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 $P(x)$.
(Adapted from the 13th United States of America Mathematical Olympiad)
[Analysis] Notice that, in the given conditions, several consecutive values of the polynomial are involved, which can be associated with differences.
Thus, by Newton's formula (2), let $x=0$, we have
$$
\Delta^{k} f(0)=\sum_{i=0}^{k}(-1)^{k-i} \mathrm{C}_{k}^{i} f(i)=0 .
$$

This gives the values of $P(0), P(1), \cdots, P(2 n+1)$ and $\Delta^{2 n+1} P(0)$.
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, $P(x)=-2 x^{2}+4 x$.
[Comment] This problem can also be solved using the Lagrange interpolation identity. However, compared to that, solving it using the difference operator identity is more concise and elegant, which is the charm of the difference operator.