Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r/Example

Theorem

 * $\dfrac {x \left({x - 2}\right) \left({x - 3}\right)} {\left({-1}\right) \left({-2}\right) \left({-3}\right)} + \dfrac {\left({x + 1}\right) \left({x - 1}\right) \left({x - 2}\right)} {\left({1}\right) \left({-1}\right) \left({-2}\right)} + \dfrac {\left({x + 2}\right) x \left({x - 1}\right)} {\left({2}\right) \left({1}\right) \left({-1}\right)} + \dfrac {\left({x + 3}\right) \left({x + 1}\right) x} {\left({3}\right) \left({2}\right) \left({1}\right)} = 1$

Proof
This is an example of Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r:
 * $\displaystyle \sum_{k \mathop = 1}^n \left({\dfrac {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \left({x + k - r}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right) = 1$

where $n = 4$ and $m = 2$.