$n$ couples are formed randomly from an initial group of $n$ men and $n$ women.
Denote men from $1$ to $n$ and let $X_i=1$ if the $i$-th man is associated to a women and $0$ otherwise. Let $X=\sum_{I=1}^{100}X_i$. The purpose of the exercise is to compute the expected value of each $X_i$ and then of $X$
We have that each $X_i$ is a Bernoulli distribution with parameter $$P(X_i=1)$$ but I feel that $P(X_i=1)$ depends on $i.$ I mean for $i=1$ we have $P(X_1=1)=\frac{n}{2n-1}$ but $P(X_2=1)$ will depend on the previous probability and I want to use $$P(X_2=1)=P(X_2=1\vert X_1=1)P(X_1=1)+P(X_2=1\vert X_1=0)P(X_1=0).$$
Is there a recurrence relation or am I missing something ?
Working out the equality in your question we find:$$P(X_2=1)=\frac{n-1}{2n-2}\frac{n}{2n-1}+\frac{n}{2n-2}\frac{n-1}{2n-1}=\frac{n}{2n-1}=P(X_1=1)$$This confirms (but does not prove) that $P(X_i=1)$ does not depend on $i$ but I can imagine that this on its own does not yet satisfy you.
In your thinking try to discard the order of associating in the following way:
There are $n$ men and each of them is labeled with a number and is associated with one of the other persons in the group.
Here resist the temptation to think of a timescale where the men are associated one by one.
Can you think of any reason at all why the man with label $13$ has more (or less) chance to be associated with a woman than the man with label $64$?....
edit:
valid proof of: $$P(X_i=1)=\frac{n}{2n-1}\text{ for arbitrary }i\in\{1,\dots,100\}$$
Concerning the association with man $i$ we discern $2n-1$ equiprobable candidates and exactly $n$ of them are women. So the probability that he will be associated with a woman equals $\frac{n}{2n-1}$.