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?
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Have you tried the techniques in Concrete Mathematics? – Kimball Aug 13 '14 at 03:41
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@Kimball:I am sorry but my major isn't Mathematics. could you tell me something in detail,please? – Roger209 Aug 13 '14 at 05:15
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I am referring to this book, which is an introductory text that explains how to deal with several types of sums involving binomial coefficients (I'm not sure about yours without checking). Note your sum is an alternating sum of the product of binomial coefficients ${3M-1-2i \choose 2M-1} {2M \choose i}$ divided by $2M$. By the way, if you provide a reference for where you found this, people here may be able to better help you. – Kimball Aug 13 '14 at 15:35
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@Kimball :Thanks. In fact, I discovered the identity origional(I mean I didn't copy the identity from any textbook or thesis). I don't think I can describle the background of it within 100 words, So I just omit it here. I confirm that there is not any typing error. – Roger209 Aug 14 '14 at 04:33
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Ah, from the wording of the question, I thought the identity was known (to someone) to be true and you found it in a book or paper. – Kimball Aug 14 '14 at 16:53
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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 $z^M$.
Markus Scheuer
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Thank you very much! it's a real wonderful solution. But it's a little difficult for me to follow. – Roger209 Sep 25 '14 at 13:58
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@Roger209: No worries! Maybe the following hints are useful. Observe that the binomial coefficient $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, so with some calculation you may see how to proceed from the first to the second line. The next you should keep in mind is $(1+u)^{2M}=\sum_{k=0}^{2M}\binom{2M}{k}u^k$, therefore $\binom{2M}{i}$ is the coefficient of $u^{i}$ in $(1+u)^{2M}$. So we can write $u^i^{2M}=\binom{2M}{i}$. – Markus Scheuer Sep 25 '14 at 18:43
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@Roger209: (continued) The last thing is $\mathop{res}_u A(u)=[u^{-1}]A(u)$. That's why we divide by $u^{i+1}$ which is merely a shift by $i+1$ to address the wanted coefficient $\binom{2M}{i}$. The rest are just some algebraic manipulations. – Markus Scheuer Sep 25 '14 at 18:43
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Wow, I get the hints, and I will try to master them. I have just made an upvote to your solution. Tks. :)~ – Roger209 Sep 26 '14 at 00:27
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