5. Given a tetrahedron $ABCD$ divided by $n+1$ (where $n$ is odd) planes $\pi_{0}$, $\pi_{1}, \cdots, \pi_{n}$ parallel to $AB$ and $CD$ and equally spaced, into $n$ parts, where $AB \subset \pi_{0}$ and $CD \subset \pi_{n}$. Let the volumes of these $n$ parts be $V_{1}, V_{2}, \cdots, V_{n}$. Then $\left(\sum_{i \text { odd }} V_{i}\right):\left(\sum_{i \in \mathbb{N}_{0}} V_{i}\right)=$ $\qquad$
5. $\left(n^{3}+1\right):\left(n^{3}-1\right)$.

As shown in Figure 1, plane $E F G H$ is parallel to edges $A B$ and $C D$. A plane $E H K$ is drawn through $E H$ parallel to plane $B C D$.

Let $A E=t A C(t \in(0,1))$. Then,
$$
\begin{array}{c}
V_{A-E H K} \\
=t^{3} V_{A-B C D}, \\
V_{B F G-N E H}=\frac{3(1-t)}{t} V_{A-E H K} \\
=3(1-t) t^{2} V_{A-B C D} .
\end{array}
$$

Thus, $V_{A B-E F C H}=\left(3 t^{2}-2 t^{3}\right) V_{A-B C D}$.
Then $\sum_{i=1}^{k} V_{i}=\left[3\left(\frac{k}{n}\right)^{2}-2\left(\frac{k}{n}\right)^{3}\right] V_{A-B C D}$.
Hence $V_{k}=\left\{\left[3\left(\frac{k}{n}\right)^{2}-2\left(\frac{k}{n}\right)^{3}\right]-\right.$
$$
\begin{array}{r}
\left.\left[3\left(\frac{k-1}{n}\right)^{2}-2\left(\frac{k-1}{n}\right)^{3}\right]\right\} V_{A-B C D} \\
=\frac{1}{n^{3}}\left[3 n(2 k-1)-2\left(3 k^{2}-3 k+1\right)\right] V_{A-B C D} .
\end{array}
$$
$$
\begin{array}{l}
=\sum_{k=0}^{\frac{n-1}{2}}\left\{\frac{3(4 k+1)}{n^{2}}-\frac{2[3(2 k+1) \times 2 k+1]}{n^{3}}\right\} V_{A-B(D)} \\
=\frac{n^{3}+1}{2 n^{3}} V_{A-B(D)} . \\
\text { Therefore, }\left(\sum_{i \in\{} V_{i}\right):\left(\sum_{i \text { i }} V_{i}\right)=\left(n^{3}+1\right):\left(n^{3}-1\right) .
\end{array}
$$