14.15. Let $P(x)$ be a polynomial of degree $n$, and $P(x) = 2^x$ for $x = 1, 2, \ldots, n+1$. Compute $P(n+2)$.
14.15. Answer: $P(n+2)=2^{n+2}-2$.

The expression $C_{m}^{x-1}=\frac{(x-1)(x-2) \ldots(x-m)}{m!}$ can be considered as a polynomial of degree $n$ in the variable $x$. Let

$$
f(x)=2\left(C_{x-1}^{0}+C_{x-1}^{1}+\ldots+C_{x-1}^{n}\right)
$$

It is clear that $C_{x-1}^{0}+C_{x-1}^{1}+\ldots+C_{x-1}^{n}=(1+1)^{x-1}=2^{x-1}$ for $x=1,2, \ldots, n+1$. Therefore, $P(x)=f(x)$, since these polynomials of degree $n$ coincide at $n+1$ different values of $x$. Consequently,

$$
\begin{aligned}
& P(n+2)=2\left(C_{n+1}^{0}+C_{n+1}^{1}+\ldots+C_{n+1}^{n}\right)= \\
& \quad=2\left(C_{n+1}^{0}+C_{n+1}^{1}+\ldots+C_{n+1}^{n}+C_{n+1}^{n+1}-C_{n+1}^{n+1}\right)= \\
& \quad=2\left(2^{n+1}-1\right)=2^{n+2}-2
\end{aligned}
$$