If \(n\) is an odd positive integer and \(\left(1+x+x^{2}+x^{3}\right)^{n}=\sum_{r=0}^{3 n} a_{r} x^{r}\), then \(a_{0}-a_{1}+a_{2}-a_{3}+\ldots-a_{3 n}\) is equal to

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:
\[
a_{0}-a_{1}+a_{2}-a_{3}+\ldots-a_{3 n}=0
\]