Probability problem involving geometric distribution and expected value We're given the following problem: 
"An experiment is conducted until it results in success: the first step has probability $\frac{1}{2}$ to be successful, the second step (only conducted if the first step had no success) has probability $\frac{1}{3}$ to be successful, the third step (only conducted if the first two steps had no success) has probability $\frac{1}{4}$ to be successful : if none of the steps were successful, we repeat the experiment until success is achieved. Assuming the first step has cost $2$, and the second step as well as the third step have cost $1$, what is the overall cost until success ? "
Here's my approach. Let $X$ be the random variable the counts the number of steps required to achieve success. Thus giving me: $$P(X=1) = \frac{1}{2} \\ P(X=2)=\frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \\ P(X=3) = \frac{1}{2} \cdot \frac{5}{6} \cdot \frac{1}{4} = \frac{5}{48}  $$
Now, to compute the expected cost, I did the following: $$E(X) = 2 \cdot 1 \cdot P(X=1) + 1 \cdot 2 \cdot P(X=2) + 1 \cdot 3 \ P(X=3)$$ (where the first coeffeicient of each term of the sum is the cost of each step). This gave me $\frac{79}{48}$ , which is far from the correct answer. 
So, I do not understand what I did wrong (I guess there must be a problem with my reasoning), and I do not know how to find the correct result (which is $\frac{34}{9}$).   
 A: Let $\mu_{0}$ denote the expectation of the cost of the steps yet
to be done if no steps have been made.
Let $\mu_{1}$ denote the expectation of the cost of the steps yet
to be done if step 1 has been made without success.
Let $\mu_{2}$ denote the expectation of the cost of the steps yet
to be done if step 1 and step 2 have been made without success.
Then we have the following equalities.
$\mu_{0}=\frac{1}{2}2+\frac{1}{2}\left[2+\mu_{1}\right]=2+\frac{1}{2}\mu_{1}$
$\mu_{1}=\frac{1}{3}1+\frac{2}{3}\left[1+\mu_{2}\right]=1+\frac{2}{3}\mu_{2}$
$\mu_{2}=\frac{1}{4}1+\frac{3}{4}\left[1+\mu_{0}\right]=1+\frac{3}{4}\mu_{0}$
Then:$$\mu_0=2+\frac12\left[1+\frac23\left[1+\frac34\mu_0\right]\right]$$
The solution of this equality is: $$\mu_0=\frac{34}9$$
A: From my reading of the question (though I can see where you may have interpreted differently), your random variable $X$ can take on values greater than $3$. If all 3 first steps fail, we precede to a 4th step which is a repetition of the first step. Thus, e.g., if $X=4$, the cost will be $2+1+1+2=6$. You must take into account that $X$ can be any natural number.
To fix this, let $Y$ be the cost, and take your equation 
$$E(Y) = 2 \cdot 1 \cdot P(X=1) + 1 \cdot 2 \cdot P(X=2) + 1 \cdot 3 \ P(X=3)$$
and modify it to 
$$E(Y) = 2 \cdot P(X=1) + 3 \cdot P(X=2) + 4 \cdot \ P(X=3) + q(4+E(Y))$$
where $q = \frac12\cdot\frac23\cdot\frac34$ is the probability of failure on all three steps, and the $(4+E(Y))$ comes from the fact that if we fail the three steps, we are back where we started, plus we have already spent a cost of $4$. (I have also modified the coefficients as per lulu's comment.)
