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Prove that $$\sum_{j=1}^r{\prod_{i=0}^{k-2}(j+i)\over (k-1)!} = \frac{\prod_{i=0}^{k-1}(r+i)}{k!}$$ where $k\in\Bbb N , k\ge2$

Example) $$k=2:\sum_{j=1}^rj= \frac{r(r+1)}{2!}$$ $$k=3:\sum_{j=1}^r\frac{j(j+1)}{2!} = \frac{r(r+1)(r+2)}{3!}$$

I don't know how to prove it. I tried with mathematical induction but it was so complicated for me. Can anyone help me?

Anne Bauval
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TH_Lee
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    Hint: $$\frac{\prod_{i=0}^{m-1}(a+i)}{m!}=\binom{a+m-1}{m}$$ – Thomas Andrews Apr 29 '24 at 13:54
  • It's certainly possible to prove it by induction on $x_k.$ It amounts to knowing $\binom{n+1}{m}-\binom nk=\binom n{m-1}.$ – Thomas Andrews Apr 29 '24 at 14:00
  • One thing that you should do is simplify by replacing $x_{k-1}$ and $x_k$ with just simple variables, $a,b.$ The equality has nothing to do with such a sequence $x_1,\dots,x_k,$ just three natural numbers $a,b,k.$ – Thomas Andrews Apr 29 '24 at 14:02
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    $\sum_{j=1}^r{\prod_{i=0}^{k-2}(j+i)\over (k-1)!} = \frac{\prod_{i=0}^{k-1}(r+i)}{k!}$ is equivalent to $\sum_{s=0}^{r-1}\binom{s+k-1}s=\binom{r-1+k}k$, which is a megaduplicate. – Anne Bauval Apr 29 '24 at 14:26
  • I used $x_{k-1}$ and ${x_k}$ because I found that equation by $$\sum_{x_{k-1}=1}^{x_k}{\sum_{x_{k-2}=1}^{x_{k-1}}{\sum_{x_{k-3}=1}^{x_{k-2}}{\dots\sum_{x_1=1}^{x_2}{x_1}}}} = \frac{\prod_{i=0}^{k-1}(x_k+i)}{k!}$$ – TH_Lee Apr 29 '24 at 14:45

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It can be easily proven with $k=2$. Aiming for the main proof, $$LHS=\sum_{j=1}^r{\prod_{i=0}^{k-2}(j+i)\over (k-1)!}=\sum_{j=1}^r \frac{j×(j+1)×(j+2)...×(j+k-2)}{(k-1)!} \\ \sum_{j=1}^r \frac{(j+k-2)!}{(j-1)!(k-1)!}\\ =\sum_{j=1}^r \binom{j+k-2}{k-1}$$ This is actually a telescopic summation. From the fact that $\binom{n+1}{r+1}-\binom{n}{r+1}=\binom{n}{r}$, we can write this as $$\binom{k-1}{k-1}+\sum_{j=2}^r \bigg[\binom{j+k-1}{k}+\binom{j+k-2}{k}\bigg]\\=1+\sum_{j=2}^r \binom{j+k-1}{k}-\sum_{j=2}^r\binom{j+k-2}{k} \\ $$ Here, the moment $j$ reaches $3$ in the second summation, it cancels out the first value of the first summation, which is $\binom{k+1}{k}$. This continues, until we reach $j=r$ where the value of the first summation persists as their will be no one to cancel it. Hence the sum is $$=1+\binom{r+k-1}{k}-\binom{k}{k}=\binom{r+k-1}{k}$$

Now time for $$RHS=\frac{\prod_{i=0}^{k-1}(r+i)}{k!}=\frac{r×(r+1)×(r+2)×...(r+k-1)}{(k)!}\\ =\frac{(r+k-1)!}{(r-1)!(k)!}\\ =\binom{r+k-1}{k}$$

Hence proof complete

Gwen
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