A new combinatorics identity-- similar to Catalan number I find a combinatorics identity during my study, but fail to prove it.$$\sum_{i=0}^{[M/2]}(-1)^i\frac{(3M-1-2i)!}{(M-2i)!i!(2M-i)!} = \frac{1}{2M}\big(_{M}^{2M}\big)$$
where $M=1,2,3\cdots$. Note than Catalan number is $C_M=\frac{1}{M+1}\big(_{M}^{2M}\big)$. Can some one give me some suggestions or just prove it?
 A: It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{i=0}^{\left\lfloor\frac{M}{2}\right\rfloor}}&\color{blue}{(-1)^i\frac{(3M-1-2i)!}{(M-2i)!i!(2M-i)!}}\\
&=\frac{1}{2M}\sum_{i=0}^{\left\lfloor\frac{M}{2}\right\rfloor}(-1)^i\binom{2M}{i}\binom{3M-1-2i}{M-2i}\\
&=\frac{1}{2M}\sum_{i=0}^{{\left\lfloor\frac{M}{2}\right\rfloor}}(-1)^i\binom{2M}{i}\binom{-2M}{M-2i}(-1)^M\tag{2}\\
&=\frac{(-1)^M}{2M}\sum_{i=0}^{{\left\lfloor\frac{M}{2}\right\rfloor}}(-1)^i\binom{2M}{i}[z^{M-2i}](1+z)^{-2M}\tag{3}\\
&=\frac{(-1)^M}{2M}[z^M](1+z)^{-2M}\sum_{i=0}^{\infty}(-1)^i\binom{2M}{i}z^{2i}\tag{4}\\
&=\frac{(-1)^M}{2M}[z^M](1+z)^{-2M}\left(1-z^2\right)^{2M}\tag{5}\\
&=\frac{(-1)^M}{2M}[z^M](1-z)^{2M}\tag{6}\\
&\,\,\color{blue}{=\frac{1}{2M}\binom{2M}{M}}\tag{7}
\end{align*}
and the claim follows.

Comment:

*

*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.


*In (3) we use the coefficient of operator according to (1).


*In (4) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$. We also set the upper limit to $\infty$ which is admissible, since powers of $z$ greater $M$ do not contribute.


*In (5) we apply the binomial theorem.


*In (6) we do some simplifications.


*In (7) we select the coefficient of $z^M$.
