5. Given the sequence $\left\{(-1)^{n} \frac{3 n+2}{n(n+1) \cdot 2^{n+1}}\right\}$ with the sum of the first $n$ terms being $S_{n}$. Then $S_{18}=$ $\qquad$
5. $\frac{1}{2^{19} \times 19}-\frac{1}{2}$.

Notice,
$$
\begin{array}{l}
\frac{3 n+2}{n(n+1) \cdot 2^{n+1}}=\frac{n+2(n+1)}{n(n+1) \cdot 2^{n+1}} \\
=\frac{1}{(n+1) \cdot 2^{n+1}}+\frac{1}{n \cdot 2^{n}} .
\end{array}
$$

Let $b^{n}=\frac{1}{n \cdot 2^{n}}$. Then
$$
\begin{array}{l}
S_{18}=\sum_{n=1}^{18}(-1)^{n}\left(b_{n}+b_{n+1}\right) \\
=\sum_{n=1}^{18}(-1)^{n} b_{n}-\sum_{n=1}^{18}(-1)^{n+1} b_{n+1} \\
=\sum_{n=1}^{18}(-1)^{n} b_{n}-\sum_{n=2}^{19}(-1)^{n} b_{n} \\
=-b_{1}+b_{19}=\frac{1}{2^{19} \times 19}-\frac{1}{2} .
\end{array}
$$