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.
 A: The key is to condition by the composition of the urn at time $n$, say $X_n$ red balls and $Y_n$ blue balls, since $$P(R_{n+1}\mid X_n,Y_n)=Z_n,\qquad Z_n=X_n/(X_n+Y_n).$$ Obviously, $X_0=R$, $Y_0=B$, and, for every $n$, $$X_n+Y_n=R+B+nc,$$ which is deterministic. Conditionally on $(X_n,Y_n)$, one adds $c$ red balls with probability $Z_n$ and zero otherwise, hence $$E(X_{n+1}\mid X_n,Y_n)=X_n+cZ_n=(X_{n+1}+Y_{n+1})Z_n,$$ which implies $$E(Z_{n+1}\mid X_n,Y_n)=Z_n.$$ In particular, for every $n$, $$P(R_{n+1})=E(Z_n)=Z_0=R/(R+B).$$
A: 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.
A: ( After reading the comments bellow and consulting with a teacher I realized that this hint, as well as the link posted bellow,  are in essence incorrect if one does not intend to solve the problem using random variables. I  will not delete this answer so it can serve for future reference, but the OP should untag this answer since it is not correct)
Hint: Suppose that just before the n-th extraction there are $ r_n $ red balls and $b_n$ blue balls. Then $ P(R_n) = \dfrac{r_n}{r_n + b_n} \ $ and  $ \ P(R_{n + 1} | R_n) = \dfrac{r_n + c}{r_n + b_n + c} \ $. Similarly you can write down the other probabilities in your sum in terms of $r_n $ and $b_n$. Now try to factor out $P(R_n)$ and use your inductive hypothesis.
If you are still stuck after trying to apply the hint I posted above, this link might be helpful:     http://everything2.com/title/Polya+urn+scheme
A: $P(R_1)=\frac{r}{r+b}$ and $P(B_1)=\frac{b}{r+b}$
Applying theorem of total probability we have :
\begin{eqnarray*}
P(R_2)&=&P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)\\
&=& \frac{r+1}{r+b+1}\frac{r}{r+b}+\frac{r}{r+b+1}\frac{b}{r+b}\\
&=&\frac{r}{r+b}
\end{eqnarray*}
Now we prove for $P(R_3)$
Again apply theorem of total probability:
\begin{eqnarray*}
P(R_3)&=&P(R_3|R_1)P(R_1)+P(R_3|B_1)P(B_1)
\end{eqnarray*}
Now question is what is $P(R_3|R_1)$ and $P(R_3|B_1)$? We will show $P(R_3|R_1)=P(R_2|R_1)=\frac{r+1}{r+b+1}$. How? -- It is as follows:
Apply theorem of total probability on conditional probability.
\begin{eqnarray*}
P(R_3|R_1)&=&P(R_3\cap R_2|R_1)+P(R_3\cap B_2|R_1)\\
&=& P(R_3|R_2\cap R_1)P(R_2|R_1) + P(R_3|B_2 \cap R_1)P(B_2|R_1)\\
&=& \frac{r+2}{r+b+2}\frac{r+1}{r+b+1}+\frac{r+1}{r+b+2}\frac{b}{r+b+1}\\
&=&\frac{r+1}{r+b+1}
\end{eqnarray*}
Same way one can show $P(R_3|B_1)=\frac{r}{r+b+1}$.
Now under the induction hypothesis we have : 
$P(R_{n-1}|R_1)=\frac{r+1}{r+b+1}$ and $P(R_{n-1}|B_1)=\frac{r}{r+b+1}$
Therefore,
\begin{eqnarray*}
P(R_{n})&=& P(R_{n-1}| R_1)P(R_1) + P(R_{n-1}| B_1)P(B_1)\\
&=& \frac{r+1}{r+b+1}\frac{r}{r+b}+\frac{r}{r+b+1}\frac{b}{r+b}\\
&=&\frac{r}{r+b}
\end{eqnarray*}
This is a classic example of Markov Chain
