The situation is easy to understand in the special case where $b=r$, since the whole situation is then symmetric between the two colors. Parts of the situation are asymmetric, for example the situation after you draw a blue ball on the first trial and add $c$ more blue balls. Obviously, blue is then favored. But there is an equally likely situation that equally favors red, so the overall situation is still symmetric.
Let me now try to explain a similar intuition in the non-symmetric case where $b\neq r$. Imagine that the $b+r$ balls initially in the urn are not only colored with two colors but also numbered, with $b+r$ distinct numbers, say numbers $1$ to $b$ for the blue balls and $b+1$ to $b+r$ for the red balls. Also, imagine that whenever a ball is drawn in a trial and is put back into the urn, the additional $c$ balls have not only the same color but also the same number as the ball that was drawn. (So, although the initial situation had different numbers on all the balls, later situations will have many balls with the same number.) Notice that, throughout the process, all balls numbered $1$ to $b$ will be blue and all balls numbered $b+1$ to $b+r$ will be red.
Now temporarily forget about colors and concentrate on numbers. The initial situation is symmetric between all $b+r$ numbers. Just as in the first paragraph above, the overall situation remains symmetric throughout the process. So all $b+r$ numbers have the same chance of being drawn in the $n$-th trial.
As before, there are asymmetries in conditional probabilities, for example the second ball drawn is more likely to have the same number as the first than to have any other particular number. But, as before, the asymmetries are equally likely to favor any of the numbers. So the overall probabilities are equal.
But (taking the colors into consideration again) this means that the probability of drawing a blue ball on the $n$th trial is $b/(b+r)$ because $b$ of the $b+r$ equally likely numbers correspond to blue balls.
Firstly, $$P(B_m \cap B_n) = P(B_n \mid B_m)\,P(B_m).$$
Now use induction to establish the value of each factor on the RHS.
Our first claim is that for all $m \geq 1$: $$P(B_m) = \dfrac{b}{b+r}$$
Initial case, $m=1$: $$P(B_1) = \dfrac{b}{b+r}$$ is obviously true since there are $b$ chances to choose a black ball out of $b+r$ balls.
Now assume the claim is true for $m=k$ for some $k \geq 1$. Then, conditioning on $B_1$,
\begin{eqnarray*}
P(B_{k+1}) &=& P(B_{k+1} \mid B_1)\,P(B_1) \,+\, P(B_{k+1} \mid B_1^c)\,P(B_1^c).
\end{eqnarray*}
Now,
\begin{eqnarray*}
P(B_{k+1} \mid B_1) &=& P(B_k) \qquad\mbox{starting with $b+c$ black and $r$ red balls} \\
&=& \dfrac{b+c}{b+r+c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
Also,
\begin{eqnarray*}
P(B_{k+1} \mid B_1^c) &=& P(B_k) \qquad\mbox{starting with $b$ black and $r+c$ red balls} \\
&=& \dfrac{b}{b+r+c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
Therefore, we have,
\begin{eqnarray*}
P(B_{k+1}) &=& \dfrac{b+c}{b+r+c} \dfrac{b}{b+r} + \dfrac{b}{b+r+c} \dfrac{r}{b+r} \\
&=& \dfrac{b}{b+r}.
\end{eqnarray*}
This proves the case for $m=k+1$ and the inductive proof of the first claim is done.
Our second claim is that for all $n \gt 1$ and $m: 1 \leq m < n$: $$P(B_n \mid B_m) = \dfrac{b+c}{b+r+c}.$$
We use induction on $m$. Initial case is for any $n \gt 1$ and with $m=1$: In proving the first claim we have already shown that $$P(B_n \mid B_1) = \dfrac{b+c}{b+r+c}.$$
Now assume that our second claim is true for any $n \gt 1$ and some $k: 1 \leq k < n$. Conditioning on $B_1$,
$$P(B_n \mid B_{k+1}) = P(B_n \mid B_1 \cap B_{k+1})\,P(B_1 \mid B_{k+1}) \,+\, P(B_n \mid B_1^c \cap B_{k+1})\,P(B_1^c \mid B_{k+1}).$$
Evaluating the four probabilities on the RHS,
\begin{eqnarray*}
P(B_n \mid B_1 \cap B_{k+1}) &=& P(B_{n-1} \mid B_k) \qquad\mbox{starting with $b+c$ black and $r$ red balls} \\
&=& \dfrac{b+2c}{b+r+2c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
$\\$
\begin{eqnarray*}
P(B_n \mid B_1^c \cap B_{k+1}) &=& P(B_{n-1} \mid B_k) \qquad\mbox{starting with $b$ black and $r+c$ red balls} \\
&=& \dfrac{b+c}{b+r+2c} \qquad\mbox{by inductive assumption.}
\end{eqnarray*}
$\\$
\begin{eqnarray*}
P(B_1 \mid B_{k+1}) &=& \dfrac{P(B_{k+1} \mid B_1) P(B_1)}{P(B_{k+1})} \\
&& \\
&=& \dfrac{b+c}{b+r+c} \dfrac{b}{b+r} \bigg/ \dfrac{b}{b+r} \\
&& \qquad\mbox{first factor is by our initial case, the second and third terms by our first claim} \\
&& \\
&=& \dfrac{b+c}{b+r+c}.
