Sejam $n\in {\mathbb{N}}^{\ast },p\in \mathbb{N}$, onde $\mathbb{N}=\{0,1,2,3,\dots \}$ e ${\mathbb{N}}^{\ast }=\{1,2,3,\dots \}$. Então $\sum _{p=0}^n(-1{)}^{p-n}(-1{)}^p(-1{)}^{n-p}(\genfrac{}{}{0ex}{}{n}{p})$ vale: