28 Let $P(x)$ be an $n$-degree polynomial, and
$$
P(x)=2^{x}, x=1,2, \cdots, n+1 .
$$

Find $P(n+3)$.
28 Consider the polynomial
$$
f(x)=2\left(C_{x-1}^{0}+C_{x-1}^{1}+\cdots+C_{x-1}^{n}\right),
$$
where $n$ is a natural number. The term with the highest degree in $f(x)$ is
$$
2 C_{x-1}^{n}=2 \frac{(x-1)(x-2) \cdots(x-n)}{n!},
$$

which is an $n$-degree polynomial in $x$, so $f(x)$ is an $n$-degree polynomial. Furthermore, when $x=1,2, \cdots, n+1$, we have
$$
f(x)=2\left[(1+1)^{x-1}\right]=2^{x},
$$
(if $m<n$, define $C_{m}^{n}=0$). Thus, $f(x) \equiv P(x)$. Therefore,
$$
P(n+3)=f(n+3)=2\left(C_{n-2}^{0}+C_{n+2}^{1}+\cdots+C_{n-2}^{n}\right)
$$
$$
\begin{array}{l}
=2\left((1+1)^{n+2}-C_{n+2}^{n+1}-C_{n+2}^{n+2}\right) \\
=2\left(2^{n+2}-n-3\right) .
\end{array}
$$