\end{eqnarray*}
$\\$
\begin{eqnarray*}
P(B_1^c \mid B_{k+1}) &=& \dfrac{P(B_{k+1} \mid B_1^c) P(B_1^c)}{P(B_{k+1})} \\
&& \\
&=& \dfrac{b}{b+r+c} \dfrac{r}{b+r} \bigg/ \dfrac{b}{b+r} \\
&& \qquad\mbox{by similar reasoning as previously} \\
&& \\
&=& \dfrac{r}{b+r+c}.
&&
\end{eqnarray*}
$\\$ $\\$
Putting these together, we have,
\begin{eqnarray*}
P(B_n \mid B_{k+1}) &=& \dfrac{b+2c}{b+r+2c} \dfrac{b+c}{b+r+c} + \dfrac{b+c}{b+r+2c} \dfrac{r}{b+r+c} \\
&& \\
&=& \dfrac{b+c}{b+r+c}.
\end{eqnarray*}
Thus the claim is true for $m=k+1$ and our induction is complete for our second claim.
Therefore, as required, we have $$P(B_m \cap B_n) = P(B_n \mid B_m)P(B_m) = \dfrac{b(b+c)}{(b+r)(b+r+c)}.$$
Best Answer
Notice that if the formula on $P(B_m \cap B_n)$, we would completely determine $P(B_m \cap R_n)$ since $P(B_m \cap B_n) + P(B_m \cap R_n)=P(B_m) $.
\begin{align} &P(B_m \cap B_{m+k+1})\\&=P(B_{m+k+1} | R_{m+k}, B_m)P(R_{m+k}|B_m)P(B_m) \\&+P(B_{m+k+1} | B_{m+k}, B_m)P(B_{m+k}|B_m)P(B_m) \end{align}
Let's compute the individual terms.
$$P(B_m) = \frac{b}{b+r}.$$
Also, by induction hypothesis,
$$P(R_{m+k}|B_m)=\frac{P(R_{m+k} \cap B_m)}{P(B_m)} = \frac{\frac{br}{(b+r)(b+r+c)}}{\frac{b}{b+r}}=\frac{r}{b+r+c}.$$
We drew a black ball at round $m$, also, we know that at round $m+k$, there are a total of $b+r+(m+k-1)c$ balls, of which $\frac{r}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are red and hence $\frac{b+c}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are blue.
At round $m+k+1$, there are a total of $b+r+(m+k)c$ balls, the number of blue ball didn't increase if we previously draw a red ball. Hence, we have
$$P(B_{m+k+1}|R_{m+k}, B_m )=\frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]}{b+r+(m+k)c}$$
We have $$P(B_{m+k}|B_m)=\frac{b+c}{b+r+c}$$
We drew a black ball at round $m$, also, we know that at round $m+k$, there are a total of $b+r+(m+k-1)c$ balls, of which $\frac{r}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are red and hence $\frac{b+c}{b+r+c}\cdot [b+r+(m+k-1)c]$ of them are blue.
At round $m+k+1$, there are a total of $b+r+(m+k)c$ balls, the number of blue ball increased by $c$ if we previously draw a red ball. Hence, we have
$$P(B_{m+k+1}|B_{m+k}, B_m )=\frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]+c}{b+r+(m+k)c}$$
Now, we have each individual term and we just have to substitute them back to compute $P(B_m \cap B_{m+k+1})$.
\begin{align} &P(B_m \cap B_{m+k+1})\\ &=\frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]}{b+r+(m+k)c} \cdot \frac{r}{b+r+c} \cdot \frac{b}{b+r}\\ &+ \frac{\left( \frac{b+c}{b+r+c}\right) [b+r+(m+k-1)c]+c}{b+r+(m+k)c} \cdot \frac{b+c}{b+r+c} \cdot \frac{b}{b+r}\\ &= \left(\frac{b(b+c)}{(b+r)(b+r+c)^2(b+r+(m+k)c)} \right)\cdot\\& \left((b+r+(m+k-1)c )r+(b+r+(m+k-1)c )(b+c) + c(b+r+c)\right) \\ &= \left(\frac{b(b+c)}{(b+r)(b+r+c)(b+r+(m+k)c)} \right)\cdot \left(b+r+(m+k-1)c + c\right) \\ &= \frac{b(b+c)}{(b+r)(b+r+c)} \end{align}