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I want to prove the following identity: if $1\leq m\leq n$, then
$$ \sum_{i=1}^{m}(-1)^{i-1}\binom{n-i}{n-m}\left[\binom{n+m}{i}-\binom{n+m}{i-1}\right]=\binom{n}{m} $$

My attempt: I wanted to use the following identity $$ \sum_{j=0}^{m}(-1)^j\binom{m}{j}\binom{n-j}{r}=\binom{n-m}{n-r} $$ where $m\leq r\leq n$, but I got nowhere.

RobPratt
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boaz
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  • General purpose hint to get started: rewrite the known identity with $m$ and $n$ replaced by other symbols (not $i$, $j$, or $r$) so that it will be easier to make explicit substitutions to apply the result to your unproven identity. – Sammy Black Jun 05 '25 at 23:42
  • My guess is to re-arrange the terms to get $\binom{n+m}i$ times a coefficient. – Thomas Andrews Jun 05 '25 at 23:43
  • What did you try? If you expand the bracket you are nearly done. – Servaes Jun 06 '25 at 09:26

3 Answers3

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We seek to show that with $1\le m\le n$

$$\sum_{q=1}^m (-1)^{q-1} {n-q\choose n-m} \left[{n+m\choose q} - {n+m\choose q-1}\right] = {n\choose m}.$$

Re-write to obtain

$$\sum_{q=1}^m (-1)^{q-1} {n-q\choose m-q} \left[{n+m\choose q} - {n+m\choose q-1}\right] = {n\choose m}.$$

We get for the first piece using the extractor to enforce the upper range

$$[z^m] (1+z)^n \sum_{q\ge 1} {n+m\choose q} (-1)^{q-1} \frac{z^q}{(1+z)^q} \\ = {n\choose m} + [z^m] (1+z)^n \sum_{q\ge 0} {n+m\choose q} (-1)^{q-1} \frac{z^q}{(1+z)^q} \\ = {n\choose m} - [z^m] (1+z)^n \left[1-\frac{z}{1+z}\right]^{n+m} \\ = {n\choose m} - [z^m] \frac{1}{(1+z)^m} = {n\choose m} - (-1)^m {2m-1\choose m}.$$

Continuing with the second piece including the sign,

$$-[z^m] (1+z)^n \sum_{q\ge 1} {n+m\choose q-1} (-1)^{q-1} \frac{z^q}{(1+z)^q} \\ = -[z^{m-1}] (1+z)^{n-1} \sum_{q\ge 1} {n+m\choose q-1} (-1)^{q-1} \frac{z^{q-1}}{(1+z)^{q-1}} \\ = -[z^{m-1}] (1+z)^{n-1} \left[1-\frac{z}{1+z}\right]^{n+m} \\ = - [z^{m-1}] \frac{1}{(1+z)^{m+1}} = - (-1)^{m-1} {2m-1\choose m-1}.$$

Collecting everything,

$${n\choose m} - (-1)^m {2m-1\choose m} + (-1)^m {2m-1\choose m} = {n\choose m}.$$

This is the claim.

