Given, $\left(1+x+x^{2}+x^{3}\right)^{n}=\sum_{r=0}^{3 n} a_{r} x^{r}$ and $n$ is an odd positive integer.
$$
\begin{aligned}
&\Rightarrow \quad\left[(1+x)\left(1+x^{2}\right)\right]^{n}=\sum_{r=0}^{3 n} a_{r} x^{r} \\
&\Rightarrow \quad(1+x)^{n}\left(1+x^{2}\right)^{n}=\sum_{r=0}^{3 n} a_{r} x^{r}
\end{aligned}
$$
If we take $n=1$,
$$
\begin{gathered}
\left(1+x+x^{2}+x^{3}\right)=\sum_{r=0}^{3} a_{r} x^{r} \\
=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}
\end{gathered}
$$
On comparing both sides,
$$
a_{0}=1, a_{1}=1, a_{2}=1, a_{3}=1 \ldots (i)
$$
If we take $n=3$,
$$
\begin{gathered}
(1+x)^{3}\left(1+x^{2}\right)^{3}=\sum_{r=0}^{9} a_{r} x^{r} \\
\left(1+x^{3}+3 x^{2}+3 x\right)\left(1+x^{6}+3 x^{4}+3 x^{2}\right) \\
=\sum_{r=0}^{9} a_{r} x^{r}\left(1+x^{3}+3 x^{2}+3 x+x^{6}+x^{9}\right. \\
\quad+3 x^{8}+3 x^{7}+3 x^{4}+3 x^{7}+9 x^{6} \\
\left.\quad+9 x^{5}+3 x^{2}+3 x^{5}+9 x^{4}+9 x^{3}\right) \\
=\sum_{r=0}^{9} a_{r} x^{r}\left(1+3 x+6 x^{2}+10 x^{3}+12 x^{4}\right. \\
\left.\quad+12 x^{5}+10 x^{6}+6 x^{7}+3 x^{8}+x^{9}\right) \\
=\sum_{r=0}^{9} a_{r} x^{r}
\end{gathered}
$$
On comparing the coefficient of $x$ on both sides; $a_{0}=1, a_{1}=3, a_{2}=6, a_{3}=10, a_{4}=12, a_{5}=12$, $a_{6}=10, a_{7}=6, a_{8}=3, a_{9}=1$
From Eq. (i), we see that,
$$
a_{0}-a_{1}+a_{2}-a_{3}=0 \text {, when } n=1
$$
From Eq. (ii), we see that,
$$
\begin{aligned}
a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+& a_{6}-a_{7} \\
&+a_{8}-a_{9}=0
\end{aligned}
$$
when $n=3$,
Similarly, for each odd terms:
$$
a_{0}-a_{1}+a_{2}-a_{3}+\ldots-a_{3 n}=0
$$