Let's assume a betting game where each bet has a win probability of $W$. In case of a win, we gain $B$, in case of a loss, we loose $A$ of our invested capital $kC$, where $k$ is the fraction of our capital $C$ we invest for each bet. We repeat the bets $n$ times.
Approach 1
The expected value of our capital after one trial is:
$$ C_2 = C_1 + k C_1 W B - k C_1 (1-W) A $$
Rearranging,
$$ C_2 = C_1 (1 + k W B - k (1-W) A) $$
If we repeat the process $n$ times, the expected value of the final capital $C_n$ is:
$$ C_n = C_1 (1 + W B k - (1-W) A k)^n $$
Approach 2
But we could also reason that since we do the bets $n$ times, we would expect to have $W n$ gains and $(1-W) n$ losses. Each gain multiplies the capital by $1 + k B$, each loss multiplies the capital by $1-k A$, therefore the final capital is:
$$ C_n = C_1 (1+k B)^{W n} (1-k A)^{(1-W) n} $$
I'm confused, which formula is the correct one?
Note:
The article https://en.wikipedia.org/wiki/Kelly_criterion states that approach 2 is the correct one (under the section Proof), but I don't know why approach 1 is incorrect.
The first one is correct for a computation of expected wealth at the end of $n$ rounds.
Your final capital is given by $$ C_1(1+kB)^{n_w} (1-kA)^{n-n_w}$$ where $n_w$ is the number of wins, but it's not generally the case that $E(f(X)) = f(E(X)),$ so you can't just replace the $n_w$'s with their expected values to get the expected value of this expression.
If you want to compute the EV the right way starting from this expression, you need to use the fact that $n_w$ is binomially distributed and compute $$ \sum_{k=0}^n W^k(1-W)^{n-k}C_1 (1+kB)^k(1-kA)^{n-k} = C_1(W(1+kB)+(1-W)(1-kA))^n$$
Alternatively, you can do as in your first method and observe that your final capital is $C_1 Z_1 \cdot Z_2\cdot\ldots\cdot Z_n$ where $Z_i$'s are independent identical random variables with mean $E(Z)=W(1+kB)+(1-W)(1-kA),$ so the expected value is just $C_1 E(Z)^n,$ which is the same thing as above. (And the same thing you got for part 1.)
Since it seems your real question is about the Kelly criterion...
The important insight of the Kelly criterion is that maximizing your expected capital is not actually a great strategy for maximizing your performance, even if you have a decent edge and full certainty of the odds.
Recall from above that we can write the problem as $$ C_n = C_1 Z_1\cdot Z_2\cdot\ldots \cdot Z_n$$ where $Z_i$ are independent, identically distributed random variables. Having an edge corresponds to having $WB-(1-W)A>0.$ As you've computed, if you opt to bet a fraction $k,$ $E(Z) = 1 + k(WB-(1-W)A),$ which is clearly maximized for $k=1$.
SO that's great, right? Your expected capital at time $n$ is $C_n = C_1E(Z)^n,$ so if $E(Z)$ is bigger than $1,$ this geometrically grows to infinity, and if you set $k$ to where $E(Z)$ is highest, it geometrically grows at the highest possible rate.
Not really! We care about what actually is going to typically happen, not about the expected value. The expected value may almost never actually be exceeded, as long as the few times it does get exceeded are astronomically lucrative. That's what's going on here. ("Expected value" is a terribly-named concept. It has absolutely nothing to do with what you can 'expect' to happen.)
The absurdity of 'betting it all' every time should be readily apparent to anyone with any kind of common sense: even if you have great odds each time, you will certainly eventually lose everything. The fact that you still have favorable EV is just a reflection of the fact that at each time $n$ there is some chance you haven't gone broke yet, and in these worlds you will be quite rich, so it brings your average up to the point your expected wealth is still better than the wealth you started with.
The correct way to analyze the situation is to take logs: $$ \log C_n = \log C_1 + \sum_{i=1}^n \log(Z_i).$$ This converts products to sums and lets us exploit the law of large numbers. Now, we see the log of our capital changes by a random sum of independent variables $\log(Z_i).$
If $E(\log(Z))>0,$ then our log capital (and hence our capital) will eventually drift deep into positive territory with certainty. Whereas if it is negative, it will eventually drift into negative territory with certainty, or equivalently our wealth will drift toward zero with certainty.
So what you really want to do is make sure that $E(\log(Z))>0,$ not $E(Z)>1.$ And hence it makes sense to maximize the former and not the latter.
So in other words maximizing this expression: $$ W\log(1+kB) + (1-W)\log(1-kA)$$ over $k,$ not maximizing the expected wealth (and hence getting $k=1$) is what we want to do. This is the Kelly Criterion.
And note, it can be quite different.
Say we have $W=.4$ and $B=2$ and $A=1$ and $k=.8,$ then $E(Z)= 1.16,$ so your expected capital grows at a healthy rate of $1.16^n$ if you bet this much. But $E(\log(Z)) \approx -.58$, so if you bet a fraction $k=.8$ of your weath on this (favorable) game over and over, you will almost certainly go broke. As I mentioned above, these don't conflict: it's just that since there's exponential growth when you win, the very few worlds in which you do get very lucky for a long time, you win so much that it brings your expected value up to that $1.16^n$ figure (which will be quite large for $n$ large), even though in almost every world, you have lost almost all of your capital.