I need to get a closed form for this series $$\sum_{x=0}^{\infty} x {z \choose x} \lambda ^ x \mu^{z-x}$$
I know that that $\sum_{x=0}^{\infty} {z \choose x} \lambda ^ x \mu^{z-x} = (\lambda + \mu)^z$ (formally) and I feel that I am supposed to proceed from here by differentiation, but I do not know how.
Let $$\sum_{x=0}^{\infty} {z \choose x} \lambda ^ x \mu^{z-x}=(\lambda + \mu)^z$$ by differentiation about $\lambda$: $$\sum_{x=0}^{\infty} x {z \choose x} \lambda ^ {x-1} \mu^{z-x}=z(\lambda + \mu)^{z-1}$$ multiple two sides with $\lambda$ $$\sum_{x=0}^{\infty} x {z \choose x} \lambda ^ x \mu^{z-x}=z\lambda(\lambda + \mu)^{z-1}$$
Here is another variation of the theme without differentiation.
We obtain \begin{align*} \sum_{x=1}^\infty x\binom{z}{x}\lambda^x\mu^{z-x} &=\mu^zz\sum_{x=1}^\infty\binom{z-1}{x-1}\left(\frac{\lambda}{\mu}\right)^x\tag{1}\\ &=\mu^{z}z\sum_{x=0}^\infty\binom{z-1}{x}\left(\frac{\lambda}{\mu}\right)^{x+1}\tag{2}\\ &=\lambda\mu^{z-1}z\left(1+\frac{\lambda}{\mu}\right)^{z-1}\tag{3}\\ &=\lambda z(\lambda+\mu)^{z-1} \end{align*}
Comment:
In (1) we start the left hand series with $x=1$ due to the factor $x$ and at the right hand side we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
In (2) we shift the index $x$ by one to start from $x=0$.
In (3) we apply the binomial series expansion.
To get something resembling the binomial theorem you have to eliminate that $x$ factor in the terms. Well, when we express the binomial coefficient as factorials, we have a factor of $x!$ in the denominator, so we can cancel except when $x=0$, but fortuitously ...
$$\begin{align}\mu^z\sum\limits_{x=0}^z\dfrac{z!x(\frac{\lambda}{\mu})^x}{x!(z-x)!}~&=~\mu^z \sum_{x=1}^z\dfrac{z!x(\frac{\lambda}{\mu})^{x}}{x!(z-x)!} &&\text{term for }x=0\text{ is }0\\[1ex] &=~ \mu^z \sum_{x=1}^z\dfrac{z!(\frac{\lambda}{\mu})^{x}}{(x-1)!(z-x)!}&&\text{cancelling common factor}\end{align}$$
And you can proceed from there until you have something familiar.
(Hint: use a change of variables.)
