Sum over k of n Choose k by p^k by (1-p)^n-k by Absolute Value of k-np

Theorem
Let $n \in \Z_{\ge 0}$ be a non-negative integer.

Then:
 * $\ds \sum_{k \mathop \in \Z} \dbinom n k p^k \paren {1 - p}^{n - k} \size {k - n p} = 2 \ceiling {n p} \dbinom n {\ceiling {n p} } p^{\ceiling {n p} } \paren {1 - p}^{n - 1 - \ceiling {n p} }$

Proof
Let $t_k = k \dbinom n k p^k \paren {1 - p}^{n + 1 - k}$.

Then:
 * $t_k - t_{k + 1} = \dbinom n k p^k \paren {1 - p}^{n - k} \paren {k - n p}$

Thus the stated summation is:


 * $\ds \sum_{k \mathop < \ceiling {n p} } \paren {t_{k + 1} - t_k} + \sum_{k \mathop \ge \ceiling {n p} } \paren {t_k - t_{k + 1} } = 2 t_{\ceiling {n p} }$