The Problem A bag contains $b$ black balls and $w$ white balls. Balls are drawn at random from the bag until the last white ball is drawn. What is the expected number of balls drawn?
My Partial Solution Suppose the balls are lined up in a line and drawn from left to right. The number of balls to the left of the rightmost white ball (say $N$) ranges from $w-1$ to $w+b-1$, the number of balls to right is given by $w+b-N-1$. So, we can calculate the probability of each value of $N$ using the hypergeometric distribution, and we have
$\text{Expected number}=\displaystyle \sum_{k=w-1}^{b+w-1}\frac{\binom{k}{w-1}\binom{w+b-k-1}{0}}{\binom{w+b-1}{w-1}}\cdot\left(k+1\right)$
which requires computation of
$\displaystyle \sum_{k=w-1}^{b+w-1}\binom{k}{w-1} \cdot \left(k+1\right)$
which I am unable to do.
Is my method even correct, and is there any easy way to do the problem or compute the sum?