Average number of trials until drawing $k$ red balls out of a box with $m$ blue and $n$ red balls 
A box has $m$ blue balls and $n$ red balls. You are randomly drawing a ball from the box one by one until drawing $k$ red balls ($k < n$)? What would be the average number of trials needed?

To me, the solution seems to be 
$$\sum_i i * \frac{\mbox{the number of cases where k-th red ball is picked at i-th trial}}{\mbox{the number of cases where k-th red ball is the last ball picked}}$$
the denometer seems to be 
$$\sum_{r=k-1}^{m-1} \binom{r}{k-1} = \binom{m}{k-1} $$
But, I have a difficulty of deriving a closed form formula for the numerator which seems to be $\sum_{i=k}^{m-1} i   \binom{i-1}{k-1}$.
I would appreciate if somebody helps me on that.

For your sum, we have
$$\sum_{i=k}^{m-1} i   \binom{i-1}{k-1}
       = \sum_{i=k}^{m-1} \frac{i!}{(k-1)!(i-k)!} = k\sum_{i=k}^{m-1} \frac{i!}{k!(i-k)!}
       = k \sum_{i=k}^{m-1} \binom{i}{k}.$$
By the hockey stick formula, this simplifies further:
$$k \sum_{i=k}^{m-1} \binom{i}{k}
      = k\sum_{i=k}^{m-1}\binom{i}{i-k}
      = k\sum_{i=0}^{m-1-k}\binom{i+k}{i} = k\binom{m}{m-k-1}
      = k\binom{m}{k+1}.$$
