As my title says …
Does someone know how to evaluate $\displaystyle \sum\limits_{k=0}^{2n}\frac{(-1)^k}{k!(2n-k)!}=0$
Thanks
PS : I did not know where to post this but it arose from an algebra problem so …
$\displaystyle \sum_{k=0}^{2n} \frac{(-1)^k}{k!(2n-k)!}=\frac{1}{(2n)!} \sum_{k=0}^{2n} \frac{(-1)^k(2n)!}{k!(2n-k)!}=\frac{1}{(2n)!} \sum_{k=0}^{2n} (-1)^k \binom{2n}{k} = \frac{1}{(2n)!}(1-1)^{2n}=0.$