# interesting combinatorial identity proof

• Nov 29th 2009, 03:52 PM
galactus
interesting combinatorial identity proof
Hi ALL;

Here is one I found interesting. If you want to give it a go.

I am sorry, I can not get those capital Pi's (product series) to display larger.

Prove for integers n and t with $n\geq 0, \;\ t\geq 1$

$\frac{1}{t!}\sum_{k=0}^{n}\prod_{j=1}^{t}(n-k+j)=\frac{1}{(t+1)!}\prod_{j=1}^{t+1}(n+j)$

In order to prove this, show the following. The first two can be done algebraically and the last inductively on n.

$\frac{1}{t!}\sum_{k=0}^{n}\prod_{j=1}^{t}(n-k+j)=\sum_{k=0}^{n}\binom{t+k}{k}$

$\frac{1}{(t+1)!}\prod_{j=1}^{t+1}(n+j)=\binom{t+n+ 1}{n}$

$\sum_{k=0}^{n}\binom{t+k}{k}=\binom{t+n+1}{n}$
• Nov 29th 2009, 05:34 PM
Chris L T521
Quote:

Originally Posted by galactus
Hi ALL;

Here is one I found interesting. If you want to give it a go.

I am sorry, I can not get those capital Pi's (product series) to display larger.

Prove for integers n and t with $n\geq 0, \;\ t\geq 1$

$\frac{1}{t!}\sum_{k=0}^{n}{\Pi}_{j=1}^{t}(n-k+j)=\frac{1}{(t+1)!}{\Pi}_{j=1}^{t+1}(n+j)$

In order to prove this, show the following. The first two can be done algebraically and the last inductively on n.

$\frac{1}{t!}\sum_{k=0}^{n}\Large{\Pi}_{j=1}^{t}(n-k+j)=\sum_{k=0}^{n}\binom{t+k}{k}$

$\frac{1}{(t+1)!}{\Pi}_{j=1}^{t+1}(n+j)=\binom{t+n+ 1}{n}$

$\sum_{k=0}^{n}\binom{t+k}{k}=\binom{t+n+1}{n}$

Hey galactus,

You can use \prod to generate $\prod$. I've edited your post and put them in there for you.