12. For $x \in \mathbf{R}, n \in \mathbf{N}^{*}$, define $\mathrm{C}_{x}^{n}=$ $\frac{x(x-1) \cdots(x-n+1)}{n!}$. Let $p(x)$ be a polynomial of degree $2m$ that satisfies $p(0)=1, p(k)=2^{k-1}(k=1,2$, $\cdots, 2 m)$. Express $p(x)=$ $\qquad$ using $\mathrm{C}_{x}^{k}(k=1,2, \cdots, 2 m)$.
12. $1+\mathrm{C}_{x}^{2}+\mathrm{C}_{x}^{4}+\cdots+\mathrm{C}_{x}^{2 m}$

Solution: From $p(0)=1$, we get that there exists a polynomial $q_{1}(x)$ such that $p(x) = 1 + x q_{1}(x)$, so $1 = p(1) = 1 + q_{1}(1)$, $q_{1}(1) = 0$, hence there exists a polynomial $q_{2}(x)$ such that $q_{1}(x) = (x-1) q_{2}(x)$, i.e., $p(x) = 1 + x(x-1) q_{2}(x)$, thus $2 = p(2) = 1 + 2 q_{2}(2)$, $q_{2}(2) = \frac{1}{2}$. Therefore, there is another polynomial $q_{3}(x)$ such that
$$
q_{2}(x) = (x-2) q_{3}(x) + \frac{1}{2},
$$

then $p(x) = 1 + \mathrm{C}_{x}^{2} + x(x-1)(x-2) q_{3}(x)$. Also, 4 $= p(3) = 1 + 3 + 3! q_{3}(3)$, $q_{3}(3) = 0$, hence $q_{3}(x) = (x - 3) q_{4}(x)$, $p(x) = 1 + \mathrm{C}_{x}^{2} + x(x-1)(x-2)(x-3) q_{4}(x)$, further we have
$$
p(x) = 1 + \mathrm{C}_{x}^{2} + \mathrm{C}_{x}^{4} + x(x-1)(x-2)(x-3)(x
$$
$-4) q_{5}(x)$.

Continuing this process and using the fact that $p(x)$ is a polynomial of degree $2m$, we get $p(x) = 1 + \mathrm{C}_{x}^{2} + \mathrm{C}_{x}^{4} + \cdots + \mathrm{C}_{x}^{2 m}$.