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:
 * $\displaystyle \sum_{k \mathop \in \Z} \dbinom n k p^k \left({1 - p}\right)^{n - k} \left\lvert{k - n p}\right\rvert = 2 \left\lceil{n p}\right\rceil \dbinom n {\left\lceil{n p}\right\rceil} p^{\left\lceil{n p}\right\rceil} \left({1 - p}\right)^{n - 1 - \left\lceil{n p}\right\rceil}$

Proof
Let $t_k = k \dbinom n k p^k \left({1 - p}\right)^{n + 1 - k}$.

Then:
 * $t_k - t_{k + 1} = \dbinom n k p^k \left({1 - p}\right)^{n - k} \left({k - n p}\right)$

Thus the stated summation is:


 * $\displaystyle \sum_{k \mathop < \left\lceil{n p}\right\rceil} \left({t_{k + 1} - t_k}\right) + \sum_{k \mathop \ge \left\lceil{n p}\right\rceil} \left({t_k - t_{k + 1} }\right) = 2 t_{\left\lceil{n p}\right\rceil}$