28. [14] Determine the value of
$$
\sum_{k=1}^{2011} \frac{k-1}{k!(2011-k)!} .
$$
Answer: $\frac{2009\left(2^{2010}\right)+1}{2011!}$ We note that
$$
\begin{aligned}
(2011!) \sum_{k=1}^{2011} \frac{k-1}{k!(2011-k)!} & =\sum_{k=1}^{2011} \frac{(2011!)(k-1)}{k!(2011-k)!} \\
& =\sum_{k=1}^{2011} \frac{k(2011)!}{k!(2011-k)!}-\sum_{k=1}^{2011} \frac{2011!}{k!(2011-k)!} \\
& =\sum_{k=1}^{2011} k\binom{2011}{k}-\sum_{k=1}^{2011}\binom{2011}{k} \\
& =(2011)\left(2^{2010}\right)-\left(2^{2011}-1\right)
\end{aligned}
$$

Thus, we get an answer of $\left(2009\left(2^{2010}\right)+1\right) /(2011!)$.
Note: To compute the last two sums, observe that
$$
\sum_{k=0}^{2011}\binom{2011}{k}=(1+1)^{2011}=2^{2011}
$$
by the Binomial Theorem, and that
$$
\sum_{k=0}^{2011} k\binom{2011}{k}=\frac{1}{2}\left(\sum_{k=0}^{2011} k\binom{2011}{k}+\sum_{k=0}^{2011}(2011-k)\binom{2011}{2011-k}\right)=2011\left(2^{2010}\right)
$$