$W$ white balls, $B$ black balls, adding $K$ of the resultant color each iteration. What is the probability of getting Black ball in nth iteration. $W$ white balls, $B$ black balls, adding $K$ of the resultant color each iteration
The problem is stated as follows. We have a box with $W$ white balls and $B$ black ones. Repeat N times: each iteration a ball is taken out (uniformly), and put back along with $K$ (constant) more balls of the same color.
I am able to calculate the probabilty but am unable to derive a formula.
Before the first addition: $$P\left(\text{Black}\right) = B/(W + B)$$
In the next iteration:
$$P(\text{black considering two possibilities in previous iteration}) = \frac{B+K}{W + B + K} + \frac{B}{W + B + K}$$
 A: I'm not exactly sure what you're asking for, so I'm going to solve for the expected value of the number of balls of each color after a number of iterations.

Let $B_n$ be the number of black balls after $n$ iterations of this addition process. Define $W_n$ similarly. $K$ is a fixed constant.
The $B$ and $W$ given are $B_0$ and $W_0$, respectively. 
We want to derive a recursive formula for the expected value of $B_{n+1}$ and $W_{n+1}$, given the expected values of $B_n$ and $W_n$. We will then attempt to derive a closed form expression from the recursion.

Let $P(A)$ denote the probability that the event $A$ occurs. We begin with the first iteration:
$$E(B_1) = B_0 \cdot P\left(\text{white ball is chosen}\right) + \left(B_0+K\right)\cdot P\left(\text{black ball is chosen}\right)$$
$$E(B_1) = B_0 \cdot \frac{W_0}{B_0 + W_0} + \left(B_0 + K\right) \cdot \frac{B_0}{B_0 + W_0}$$
$$E(B_1) = \frac{\left(B_0\right)\left(B_0 + W_0 + K\right)}{B_0 + W_0}$$
$$E(W_1) = W_0 \cdot P\left(\text{black ball is chosen}\right) + \left(B_0+K\right)\cdot P\left(\text{white ball is chosen}\right)$$
$$E(W_1) = W_0 \cdot \frac{B_0}{B_0 + W_0} + \left(W_0 + K\right) \cdot \frac{W_0}{B_0 + W_0}$$
$$E(W_1) = \frac{\left(W_0\right)\left(B_0 + W_0 + K\right)}{B_0 + W_0}$$
Note that $E(B_1) + E(W_1) = B_0 + W_0 + K$, as it should.
Now, assume that $E(B_n)$ and $E(W_n)$ are known. Because I'm lazy, I'll use $B_n$ and $W_n$ for their expected values.
$$E\left(B_{n+1}\right) = B_n \cdot P\left(\text{white ball is chosen}\right) + \left(B_n + K\right) \cdot P\left(\text{black ball is chosen}\right)$$
$$E\left(B_{n+1}\right) = B_n \cdot \frac{W_n}{B_n+W_n}+ \left(B_n + K\right) \cdot \frac{B_n}{B_n+W_n}$$
Note that $B_n + W_n = B_0 + W_0 + nK$. Substitute this in to get:
$$E\left(B_{n+1}\right) = B_n \cdot \frac{W_n}{B_0+W_0 + nK}+ \left(B_n + K\right) \cdot \frac{B_n}{B_0+W_0 + nK}$$
$$E\left(B_{n+1}\right) = \frac{\left(B_n\right)\left(B_n+W_n+ K\right)}{B_0+W_0 + nK}$$
$$E\left(B_{n+1}\right) = B_n \cdot \frac{B_0+W_0 + \left(n+1\right)K}{B_0+W_0 + nK}$$
Similarly, 
$$E\left(W_{n+1}\right) = W_n \cdot \frac{B_0+W_0 + \left(n+1\right)K}{B_0+W_0 + nK}$$
Note that $E(B_{n+1}) + E(W_{n+1}) = B_0 + W_0 + \left(n+1\right)K$, as should intuitively be true (after $n+1$ iterations, there should be $(n+1)K$ more balls than the starting amount). 

The last two centered formulas above are all the information needed to derive a closed form expression for each of the expected values.
EDIT: I think solving the recursion might be a bit tricky :(
