How can I find the expected value due to the trials instead of simply expected value of trials of a geometric distribution? Suppose in a game, if I win in the $j$th round, I gain $+\$2^{j-1}$ and if I don't win in the $j$th, I lose $-\$2^{j-1}$. If I lose, I will keep playing until I win. Once I win, I leave the game. Otherwise, I continue to play until 30 rounds and leave the game even if I don't win. In other words, I will just stop at the $30$th round. Each round is independent and the probability of winning in each round is $\frac{9}{13}$.
I let $X$ be a random variable of my winnings. I want to find my expected winnings.
Since it works in a way that I would stop during my first win, I have a feeling that $X$ should be distributed over the Geometric Distribution. The game is either a win or a lose, so it is pretty much like a Bernoulli trial. But since I am not getting the expected number of rounds played, the Bernoulli trials cannot be just $0$ or $1$. So I thought I could modify it to become this way:
$$
{ X }_{ j }=\left\{\begin{matrix}
+2^{j-1} & if\;  win\\ 
-2^{j-1} & if\;  lose
\end{matrix}\right.
$$
Then, $E(X)=E(X_1+X_2+\cdots +X_{30})=E(X_1)+E(X_2)+\cdots +E(X_{30})$
However, because I thought this is a Geometric Distribution, the expected value is just $\frac{1}{p}=\frac{13}{9}$. But I don't think the expected winnings is $\frac{13}{9}$ and is wrong.
Is what I have done correct? 
So, instead of the usual finding of the expected number of trials of a standard geometric distribution, how can I find an expected number of another factor due to the trials (in this case, the expected number of winnings from the trials)?
Edit:
What I attempted to do was to make use of an indicator function to determine the expectation. But it doesn't seem successful. 
 A: The ideas in your post can be used to produce a nice solution, which will be given later. But the expected length of the game, which is almost but not quite $\frac{13}{9}$, because of the cutoff at $30$, has I think not much bearing on the problem. 
First Solution: Let's do some calculations. If we win on the first round, we win $1$ dollar and leave.  If we lose on the first and win on the second, we lost $1$ dollar but won $2$, for a net of $1$. If we lose on the first two rounds, and win on the third, we have lost $1+2$ dollars, and won $4$, for a net of $1$.  
Reasoning in the same way, we can see that we either leave with $1$ dollar, or lose a whole bunch of money, namely $1+2+4+\cdots+2^{29}$, which is $2^{30}-1$.  The reason that if we win, our net gain is $1$, is that if we win on the $i$-th round, we have lost $1+2+\cdots +2^{i-2}$ and won $2^{i-1}$. Since $1+2+\cdots +2^{i-2}=2^{i-1}-1$, our net gain is $1$. 
The probability we lose an enormous amount of money is $\left(\frac{4}{13}\right)^{30}$ ($30$ losses in a row).  Now the expectation is easy to find.  It is
$$(1)\left(1-\left(\frac{4}{13}\right)^{30}\right)- (2^{30}-1)\left(\frac{4}{13}\right)^{30}.$$
There is a bit of cancellation. The expectation simplifies to 
$$1-2^{30}\left(\frac{4}{13}\right)^{30}.$$
Second Solution: We give a much more attractive solution based on the idea of your post. For $j=1$ to $30$, let $X_j$ be the amount "won" on the $j$-th trial  Then the total amount $X$ won is $X_1+\cdots+X_{30}$. Thus, by the linearity of expectation, 
$$E(X)=\sum_{j=1}^{30} E(X_j).$$
The $X_j$ are not independent, but that is irrelevant.
Let $p=4/13$. We win $2^{j-1}$ at stage $j$ with probability $p^{j-1}(1-p)$, and lose $2^{j-1}$ with probability $p^{j-1}p$. So 
$$E(X_j)=2^{j-1}p^{j-1}(1-2p).$$
Now sum, from $j=1$ to $j=30$. The sum is
$$(1-2p)\frac{1-(2p)^{30}}{1-2p}, \quad\text{that is,}\quad 1-\left(\frac{8}{13}\right)^{30}.$$