Marko Riedel
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    Thanks Marco. Can you explain what do you mean by "extractor"? What the meaning of the bracket notation $[z^n]$ ? – boaz Jun 06 '25 at 05:48
  • I guess [z^m] is the extractor, which I suppose extracts the coefficient of z^m from a formal series; so that [z^m](1 + z)^n = the binomial coefficient (n m) – Fred vdP Jun 06 '25 at 14:23
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You want to show that for all positive integers $m$ and $n$ with $m\leq n$ you have $$\sum_{i=1}^{m}(-1)^{i-1}\binom{n-i}{n-m}\left[\binom{n+m}{i}-\binom{n+m}{i-1}\right]=\binom{n}{m},\tag{1}$$ using the identity $$\sum_{i=0}^a(-1)^i\binom{a}{i}\binom{b-i}{c}=\binom{b-a}{b-c}.\tag{2}$$ This is a matter of simply expanding the bracket to split the sum: \begin{eqnarray} \sum_{i=1}^{m}(-1)^{i-1}&&\binom{n-i}{n-m}\left[\binom{n+m}{i}-\binom{n+m}{i-1}\right]\\ &&= \sum_{i=1}^{m}(-1)^{i-1}\binom{n+m}{i}\binom{n-i}{n-m} -\sum_{i=1}^m(-1)^{i-1}\binom{n+m}{i-1}\binom{n-i}{n-m}\\ &&= -\sum_{i=1}^{m}(-1)^i\binom{n+m}{i}\binom{n-i}{n-m} -\sum_{i=0}^{m-1}(-1)^i\binom{n+m}{i}\binom{n-1-i}{n-m}. \end{eqnarray} We already see that the summand of both sums are of the same form as those on left hand side of idendity $(2)$. All that is left to do is to extend both sums to the range $0\leq i\leq m$, so that we can use the identity. With the convention that $\binom ab=0$ if $a<b$ we get \begin{eqnarray} -\sum_{i=1}^{m}(-1)^i\binom{n+m}{i}\binom{n-i}{n-m}&=&\binom{n}{n-m}&-\sum_{i=0}^{m}(-1)^i\binom{n+m}{i}\binom{n-i}{n-m}\\ -\sum_{i=0}^{m-1}(-1)^i\binom{n+m}{i}\binom{n-1-i}{n-m}&=&&-\sum_{i=0}^m(-1)^i\binom{n+m}{i}\binom{n-1-i}{n-m}. \end{eqnarray} Use identity $(2)$ with $a=n+m$, $c=n-m$ and $b=n,n+1$ we find that \begin{eqnarray} \sum_{i=0}^{m}(-1)^i\binom{n+m}{i}\binom{n-i}{n-m}&=&\binom{-m}{\hphantom{-}m}=0,\\ \sum_{i=0}^{m}(-1)^i\binom{n+m}{i}\binom{n-1-i}{n-m}&=&\binom{-m-1}{\hphantom{-}m-1}=0. \end{eqnarray} Now all that is left of the entire sum is $\binom{n}{n-m}$, and of course $\binom{n}{n-m}=\binom{n}{m}$.

Servaes
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  • Thanks @Servas. But you used identity (2) when $b<a$, where the identity is true when $a\leq c\leq b$. Is (2) is true for any $a,b,c$? – boaz Jun 06 '25 at 15:58
  • Normally, you define $\binom nk:=\frac{\prod_{i=0}^{k-1}(n-i)}{k!}$ to extend the binomial coefficient to allow negative numbers in the upper index. With this convention, $\binom{-m}{m}$ is not zero, but is instead $(-1)^m\binom{2m-1}{m}$. Similarly, $\binom{-m-1}{m-1}=(-1)^{m-1}\binom{2m-1}{m-1}$. If you make those replacements in your last two displayed equations, they still cancel each other and you get the same answer. I think your proof is still correct, but only when you use the convention $\binom nk:= \frac{\prod_{i=0}^{\color{red}{n-k}}(n-i)}{k!}.$ – Mike Earnest Jun 06 '25 at 18:45
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Summation by parts with $f_i=(-1)^{i-1}\binom{n-i}{n-m}$ and $g_i=\binom{n+m}{i-1}$ yields \begin{align} &\sum_{i=1}^m (-1)^{i-1} \binom{n-i}{n-m}\left[\binom{n+m}{i}-\binom{n+m}{i-1}\right] \\ &= \sum_{i=1}^m f_i(g_{i+1}-g_i) \\ &= f_{m+1} g_{m+1} - f_1 g_1 - \sum_{i=1}^m g_{i+1}(f_{i+1}-f_i) \\ &= (-1)^m \binom{n-m-1}{n-m} \binom{n+m}{m} - (-1)^0 \binom{n-1}{n-m} \binom{n+m}{0}\\ &\quad- \sum_{i=1}^m \binom{n+m}{i}\left[(-1)^i \binom{n-i-1}{n-m}-(-1)^{i-1}\binom{n-i}{n-m}\right] \\ &= 0-\binom{n-1}{n-m} - \sum_{i=1}^m (-1)^i \binom{n+m}{i} \binom{n-1-i}{n-m} \\ &\quad- \sum_{i=1}^m (-1)^i \binom{n+m}{i} \binom{n-i}{n-m} \\ &= -\binom{n-1}{n-m} - \left[\binom{n-1-(n+m)}{n-1-(n-m)} - (-1)^0 \binom{n+m}{0} \binom{n-1-0}{n-m}\right] \\ &\quad - \left[\binom{n-(n+m)}{n-(n-m)} - (-1)^0 \binom{n+m}{0} \binom{n-0}{n-m}\right] \\ &= -\color{red}{\binom{n-1}{n-m}} - \left[\color{blue}{\binom{-m-1}{m-1}} - \color{red}{\binom{n-1}{n-m}}\right] - \left[\color{blue}{\frac{-m}{m}\binom{-m-1}{m-1}} - \binom{n}{n-m}\right] \\ &= \binom{n}{n-m} \\ &= \binom{n}{m}. \\ \end{align}

RobPratt
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