binomial series simplification

Question

I am stuck with a binomial series that i am trying to simplify $$\frac{1}{S(1-z)^{S-1}}\sum\limits_{m=0}^{S-1} x^m(1-x-z)^{S-1-m}{S-1\choose m}m$$

The answer apparently is $$\frac{x}{1-z}(1-\frac{1}{S})$$ I am unable to proceed. The original sum can be written as $$\sum\limits_{m=0}^{S-1} x^{m-1}(1-x-z)^{S-m}{S-1\choose m-1}\frac{x}{1-x-z}(\frac{1}{S-1-m})$$

I am not sure if this really helps, since i cant seem to go forward from there

Answer 1

You can use

• $m\binom nm = n\binom{n-1}{m-1}$ for $1\leq m\leq n$ and using this you can quickly show that
• $\sum_{m=0}^{n}m\binom nma^mb^{n-m} = n\sum_{m=0}^{n-1}\binom {n-1}{m-1}a^{n+1}b^{n-1-m} = na(a+b)^{n-1}$

Now, setting $n=S-1, a=x, b= 1-x-z$ you get

\begin{eqnarray} \frac{1}{S(1-z)^{S-1}}\sum\limits_{m=0}^{S-1} x^m(1-x-z)^{S-1-m}{S-1\choose m}m & = & \frac{1}{S(1-z)^{S-1}}\cdot(S-1)x(1-x-z+x)^{S-2} \\ & = & \frac{(S-1)x(1-z)^{S-2}}{S(1-z)^{S-1}}\\ & = & \frac{(S-1)x}{S(1-z)} \\ & = & \frac{x}{1-z}\left(1-\frac 1S\right) \\ \end{eqnarray}