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.

  • 1
    $\begingroup$ Note: that is $$\mu^z \sum_{x=0}^{\color{blue}z}\binom z x(\frac\lambda\mu)^x~=~ \mu^z(1-\frac\lambda\mu)^z$$ $\endgroup$ – Graham Kemp Feb 4 '17 at 4:45

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*}


  • 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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.