$\sum_{r=0}^{2024} r \binom{2024 + r}{r}$

Question

$$ S=\sum_{r=0}^{2024} r \binom{2024 + r}{r} $$

I tried a lot of algebraic manipulations but no nice expression comes out.

Then I tried for a combinatorial arguments

I did figure out some situations but they were also not so simple and even with those situations I wasn't able to solve

the simplest situation among the ones I framed is , 2024 people in which each have 1 chocolate C with them and the other 2024 people don't have any C now some people (say r of them) come to put their C in a bag (and exactly one of them does this job to put all C in the bag) and now those r Chocolates are redistributed , thus the number of cases are represented by S , but still I wasn't able to solve

Please provide any algebraic or combinatorial solution , any help is appreciated

Thanks in Advance!

Answer 1

Here is what I would try. Namely, the $r=0$ term doesn't matter, so let's look at $$ \sum_{r=1}^n r\binom{n+r}{r}, $$ we want a solution for $n=2024$, so let's just look at small $n$ first. The first interesting case is $n=2$ which given $$ \binom{3}{1}+2\binom{4}{2}. $$ Now, I'd like to combine this into a single term. In particular, I notice that $$ 2\binom{4}{2} = 2\frac{4!}{2!}{2!} = 4\frac{3!}{1!2!} = 4\binom{3}{1}. $$ Thus, the $n=2$ case leads to getting a solution of $(1+4)\binom{3}{1}$.

Now this is $n=2$, but let's see if $n=3$ can give us more insight. We end up looking at the sum $$ \binom{4}{1}+2\binom{5}{2}+3\binom{6}{3}. $$ Now my thought is what if we tackle this like an onion. We observe that $$ 3\binom{6}{3} = 3\frac{6!}{3!3!}= 6\frac{5!}{2!3!} = 6\binom{5}{2}. $$ Thus, we end up looking at $$ \binom{4}{1} + (2+6)\binom{5}{2} = \binom{4}{1} + (1+3)2\binom{5}{2}. $$ Then, similarly, we can similarly, do $$ 2\binom{5}{2} = 5\binom{4}{1}. $$ Thus, we see that the $n=3$ case gives us the solution of $$ (1 + 5 + 5\cdot\frac{6}{2})\binom{4}{1}. $$

Looking at this, makes me conjecture, that you can get rearrange this interms of a single sum. multiplied by $\binom{n+1}{1}$. In particular, I conjecture that it will be $$ \left(\sum_{r=1}^{n} \frac{(r+n)!}{r!}\right)\binom{n+1}{1}. $$ I don't know if this is the formula you are looking for (or even if my conjecture is correct), but it might give you a hint.