I am trying to find the Fisher information of a binomial distribution where $n=2$ and $n=\theta$. I have the log-likelihood function as $$n\ln2 + \sum^{n}_{x=1}x_i\ln \theta + (2n-\sum^{n}_{x=1}x_i)(ln(1-\theta))-\sum^{n}_{x=1}(\ln x_i!+\ln((2-x_i)!))$$I have the second partial derivative of the log-likelihood function as $$\frac{-\sum^{n}_{x=1}x_i}{\theta^2}-\frac{2n-\sum^{n}_{x=1}x_i}{(1-\theta)^2}$$
When I go to take the negative expectation of this, I get $$\frac{n}{\theta}+\frac{2n-n\theta}{(1-\theta)^2}$$ which simplifies to $\frac{n}{\theta(1-\theta)^2}$. I know that the Fisher information is supposed to be $\frac{n}{\theta(1-\theta)}$, but I cannot figure out what I'm doing wrong that gets rid of the square. I think it has to do with the $2n$ somewhere along the line, but I cannot figure out where. Does my log-likelihood function look right? What about the negative expectation?