Vtyjkyuk

If I have a equipment which can detect the arrival of particles with probability $p$, and the number of particles generated in a second is governed by a Poisson distribution.$$P_r[N=n]=fraclambda^n e^-lambdan!$$
I can calculate the probability distribution of number of detected particles.$$P_d[N=n]=sum_k=0^n fraclambda^n e^-lambda(n+k)!p^n(1-p)^k$$
Does this series $$sum_k=0^n fraclambda^n e^-lambda(n+k)!p^n(1-p)^k $$
have closed form formula?

There are three errors in your sum. The sum should extend to infinity; the exponent of $lambda$ should be $n+k$ instead of $n$; and you're missing the binomial coefficient $binomn+kn$ that counts the number of ways of selecting $n$ detected particles from $n+k$ arriving particles. If you correct these errors, the result is

begineqnarray*
P_d[N=n]
&=&
sum_k=0^inftyfraclambda^n+kmathrm e^-lambda(n+k)!binomn+knp^n(1-p)^k
\
&=&
fraclambda^nmathrm e^-lambdap^nn!sum_k=0^inftyfraclambda^k(1-p)^kk!
\
&=&
fraclambda^nmathrm e^-lambdap^nn!mathrm e^lambda(1-p)
\
&=&
frac(lambda p)^nmathrm e^-lambda pn!;.
endeqnarray*

This is a Poisson distribution with parameter $lambda p$. This result could be obtained more directly by noting that a Poisson process with rate $lambda$ from which a fraction $p$ of events are selected is again a Poisson process, with rate $lambda p$.