How to prove $\sum^{m}_{k=0}(-2)^k\binom{2n-k}{n}\binom{m}{k}=0$ for $m=1,3,5,\dots,2n-1$?

Question

I would like to prove that $$ \sum^{m}_{k=0}(-2)^k\binom{2n-k}{n}\binom{m}{k}=0,\quad m=1,3,5,\cdots,2n-1,n\in\mathbb{N}^*. $$ Currently I have no idea of this: expanding $\displaystyle\binom{2n-k}{n}\binom{m}{k}$ yields $\dfrac{(2n-k)!m!}{n!(n-k)!(m-k)!k!}$, which does not simplify to anything. I tried also considering $P_n(x):=\displaystyle\sum^{n}_{k=0}\binom{2n-k}{n}x^k$, then $\displaystyle\sum^{m}_{k=0}(-2)^k\binom{2n-k}{n}\binom{m}{k}$ is the $x^m$-coefficient of $(1+x)^mP_n(-2x)$, which does not seem to help because I can't simplify $P_n(x)$. I don't know how $m$ being odd plays a role either. Could anybody please help me with this identity? Thank you in advance.

Answer 1

Let's try the Snake Oil Method with $m=2l-1$ for a positive integer $l$. Before we start, let us recall a few well-known functions and relevant identities involving those. Let $$ B(x)=\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}{n}x^n, \quad C(x)=\frac{1-\sqrt{1-4x}}{2x}=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n. $$ Then $$ B(x)=\frac{1}{1-2xC(x)} \quad \text{and} \quad \sum_{n=0}^{\infty}\binom{2n-k}{n}x^n=x^kB(x)C(x)^k. $$ Now we can proceed. $$ \begin{split} \sum_{n=0}^{\infty}\sum_{k=0}^{m}(-2)^k\binom{2n-k}{n}\binom{m}{k}x^n &=\sum_{k=0}^{m}\sum_{n=0}^{\infty}(-2)^k\binom{2n-k}{n}\binom{m}{k}x^n\\ &=\sum_{k=0}^{m}\left(\sum_{n=0}^{\infty}\binom{2n-k}{n}x^n\right)\binom{m}{k}(-2)^k\\ &=\sum_{k=0}^{m}\left(x^kB(x)C(x)^k\right)\binom{m}{k}(-2)^k\\ &=B(x)\sum_{k=0}^{m}\binom{m}{k}\left(-2xC(x)\right)^k\\ &=\frac{1}{1-2xC(x)}\left(1-2xC(x)\right)^m\\ &=\left(1-2xC(x)\right)^{m-1}\\ &=\left(\frac{1}{B(x)}\right)^{2l-2}=\left(\frac{1}{B(x)^2}\right)^{l-1}=(1-4x)^{l-1}. \end{split} $$ In our problem, we want $1\le l\le n$, i.e. $0\le l-1<n$. Therefore, $$ \sum_{k=0}^{m}(-2)^k\binom{2n-k}{n}\binom{m}{k}=[x^n](1-4x)^{l-1}=0. $$ In fact, more generally, for $m=2l+1$, where $l$ is a nonnegative integer, we have $$ \sum_{k=0}^{2l+1}(-2)^k\binom{2n-k}{n}\binom{2l+1}{k}=(-4)^n\binom{l}{n}. $$

Answer 2

Note that $\binom{2n-k}{n} = [x^n](1+x)^{2n-k}$, hence

$$\begin{align} \sum\limits_{k=0}^m (-2)^k \binom{2n-k}{n}\binom{m}{k} &= [x^n] (1+x)^{2n}\sum\limits_{k=0}^m (-2)^k (1+x)^{-k} \binom{m}{k} \\ &= [x^n] (1+x)^{2n} \left(1-\frac{2}{1+x}\right)^m \\ &= [x^n](x+1)^{2n-m} (x-1)^m. \end{align}$$ Let $f(x) = (x+1)^{2n-m}(x-1)^m$, we can notice that $[x^{k}]f(x) = (-1)^m [x^{2n-k}] f(x)$, as follows from the fact that $x^{2n} f(x^{-1}) = (-1)^m f(x)$. But for $k=n$ and odd $m$ it means that $[x^n] f(x) = -[x^n]f(x)$, which can only be true if $[x^n] f(x) = 0$, QED.