# 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:

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

$\blacksquare$

## Historical Note

This result was stated by Abraham de Moivre in his $1730$ work Miscellanea Analytica for the case where $n p$ is an integer.

The general case was proved by Henri Poincaré in his Calcul des Probabilités of $1896$.