Pólya's urn scheme, proof using conditional probability and induction Problem 
An urn contains $B$ blue balls and $R$ red balls. Suppose that one extracts successively $n$ balls at random such that when a ball is chosen, it is returned to the urn again along with $c$ extra balls of the same color. For each $n \in \mathbb N$, we define $R_n=\{\text{the n-th ball extracted is red}\}$, and $B_n=\{\text{the n-th ball extracted is blue}\}.$
Prove that $P(R_n)=\dfrac{R}{R+B}$.
I thought of trying to condition the event $R_n$ to another event in order to use induction. For example, if $n=2$, I can express $$P(R_2)=P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)$$$$=\dfrac{R+c}{R+B+c}\dfrac{R}{R+B}+\dfrac{R}{R+B+c}\dfrac{B}{R+B}$$$$=\dfrac{R}{R+B}.$$
Now, suppose the formula is true for $n$, I want to show it is true for $n+1$.
So, $P(R_{n+1}=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)P(B_n)$$$=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)(1-P(R_n)$$$$=P(R_{n+1}|R_n)\dfrac{R}{R+B}+P(R_{n+1}|B_n)(1-\dfrac{R}{R+B}).$$
I am having some difficulty trying to calculate $P(R_{n+1}|R_n)$ and $P(R_{n+1}|B_n)$. I would appreciate if someone could complete my answer or suggest me how can I finish the proof if what I've done up to now is correct.

I would actually still advocate the approach suggested here with a small change in the way it is presented:
$P(R_1)=\frac{R}{R+B}$, now we need to prove that $P(R_n)=P(R_{n+1})$.
$P(R_{n+1})=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)(1-P(R_n))$
$X_n$, the number of red balls in the urn at step $n$, is $P(R_n)T_n$, where $T_n$ is the total number of balls on step $n$ which is deterministic.
$P(R_{n+1}|R_n)=\frac{T_nP(R_n)+c}{T_n+c}$
$P(R_{n+1}|B_n)=\frac{T_nP(R_n)}{T_n+c}$
$P(R_{n+1})=\frac{T_nP(R_n)+c}{T_n+c}P(R_n)+\frac{T_nP(R_n)}{T_n+c}(1-P(R_n))=P(R_n)$.
The approach does not use mathematical expectations, it can be considered as an advantage because this problem is often given to students before the study mathematical expectations.
