{equation*}
\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^{k}}=2^n
\end{equation*}
in Item (1.79) on page 35 of the monograph
R. Sprugnoli, Riordan Array Proofs of Identities in Gould’s Book, University of Florence, Italy, 2006. (Has this monograph been formally published somewhere?)
I proved the combinatorial identity
$$
\sum_{k=1}^{n}\binom{2n-k-1}{n-1}k2^k=n\binom{2n}{n}, \quad n\in\mathbb{N}.
$$
My question is: how to prove the more general combinatorial identity
$$
\sum_{k=\ell}^{n}\binom{2n-k-1}{n-1}k2^k=2^\ell n\binom{2n-\ell}{n}
$$
for $n\ge\ell\ge0$?
 A: For $n\ge\ell=0$, it follows that
\begin{align*}
\sum_{k=0}^{n}\binom{2n-k-1}{n-1}2^kk
&=\sum_{k=0}^{n-1}\binom{n+k-1}{n-1}(n-k)2^{n-k}\\
&=n2^n\sum_{k=0}^{n-1}\binom{n+k-1}{n-1}\frac{1}{2^{k}}
-2^{n}\sum_{k=1}^{n-1}\binom{n+k-1}{n-1}\frac{k}{2^{k}}\\
&=n2^n\sum_{k=0}^{n-1}\binom{n+k-1}{k}\frac{1}{2^{k}}
-2^{n}n\sum_{k=0}^{n-2}\binom{n+k}{k}\frac{1}{2^{k}}\\
&=n\binom{2n}{n},
\end{align*}
where we used the identity
\begin{equation}
\sum_{k=0}^{n}\binom{n+k}{k}\frac1{2^{k}}=2^n, \quad n\ge0,
\end{equation}
which has been mentioned in the question.
Assume that the claimed identity in the question is valid for some $n>\ell>0$. Then it is easy to see that
\begin{align*}
\sum_{k=\ell+1}^{n}\binom{2n-k-1}{n-1}2^kk
&=\sum_{k=\ell}^{n}\binom{2n-k-1}{n-1}2^kk-\binom{2n-\ell-1}{n-1}\ell2^\ell\\
&=2^\ell n\binom{2n-\ell}{n}-\binom{2n-\ell-1}{n-1}\ell2^\ell\\
&=2^{\ell+1}n\binom{2n-\ell-1}{n}.
\end{align*}
Inductively, we conclude that the claimed identity in the question is valid for all $n\ge\ell\ge0$. That is, the identity
\begin{equation}\label{sum-central-binom-ell-eq}
\sum_{k=\ell}^{n}\binom{2n-k-1}{n-1}2^kk
=\binom{2n-\ell}{n}2^\ell n, \quad n\ge\ell\ge0
\end{equation}
is valid.
A: The classical Euler's gamma function $\Gamma(z)$ can be defined by
\begin{equation}
\Gamma(z)=\lim_{n\to\infty}\frac{n!n^z}{\prod_{k=0}^n(z+k)}, \quad z\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}.
\end{equation}
The extended binomial coefficient $\binom{z}{w}$ for $z,w\in\mathbb{C}$ is defined by
\begin{equation}
\binom{z}{w}=
\begin{cases}
\dfrac{\Gamma(z+1)}{\Gamma(w+1)\Gamma(z-w+1)}, & z\not\in\mathbb{N}_-,\quad w,z-w\not\in\mathbb{N}_-;\\
0, & z\not\in\mathbb{N}_-,\quad w\in\mathbb{N}_- \text{ or } z-w\in\mathbb{N}_-;\\
\dfrac{\langle z\rangle_w}{w!},& z\in\mathbb{N}_-, \quad w\in\mathbb{N}_0;\\
\dfrac{\langle z\rangle_{z-w}}{(z-w)!}, & z,w\in\mathbb{N}_-, \quad z-w\in\mathbb{N}_0;\\
0, & z,w\in\mathbb{N}_-, \quad z-w\in\mathbb{N}_-;\\
\infty, & z\in\mathbb{N}_-, \quad w\not\in\mathbb{Z},
\end{cases}
\end{equation}
where
\begin{align}
\mathbb{Z}&=\{0,\pm1,\pm2,\dotsc\}, & \mathbb{N}&=\{1,2,\dotsc\},\\
\mathbb{N}_0&=\{0,1,2,\dotsc\}, & \mathbb{N}_-&=\{-1,-2,\dotsc\}
\end{align}
and
\begin{align}
\langle\alpha\rangle_n&=\prod_{k=0}^{n-1}(\alpha-k)\\
&=
\begin{cases}
\alpha(\alpha-1)\dotsm(\alpha-n+1), & n\ge1\\
1, & n=0
\end{cases}
\end{align}
is called the falling factorial.
By the idea of descending induction from Darij Grinberg, we can obtain the following more general identity.
Let $\ell,n\in\mathbb{Z}$ such that $\ell\le n$. Then
\begin{equation}
\sum_{k=\ell}^{n}\binom{2n-k-1}{n-1}2^kk
=\binom{2n-\ell}{n}2^\ell n.
\end{equation}
If an empty sum is understood to be $0$, then this identity holds for all integers $\ell,n\in\mathbb{Z}$, without the restriction $\ell\le n$.
A: For $k\in\mathbb{N}$, let $s_k$ and $S_k$ be two sequences independent of $n$ such that $n\ge k\in\mathbb{N}$. The inversion theorem, Theorem 4.4 on page 528 in the freely downloading paper [1] below, reads that
\begin{equation}\label{Qi-Zou-Guo-Inversion-thm}\tag{1}
s_n=\sum_{k=1}^{n}\binom{k}{n-k}S_k
\quad\text{if and only if}\quad
(-1)^nnS_n=\sum_{k=1}^{n}\binom{2n-k-1}{n-1}(-1)^kks_k.
\end{equation}
Applying the inversion theorem in \eqref{Qi-Zou-Guo-Inversion-thm} and considering the identity
\begin{equation*}
\sum_{k=1}^{n}\binom{2n-k-1}{n-1}2^kk
=2n\binom{2n-1}{n},
\end{equation*}
which can be deduced from letting $\ell=1$ in the resulted identity of the above answers, we conclude
\begin{equation*}
\sum_{k=1}^{n}(-1)^k\binom{k}{n-k}\binom{2k-1}{k}=(-1)^n2^{n-1}, \quad n\in\mathbb{N}.
\end{equation*}
See Remark 3.4 on page 11 in the paper [2] below.
References

*

*Feng Qi, Qing Zou, and Bai-Ni Guo, The inverse of a triangular matrix and several identities of the Catalan numbers, Applicable Analysis and Discrete Mathematics 13 (2019), no. 2, 518--541; available online at https://doi.org/10.2298/AADM190118018Q.

*Feng Qi and Mark Daniel Ward, Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function, arXiv (2021), available online at https://arxiv.org/abs/2110.08576v1.

