# How to calculate/approximate expectation of function of a binomial random variable?

I am stuck at following problem in my research.

Suppose that $M=m$ is a random variable with binomial distribution and parameters $n,p$. The constants $r$ and $\gamma$ are greater than zero. $\mathcal{E} _M$ is the expectation operator with respect to $M$. I need to calculate following expectations: $$\mathcal{E} _M \left[ \frac{m}{m+1} e^{ - \frac{2^{\frac{m+1}{m}\cdot r } - 1}{\gamma}} \right]$$ and $$\mathcal{E} _M \left[ \frac{m}{m+1} \left(e^{ - \frac{2^{\frac{m+1}{m}\cdot r } - 1}{\gamma}}\right)^{m-1} - \frac{m}{m+1} \left(e^{ - \frac{2^{\frac{m+1}{m}\cdot r } - 1}{\gamma}}\right)^{m} \right]$$ The values of $r$ and $\gamma$ are such that the exponential function cannot be approximated.

Can anyone please provide any guidance/reference for how to go about solving/approximating above expectations? Thanks in advance.

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I doubt that there are closed forms. If $n$ is not too large, you can just calculate the expected values as explicit sums. If $n$ is large and $p$ is not too close to $0$ or $1$, you could approximate the binomial distribution with a normal distribution, so your expected value is approximated by an integral. Again I doubt that there will be closed forms, but you might get asymptotics. Note that if you write $m = \mu + \sigma Z$ where $\mu = np$ and $\sigma = \sqrt{np(1-p)}$, you can expand a function $g(m)$ as a series $g(m) = g(\mu) + \sum_j c_j Z^j$ and try approximating $E[g(m)] = g(\mu) + \sum_j c_j E[Z^j]$ using a partial sum of the series.