I want to differentiate a function with respect to m, and I know that I have to first find a representation for the whole sum to do that since m is discrete. However, I could not find one yet. Any help?
$ f(x) = \sum_{k=0}^{m} \binom{m}{k} \big({(1-p)}^{(m-k)}p^k - {(1-p)}^{(m-k)}p^k\big) \ln\big(x {(1-p)}^{(m-k)}p^k+ (1-x){(1-p)}^{(m-k)}p^k)\big)$
Any comment on how to simplify the sum is also appreciated.
On
You can take the gamma function $\Gamma(x)$ with $\Gamma(n+1)=n!$ for each $n\in\mathbb N$. So you can write $$\binom m k = \frac{m!}{k!(m-k)!} = \frac{\Gamma(m+1)}{\Gamma(k+1)\Gamma(m-k+1)}$$ The gamma function is differentiable (which is not true for the factorial because its a discrete function).
Are you sure you you want to differentiate wrt $m$? You have an expression $m!$, it's not so easy to take a derivative of a factorial function. Your expression can be rewritten as $$ f(x) = \sum_{k=0}^{m} \binom{m}{k} \alpha \log(\alpha x + (1-x) \beta) - \sum_{k=0}^{m} \binom{m}{k} \beta \log(\alpha x + \beta (1-x)) $$ For some $\alpha$ and $\beta$. I suggest you start here.