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 $n\ge\ell\ge0$?
Best Answer
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.