Show that
$$
W[t]=1-\sum_{n=0}^{K-1} q_n\left(e^{-\mu t} \sum_{k=0}^n \frac{(\mu t)^k}{k !}\right)
$$
for the $\mathrm{M} / \mathrm{M} / 1 / \mathrm{K}$ queueing system, where
$$
q_n=\frac{p_n}{1-p_K}
$$
[Hint: Write
$$
\begin{aligned}
W[t] & =\sum_{n=0}^{K-1}\left\{\int_0^t \frac{\mu(\mu x)^n e^{-\mu x}}{n !} d x\right\} q_n \\
& =\sum_{n=0}^{K-1}\left\{1-\int_t^{\infty} \frac{\mu(\mu x)^n e^{-\mu x}}{n !} d x\right\} q_n \\
& =1-\sum_{n=0}^{K-1} q_n \int_t^{\infty} \frac{\mu(\mu x)^n e^{-\mu x}}{n !} d x .
\end{aligned}
$$
Then make the change of variable $y=x-t$ in each of the integrals. By recognizing the integral form of the gamma function
$$
\Gamma(t)=\int_0^{\infty} x^{t-1} e^{-x} d x, \quad t>0,
$$
and using the property of the gamma function expressed as
$$
\Gamma(n+1)=n ! \quad n=0,1, \ldots,
$$
deduce that
$$
\int_t^{\infty} \frac{\mu(\mu x)^n e^{-\mu x}}{n !} d x=e^{-\mu t} \sum_{k=0}^n \frac{(\mu t)^k}{k !},
$$
for $n=0,1, \ldots, K-1$.]