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My friend and I co-wrote this proof to the claim that $\binom{a+b}{b+1}=\sum^{a}_{i_0=1}\sum^{i_0}_{i_1=1}\sum^{i_1}_{i_2=1}\cdots\sum^{i_{b-1}}_{i_b=1}1$. I'm interested to know if this has been proven before and any feedback on the proof itself. Thanks!

PS: I'm a third year Computer Science and Math major at the University of Virginia, if that helps you understand my level in math.

Peter
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    Generally the answer is "yes, this has been proven before" and one perspective on this topic would be to consider the "calculus of finite differences." Another basic perspective would be to consider Pascal's Triangle if you haven't already. – abiessu Dec 27 '23 at 05:15
  • The sum basically counts the number of $(b+1)$-tuples $(i_0,\dots,i_b)$ such that $1\le i_b\le\dots\le i_1\le i_0\le a$. This can be counted by the well-known dots-and-bars method, and the result is $\binom{a+(b+1)-1}{b+1}=\binom{a+b}{b+1}$. – Alexander Burstein Dec 27 '23 at 08:59
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    @Peter: See for instance this answer. – Markus Scheuer Dec 27 '23 at 21:02

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First, this is a seemingly random identity, so who knows, but if it has a motivation probably. Also, if it has a motivation, you should include it in the paper.

Second, This is best suited for mathoverflow, which is focused on research.

Third, this line from your paper:

$$f(a,b) = \sum_{i=1}^a f(i, b-1) = \sum_{i=1}^a { i \choose n} = {a + 1 \choose n + 1}$$

Needs to be formalized two lines later that $$f(a,b) = {a + b \choose b + 1}$$ Because this is not inherently obvious.

Robert Murray
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