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.
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 $n$, $$P(R_{n+1})=E(Z_n)=Z_0=R/(R+B).$$
$P(R_1)=\frac{r}{r+b}$ and $P(B_1)=\frac{b}{r+b}$

Applying theorem of total probability we have :

\begin{eqnarray*}
P(R_2)&amp;=&amp;P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)\\
&amp;=&amp; \frac{r+1}{r+b+1}\frac{r}{r+b}+\frac{r}{r+b+1}\frac{b}{r+b}\\
&amp;=&amp;\frac{r}{r+b}
\end{eqnarray*}

Now we prove for $P(R_3)$

Again apply theorem of total probability:

\begin{eqnarray*}
P(R_3)&amp;=&amp;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)&amp;=&amp;P(R_3\cap R_2|R_1)+P(R_3\cap B_2|R_1)\\
&amp;=&amp; 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)\\
&amp;=&amp; \frac{r+2}{r+b+2}\frac{r+1}{r+b+1}+\frac{r+1}{r+b+2}\frac{b}{r+b+1}\\
&amp;=&amp;\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})&amp;=&amp; P(R_{n-1}| R_1)P(R_1) + P(R_{n-1}| B_1)P(B_1)\\
&amp;=&amp; \frac{r+1}{r+b+1}\frac{r}{r+b}+\frac{r}{r+b+1}\frac{b}{r+b}\\
&amp;=&amp;\frac{r}{r+b}
\end{eqnarray*}

This is a classic example of Markov Chain