Fisher information of a Binomial distribution

So, you have $X$ ~ Binomial($n$, $p$), with pmf $f(x)$:

You seek the Fisher Information on parameter $p$. Here is a quick check using mathStatica's FisherInformation function:

which is what you got :)

Fisher information: $I_n(p) = nI(p)$, and $I(p)=-\mathbb{E_p}\Bigg( \frac{\partial^2 \log f(p,x)}{\partial p^2} \Bigg)$, where $f(p,x)={{1}\choose{x}} p^x (1-p)^{1-x}$ for a Binomial distribution. We start with $n=1$ as single trial to calculate $I(p)$, then get $I_n(p)$.

$\log f(p,x) = x \log p + (1-x) \log p$

$\frac {\partial \log f(p,X)}{\partial p} = \frac {X}{p} - \frac {1- X}{1 - p}$

$\frac {\partial^2 \log f(p,X)}{\partial p^2} = -\frac {X}{p^2} - \frac {1- X}{(1 - p)^2}$

$I(P) = -\mathbb{E_p}\Bigg( \frac{\partial^2 \log f(p,x)}{\partial p^2} \Bigg) = -\mathbb{E_p}\Bigg(-\frac {X}{p^2} - \frac {1- X}{(1 - p)^2}\Bigg) = \frac {p}{p^2} + \frac {1-p}{(1-p)^2} = \frac {1}{p} + \frac {1}{(1-p)} = \frac {1}{p(1-p)} $

As a result, $I_n(p) = n I(p) = \frac {n}{p(1-p)} $