Example 1 Let $P(x)$ be a polynomial of degree $2n$, such that
$$
\begin{array}{l}
P(0)=P(2)=\cdots=P(2 n)=0, \\
P(1)=P(3)=\cdots=P(2 n-1)=2, \\
P(2 n+1)=-6 .
\end{array}
$$

Determine $n$ and $P(x)$.
From Theorem 3 (2), we get
$$
\begin{array}{l}
\Delta^{2 n+1} P(0)=\sum_{k=0}^{2 n+1}(-1)^{k} \mathrm{C}_{2 n+1}^{k} P(2 n+1-k) \\
=0+\sum_{j=1}^{n} \mathrm{C}_{2 n+1}^{2 j}+(-6) \\
=2 \times \frac{1}{2}\left(\sum_{j=1}^{n} \mathrm{C}_{2 n+1}^{2 j}+\sum_{j=1}^{n} \mathrm{C}_{2 n+1}^{2 j-1}\right) \\
=(-6)+2 \times \frac{1}{2}\left(2^{2 n+1}-2\right) .
\end{array}
$$

From Theorem 4 (3), we get $\Delta^{2 n+1} P(0)=0$.
Thus, $(-6)+2 \times \frac{1}{2}\left(2^{2 n+1}-2\right)=0 \Rightarrow n=1$.
Let $P(x)=a x^{2}+b x+c$.
Then $\left\{\begin{array}{l}P(0)=c=0, \\ P(1)=a+b+c=2, \\ P(2)=4 a+2 b+c=0 .\end{array}\right.$
Solving, we get $a=-2, b=4, c=0$.
Therefore, $P(x)=-2 x^{2}+4 x$.