More Coin Tossing

You have the opportunity to play a game with a biased coin, where $P(H)=\frac{1}{2}+b$ and $P(T)=\frac{1}{2}-b$. In each round of the game, you toss the coin and win or lose one dollar. The game is played as follows. 

On the first toss: receive a dollar regardless of the outcome. 
On all other tosses: if you won (lost) a dollar on the prior toss, you win (lose) a dollar if the current toss comes out heads and lose (win) a dollar if it comes out tails.

What would you pay to play this game with a coin of bias $b$ for exactly $n$ rounds? Why?


using recursive relationships:

$P(W)=P$
$P(n)=P(n-1)*(\frac{1}{2}+b)+(1-P(n-1))*(\frac{1}{2}-b)$
$P(n)=\frac{1+(2*b)^n}{2}$

$E|W=(\frac{1}{2}+b)*1+(\frac{1}{2}-b)*-1=2*b$
$E|L=(\frac{1}{2}+b)*-1+(\frac{1}{2}+b)*1=-2*b$
$E(n)=E(n-1)+P(n-1)*E|W+(1-P(n-1))*E|L$
$E(n)=E(n-1)+P(n-1)*2*b-(1-P(n-1))*2*b$
$E(n)=\sum_{i=0}^{n}(2*b)^i$
$E(n)=\frac{(2*b)^{n+1}-1}{2b-1}$