Expected value of the shifted inverse of a binomial random variable, and application Here is an exercise given by a colleague to a student :

Let $X\hookrightarrow B(n,p)$ and $Y=\frac{1}{X+1}$. Find ${\rm E}(Y)$.

It is not very difficult to prove that the answer is
$${\rm E}(Y) = \frac{1-q^{n+1}}{p(n+1)}$$
where $q=1-p$. But the answer can also be written
$${\rm E}(Y) = \frac{1+q+q^2+\dots+q^n}{n+1}$$

First question: Is there any meaning to this form, which looks very much like a mean value of some sort? Or maybe another proof of this result which explains it in a more direct way?
Second question : Is there some context which could make this exercise more "concrete"?


Here is an application of your question to a setting that highly interests
me. In queueing there is the notion of utilization which is the long-run
fraction of time a server is busy serving demand. Consider a Markovian service
setting where demand arrives according to Poisson with $\lambda=1$
and there are $N+1$ servers with Exponential service time and mean rate $\mu=1$, where $N\sim\text{Bin}\left(n,p\right)$.
An application of this is the utilization of an Uber driver; the number
of Uber drivers on a given instance is uncertain as it cannot be mandated or enforced by
the firm. Given $k$ drivers choose to drive on the streets, their
utilization would be $\frac{\lambda}{\mu\cdot k}=\frac{1}{k}$. So,
in this setting if we assume the above model by the law of total probability the utilization of
an Uber driver would be $E\left[\frac{1}{1+N}\right]$. Given this,
I would be interested to know of an interpretation of the RHS? Any
ideas?
