The number of up moves of the stock $S$ after 100 days follows binomial distribution.
To calculate expected value of the stock we have to weight values by probability mass function.
After 100 days we have $k$ up moves of $1+10\%$ and $100-k$ down moves of $1-10\%$ i.e. the value of the stock is $S_0*(1+10\%)^k*(1-10\%)^{(100-k)}$
with probability ${100}\choose{k}$$*p^k*p^{(100-k)}$ where $p=\frac{1}{2}$.
The expected value at T=100 is:
$$E[S_{100}]= \sum_{k=0}^{100} S_0*(1+10\%)^k*(1-10\%)^{(n-k)}{{100}\choose{k}}*p^k*p^{(n-k)}$$
which is equal to $S_0$
but the expected geometric return is:
$$E[Return]= \sum_{k=0}^{100} (1+10\%)^k*(1-10\%)^{(n-k)}{{100}\choose{k}}*p^k*p^{(n-k)} - 1$$
which when calculated is equal to 0.
Therefore mean price of the stock is equal to initial value and the mean geometric return is 0.
EDIT:
If we consider mean geometric return i.e. return of the form $\sqrt[n]{x_1*x_2*...*x_n} -1$ then we have:
$$E[Mean GeomReturn]= \sum_{k=0}^{100} [\sqrt[n]{(1+10\%)^k*(1-10\%)^{(n-k)}} -1]{{100}\choose{k}}*p^k*p^{(n-k)}$$
Then because mean geometric return is concave function we have:
$$E[Mean GeomReturn] \leq E[Return]$$
It is the same for logarithmic returns (concave) i.e.
$$E[MeanLogReturn]= \sum_{k=0}^{100} Log{(1+10\%)^k*(1-10\%)^{(n-k)}} ]{{100}\choose{k}}*p^k*p^{(n-k)}$$
we have:
$$E[MeanLogReturn] \leq E[Return]$$
In fact for the stock process you described, the expected log return is negative (after 100 periods equals to -50%). You should be really careful with that, as you mentioned expected stock price is equal to its initial value - and that's true, but both expected log return and mean geometric return are negative. This is due to the fact that logarithm flattens high returns and inflates negative returns. For example if the stock is worth 100, next period 199 (geom return of 99%) or 1 (geom return of -99%), then what are the log returns? +68% and -460% respectively. What is expected log return? -390%. It doesn't make much sense to say that on average you earn -390% log return. But if you would calculate average stock prices next period $(0.5*199 + 0.5*1)$ and calculate log return on average prices you would get a number that is easily interpretable.
$$log((0.5*199 + 0.5*1)/100)=0%$$
Therefore the order of calculations is crucial here. For ease of intepretation I would advise to firstly calculate expected Stock price $E[S_{100}]$ and then calculate logarithm return on that. You should get 0%.
EDIT 2:
There is nice interpretation of expected log returns (rather than log of expected return) (thanks @fesman) - if they expected log returns are negative that means in the long run there is higher probability of losing money than earning money.
In our example we have one period return of $(1+X)$ where $X$ can take value of +10% and -10% with equal probability. After N periods we have:
$$(1+x_1)*(1+x_2)*...*(1+x_n)$$
the log return is
$$ln((1+x_1)*...*(1+x_N))=ln(1+x_1)+ln(1+x_2)+...(1+x_N)$$
If $N$ is sufficiently large, then this sum can be approximated by normal distribution (due to CLT).
The mean of $ln(1+X)$ is $$\hat x=E[ln(1+X)]$$
the variance is:
$$\sigma^2=E[(ln(1+X))^2]-E[(ln(1+X))]^2$$
Therefore the sum is approximately normally distributed with mean $N*\hat x$ and variance $N*\sigma^2$. This can be represented as:
$$N\hat x + \sqrt N*\sigma*Z$$ where $Z$ is standard normal rv.
Let's calculate the probability that we actually lose money on that investment:
$$P(N\hat x + \sqrt N*\sigma*Z < 0)=P(Z<\frac{-N*\hat x}{{\sqrt N}*\sigma})=\Phi(\frac{-N*\hat x}{{\sqrt N}*\sigma})$$
where $\Phi$ is CDF of standard normal distribution.
After substituting the numbers into equation we get that after $N=100$ periods the probability of losing money is ~69% (high risk of losing money) and increases when N increases. Therefore risk averse investor wouldn't take this deal. In order to get probability of 50% it is easy to check that the expected log return has to be zero, not negative.
