5. Calculate:
$$
\sum_{k=0}^{2013}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2014}^{k}}=
$$
$\qquad$
5. 0 .

Let $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}$.
And $\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n}-k-1}$, so $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}}$.
Perform a summation index transformation $l=2 n-k-1$, then
$$
\begin{array}{l}
a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \\
=\sum_{l=0}^{2 n-1}(-1)^{\prime} \frac{l+1}{\mathrm{C}_{2 n}^{l}}=-a_{n} . \\
\text { Hence } a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}=0 .
\end{array}
$$