Similar question was already asked (e.g. this page), but I have tried another approach (and failed). So I wanted to ask if my approach is sensible or completely wrong.
Here is an exercise from Introduction to Probability (Blitzstein, Hwang):
A couple decides to keep having children until they have at least one boy and at least one girl, and then stop. Assume they never have twins, that the "trials" are independent with probability 1/2 of a boy, and that they are fertile enough to keep producing children indefinitely. What is the expected number of children?
This solution has answer and it is E(X)=3. The explanation could be found in there.
But I was thinking in a different manner, and here is my approach. I've thought that maybe we can approach this problem like a two First Success problems joined with logical OR, i.e. the probability to have k children in total is equal to:
$$P(X=k)=q^{k-1}p + p^{k-1}q=2p^k$$
So they have k-1 children of one gender and the last one of another (or vice versa). Then I've tried to calculate mean as follows (where p=0.5):
$$E[X]=\sum_{k=2}^{\infty}{k\times2p^k}=2$$
Therefore my question is, could someone tell me if I have failed somewhere in calculations/formulas or the whole approach (like concatenation of two first success problems) is wrong?
Note that $q^{k-1}p + p^{k-1}q = 2p^k$ only works for $p = \frac{1}{2}$.
Anyway, your idea is good, the problem is with your final calculation:
$$E[X] = \sum_{k=2}^\infty \frac{2k}{2^k} = 2 \sum_{k=1}^\infty{ \frac{k}{2^k}} - 1 = 4 - 1 = 3,$$ in particular, you seem to have forgotten to multiply by $2$ and the fact that the sum starts from $k=2$.